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\newcommand\dukazNaznak{\begin{proof}Idea of the proof was presented, but certain parts were omitted, those can be used as an additional exercise during the exams.\end{proof}}
\newcommand\sDukazem{\begin{proof}The proof was presented, it will be examined.\end{proof}}
\newcommand\dukazLehky{\begin{proof}The proof is an easy exercise, it can be used as an additional exercise during the exams.\end{proof}}
\newcommand\bezDukazu{\begin{proof}Proof was omitted and will not be a part of the exam.\end{proof}}

\title{P\v{r}edn\'a\v{s}ka - Obecná topologie 2}
\author{Marek C\'uth}

\begin{document}
\chapter{Basic Topological Concepts}
\section{Topological Space, Base, Subbase}
\begin{defin}
Let $X$ be a set. We say that $\tau\subset \P(X)$ is a \emph{topology} on $X$ if
\begin{itemize}
\item[(T1)] $\emptyset, X\in \tau$,
\item[(T2)] if $U, V\in \tau$, then $U\cap V\in\tau$,
\item[(T3)] if $\mathcal U\subseteq \tau$, then $\bigcup \mathcal U\in\tau$.
\end{itemize}
The pair $(X,\tau)$ is then called a \emph{topological space}. Elements of $X$ are called \emph{points}, elements of $\tau$ are called \emph{open} sets.
A set $V\subseteq X$ is called a \emph{neighborhood} of a point $x\in X$ if there exists $U\in\tau$ such that $x\in U\subseteq V$. The family of all neighborhoods of the point $x$ is denoted by $\mathcal U(x)$.
A family of sets $\mathcal B\subseteq\tau$ is called a \emph{base} of the space $X$ if every $U\in\tau$ can be expressed as a union of some $\mathcal U\subseteq \mathcal B$. A family $\mathcal S\subseteq\tau$ is called a \emph{subbase} if the system
\[
\{\bigcap\mathcal F\colon \mathcal F\subseteq \mathcal S,\ \mathcal F \text{ is finite}\}
\]
is a base.
\end{defin}

\begin{examples}
\begin{itemize}
    \item If $(X,\rho)$ is a metric space, then the family of all $\rho$-open sets forms a topology on $X$. We then say that this \emph{topology is generated by the metric $\rho$}. A base of the topology generated by a metric is for example $\{B(x,r)\colon x\in X,\;r>0\}$. A topological space is called \emph{metrizable} if its topology is generated by some metric.
    \item On $\mathbb{R}$ with the usual metric, a base of the topology consists of open intervals; a subbase is for example the system $\{(-\infty, b), (a, \infty)\colon a, b\in\R\}$. On $\mathbb{R}^2$ with the usual metric, a base of the topology is for example $\{(a,b)\times (c,d)\colon a<b,\;c<d\}$; a subbase of the topology of the plane is for example $\{\er\times (a,b), (a,b)\times \er\colon a<b\}$.
    \item On every set $X$ there are two trivial topologies. The pair $(X, \P(X))$ is called a \emph{discrete space} ($\P(X)$ is the discrete topology). The pair $(X, \{\emptyset, X\})$ is called an \emph{indiscrete space} ($\{\emptyset, X\}$ is the indiscrete topology). Every discrete space is metrizable. An indiscrete space with at least two points is not metrizable.
\end{itemize}
\end{examples}

\begin{prop}[Properties of a base]
If $(X,\tau)$ is a topological space and $\mathcal B$ is its base, then
\begin{itemize}
\item[(B1)] for every $U, V\in\mathcal B$ and $x\in U\cap V$ there exists $W\in\mathcal B$ such that $x\in W\subseteq U\cap V$,
\item[(B2)] $\bigcup\mathcal B=X$.
\end{itemize}
If $X$ is a set and $\mathcal B\subseteq \mathcal P(X)$ is a family of sets satisfying conditions (B1) and (B2), then there exists a unique topology on $X$ whose base is $\mathcal B$.
\end{prop}
\sDukazem

\begin{corollary}
If $X$ is a set and $\mathcal S\subset \P(X)$ satisfies $\bigcup \mathcal S=X$, then there exists exactly one topology on $X$ whose subbase is $\mathcal S$.
\end{corollary}
\sDukazem

\begin{examples}
\begin{itemize}
    \item Let $(X,<)$ be a linearly ordered set. Then the \emph{topology generated by the order $<$} is the topology $\tau$ generated by the base
\[
\mathcal B=\{(a,b)\colon a, b\in X, a<b\}\cup\{(\leftarrow, b)\colon b\in X\}\cup\{(a,\rightarrow)\colon a\in X\}\cup\{X\}.
\]
(Elements of $\mathcal B$ are called open intervals.) We then say that $(X,\tau)$ is an \emph{ordered topological space}.
    \item The \emph{Sorgenfrey line} is the set $\er$ with the topology generated by the base
    $\mathcal B = \{[a,b)\colon a<b\}$.
\end{itemize}
\end{examples}

\begin{prop}[Characterization of open sets via neighborhoods]
Let $(X,\tau)$ be a topological space and let $U\subseteq X$. Then $U\in\tau$ if and only if
\[
\forall x\in U\ \exists V\in\mathcal U(x):\ V\subseteq U.
\]
\end{prop}
\sDukazem

\begin{prop}[Properties of the system of all neighborhoods]\label{TvrzOkoli}
If $(X,\tau)$ is a topological space, then for every $x\in X$ the family $\mathcal U(x)$ satisfies:
\begin{itemize}
\item[(U1)] $\mathcal U(x)\neq \emptyset$ and $x\in\bigcap\mathcal U(x)$,
\item[(U2)] if $U\in \mathcal U(x)$ and $U\subseteq V\subseteq X$, then $V\in\mathcal U(x)$,
\item[(U3)] if $U, V\in \mathcal U(x)$, then $U\cap V\in\mathcal U(x)$,
\item[(U4)] for every $U\in\mathcal U(x)$ there exists $V\in\mathcal U(x)$ such that for every $y\in V$ we have $U\in \mathcal U(y)$.
\end{itemize}
If $X$ is an arbitrary set and families $\mathcal U(x)\subseteq\mathcal P(X)$, $x\in X$, satisfy conditions (U1)--(U4), then there exists a unique topology $\tau$ on $X$ whose neighborhood systems are $\{\mathcal U(x)\}_{x\in X}$.
\end{prop}
\sDukazem

\begin{defin}
Let $(X,\tau)$ be a topological space. A family of sets $\mathcal B(x)$ is called a \emph{neighborhood base} at the point $x$ if $\mathcal B(x)\subseteq \mathcal U(x)$ and for every $V\in\mathcal U(x)$ there exists $U\in\mathcal B(x)$ such that $U\subseteq V$. An indexed family $\{\mathcal B(x)\}_{x\in X}$ is called a \emph{neighborhood base of the space $X$}.
\end{defin}

\begin{prop}[Properties of a neighborhood base]
If $(X,\tau)$ is a topological space and $\{\mathcal B(x)\}_{x\in X}$ is a neighborhood base, then for every $x\in X$ the family $\mathcal B(x)$ satisfies
\begin{itemize}
\item[(O1)] $\mathcal B(x)\neq \emptyset$ and $x\in\bigcap\mathcal B(x)$,
\item[(O2)] for every $U, V\in \mathcal B(x)$ there exists $W\in\mathcal B(x)$ such that $W\subseteq U\cap V$,
\item[(O3)] for every $U\in \mathcal B(x)$ there exists $V\in\mathcal B(x)$ such that for every $y\in V$ there exists $W\in \mathcal B(y)$ with $W\subseteq U$.
\end{itemize}
If $X$ is an arbitrary set and families $\mathcal B(x)\subseteq\mathcal P(X)$, $x\in X$, satisfy conditions (O1), (O2), and (O3), then there exists a unique topology $\tau$ on $X$ with neighborhood base $\{\mathcal B(x)\}_{x\in X}$.
\end{prop}
\dukazNaznak

\begin{defin}
The \emph{weight} of a topological space $X$, denoted by $w(X)$, is the smallest cardinality of a base. The more precise notation $w(X,\tau)$ is used when the topology $\tau$ is not clear from the context.
The \emph{character} at a point $x\in X$, denoted by $\chi(x, X)$, is the smallest cardinality of a neighborhood base at the point $x$. The character of the space $X$ is the supremum of the characters at its points.
\end{defin}

\begin{examples}
\begin{itemize}
    \item If $X$ is metrizable, then it has countable character. Moreover, $w(X)\leq \omega$ if and only if $X$ is separable.
    \item If $X$ is discrete, then $\chi(X)=1$ and $w(X)=|X|$.
    \item If $X$ is indiscrete, then $w(X)=\chi(X)=1$.
\end{itemize}
\end{examples}
\konecPrednasky{29. 9. 2025}

\begin{prop}
Let $X$ be a topological space. Then $\chi(X)\leq w(X)\leq 2^{|X|}$.
\end{prop}
\sDukazem

\section{Interior, Closure and Boundary, Dense and Nowhere Dense Sets, Accumulation and Isolated Points, Continuous Mappings}

\begin{defin}
A set $F$ in a topological space $(X,\tau)$ is called \emph{closed} if $X\setminus F$ is open. A set is called \emph{clopen} if it is both open and closed.
If $A\subseteq X$, then the \emph{closure} of the set $A$ is the set
\[
\cl{A}:=\bigcap\{F\colon F \text{ is closed},\ A\subseteq F\}.
\]
The \emph{interior} of the set $A$ is the set
\[
\int A= \bigcup\{ U\in \tau\colon U\subseteq A\}.
\]
The \emph{boundary} of the set $A$ is $\partial A=\cl{A}\cap \cl{X\setminus A}$.
\end{defin}

\begin{pozn}
It is easy to see that $\emptyset$ and $X$ are clopen sets and that the family of closed sets is closed under finite unions and arbitrary intersections. The closure and the boundary are closed sets, while the interior is an open set. A set is closed if and only if it equals its closure, and it is open if and only if it equals its interior.
\end{pozn}

\begin{prop}[Relation between interior and closure]
Let $X$ be a topological space. Then
\[
X\setminus\cl{A}=\int{(X\setminus A)}\quad\text{and}\quad X\setminus\int{A}=\cl{(X\setminus A)}.
\]
\end{prop}
\dukazLehky

\begin{prop}[Characterization of closure]
Let $X$ be a topological space, $x\in X$, $A\subseteq X$, and let $\mathcal B(x)$ be a neighborhood base at the point $x$. Then the following conditions are equivalent:
\begin{itemize}
\item[(1)] $x\in\cl A$,
\item[(2)] for every $U\in\mathcal U(x)$ we have $U\cap A \neq\emptyset$,
\item[(3)] for every $U\in \mathcal B(x)$ we have $U\cap A\neq\emptyset$.
\end{itemize}
In particular, if $U\subset X$ is open, then
\[
U\cap A=\emptyset\Leftrightarrow U\cap \cl A=\emptyset.
\]
\end{prop}
\sDukazem

\begin{prop}[Characterization of boundary]
Let $X$ be a topological space, $A\subseteq X$, and $x\in X$. Then
\[
x\in\partial A\Leftrightarrow \forall U\in\mathcal{U}(x):\; \big(U\cap A\neq \emptyset\ \&\  U\cap(X\setminus A)\neq\emptyset\big).
\]
\end{prop}
\dukazLehky

\begin{defin}
Let $X$ be a topological space and $A\subset X$.
The set $A$ is called \emph{dense} if $\cl A=X$. The set $A$ is called \emph{nowhere dense} if $X\setminus\cl A$ is dense in $X$. The \emph{density} of the space $X$ is the smallest cardinality of a dense subset; it is denoted by $d(X)$. We say that $X$ is \emph{separable} if $d(X)\leq \omega$.
\end{defin}

\begin{prop}[Characterization of dense and nowhere dense sets]
Let $X$ be a topological space and $A\subset X$.
\begin{itemize}
    \item $A$ is dense in $X$ if and only if every nonempty open set $U\subseteq X$ intersects $A$,
    \item $A$ is nowhere dense in $X$ if and only if $\int(\cl A)=\emptyset$,
    \item $A$ is nowhere dense in $X$ if and only if every nonempty open set $V$ contains a nonempty open set $U$ that is disjoint from $A$.
\end{itemize}
\end{prop}
\sDukazem

\begin{prop}[Relation between weight and density]
For every topological space $X$ we have $d(X)\leq w(X)$. If $X$ is metrizable, then $d(X)=w(X)$.
\end{prop}
\dukazNaznak

\begin{defin}
A point $x\in A\subseteq X$ is called an \emph{isolated point} of the set $A$ if there exists an open set $U\subseteq X$ such that $U\cap A=\{x\}$. A point $x\in X$ is called an \emph{accumulation point} of the set $A$ if every neighborhood $U$ of the point $x$ intersects $A\setminus \{x\}$. The set of all accumulation points of the set $A$ is denoted by $A'$ (the so-called derivative of the set $A$).
\end{defin}

\begin{examples}
In a discrete space all points are isolated and there are no accumulation points.
If $X=\R$ and $A=\mathbb Q$, then $A'=\R$ and no point of $A$ is isolated.
\end{examples}

\begin{prop}[Properties of the derivative]
Let $X$ be a topological space and $A\subset X$. Then
\[
\cl A=A\cup A'\quad\text{and}\quad (A\cup B)'=A'\cup B'.
\]
\end{prop}
\dukazLehky

\begin{defin}
Let $(X, \tau)$ and $(Y, \sigma)$ be two topological spaces and let $f\colon X\to Y$ be a function. The function $f$ is called \emph{continuous} if for every $U\in\sigma$ we have $f^{-1}(U)\in\tau$.
The mapping $f$ is called a \emph{homeomorphism} if $f$ is a bijection and both $f$ and $f^{-1}$ are continuous. The mapping $f$ is called an \emph{embedding} if its restriction $f:X\to f(X)$ is a homeomorphism onto the subspace $f(X)\subset Y$.
We say that $f$ is \emph{continuous at the point $x$} if for every neighborhood $V$ of the point $f(x)$ there exists a neighborhood $U$ of the point $x$ such that $f(U)\subseteq V$.
The mapping $f$ is called \emph{open/closed} if the set $f(M)$ is open/closed in $Y$ for every open/closed set $M\subseteq X$.
\end{defin}

\begin{pozn}
If $X,Y,Z$ are topological spaces and $x\in Z$, then if a function $f:X\to Y$ is continuous (at the point $x$), its restriction $f|_Z$ is also continuous (at the point $x$).
\end{pozn}

\begin{prop}[Characterization of continuous mappings]
Let $(X,\tau)$ and $(Y,\sigma)$ be topological spaces and let $f\colon X\to Y$ be a mapping. Let $\mathcal B$ be a base of $Y$ and $\mathcal S$ a subbase of $Y$.
Then the following conditions are equivalent:
\begin{enumerate}
\item[(1)] $f$ is continuous,
\item[(2)] preimages of sets from the subbase $\mathcal S$ are open,
\item[(3)] $f$ is continuous at every point,
\item[(4)] preimages of sets from the base $\mathcal B$ are open,
\item[(5)] preimages of closed sets are closed,
\item[(6)] for every $A\subseteq X$ we have $f(\cl A)\subseteq \cl{f(A)}$.
\end{enumerate}
\end{prop}
\dukazLehky

\begin{pozn}
    The composition of continuous mappings is continuous.
\end{pozn}

\konecPrednasky{6. 10. 2025}

\begin{defin}
If $(X,\tau)$ is a topological space and $Y\subset X$, let $\sigma:=\{U\cap Y\colon U\in \tau\}$. Then $\sigma$ is a topology on $Y$, and we say that $(Y,\sigma)$ is a \emph{subspace} of $X$.
\end{defin}

\begin{examples}
For metrizable spaces it holds that the weight equals the density and that a subspace of a separable space is separable. For non-metrizable spaces these statements need not hold. The following examples will be discussed gradually in exercises.
    \begin{itemize}
        \item If $X$ is the double arrow space or the Sorgenfrey line, then $d(X) = \omega < 2^\omega = w(X)$. Moreover, every subspace of $X$ is separable.
        \item If $\A\subset \P(\en)$ is an uncountable family of infinite sets that is almost disjoint, then for the corresponding Mrówka--Isbell space $X:=\psi(\A)$ we have $d(X) = \omega < |\A| = w(X)$. Moreover, $X$ is separable but contains a non-separable subspace (more precisely, $\A\subset \psi(\A)$ is an uncountable closed subspace that is discrete).
    \end{itemize}
\end{examples}

\section{Nets and Convergence}

\begin{defin}
A pair $(I, \leq)$ is called an \emph{upward directed} set if $\leq$ is a partial order on $I$ (i.e.\ a binary reflexive, transitive, and weakly antisymmetric relation) with the property that for $i, j\in I$ there exists $k\in I$ such that $i\leq k$ and $j\leq k$.
By the symmetric symbol $\geq$ we naturally mean the inverse relation.
A \emph{net} in a topological space $X$ is a mapping from some nonempty directed set into $X$. A special case is a \emph{sequence} (which is a net indexed by the natural numbers).
We say that a net $(x_i)_{i\in I}$ \emph{converges} to a point $x\in X$ if for every $U\in\mathcal{U}(x)$ there exists $i\in I$ such that for every $j \geq i$ we have $x_j\in U$. The point $x$ is then called the \emph{limit of the net $(x_i)_{i\in I}$}.
\end{defin}

\begin{prop}[Characterization of closure via convergence of nets]\label{prop:uzaverNety}
Let $X$ be a topological space and $A\subseteq X$. Then
\begin{itemize}
\item $x\in \cl A$ if and only if $x$ is the limit of a net consisting of points from $A$,
\item $A$ is closed if and only if every convergent net consisting of points in $A$ has its limit in $A$.
\end{itemize}
\end{prop}
\sDukazem

\begin{prop}[Characterization of continuity via convergence of nets]\label{prop:spojitostNety}
Let $X$ and $Y$ be topological spaces. A mapping $f\colon X\to Y$ is continuous at the point $x$ if and only if for every net $(x_i)_{i\in I}$ converging to $x$ the net $(f(x_i))_{i\in I}$ converges to $f(x)$.
\end{prop}
\sDukazem

\begin{example}
Statements~\ref{prop:uzaverNety} and~\ref{prop:spojitostNety} do not hold if we replace the word ``net'' by the word ``sequence''. Indeed, let $X$ be an uncountable set with the topology
\[
\tau:=\{\emptyset\}\cup \{A\subset X\colon X\setminus A\text{ is countable}\}.
\]
This topology $\tau$ is called the \emph{co-countable topology}. Then for any uncountable set $A\subsetneq X$ we have that $A$ is not closed, but at the same time every convergent sequence consisting of points from $A$ is eventually constant. Furthermore, if we consider the discrete topology on $\{0,1\}$ and define a function $f:X\to \{0,1\}$ by $f(x)=1$ for $x\in A$ and $f(x)=2$, then $f$ is not continuous, but whenever $(x_n)_{n\in\en}$ is a convergent sequence in $A$, we have
\[
\lim_{n\to\infty} f(x_n) = f\!\left(\lim_{n\to\infty} x_n\right).
\]
\end{example}
\sDukazem

\section{Separation Axioms}

\begin{defin}
A topological space $X$ is called
\begin{itemize}
\item $T_0$ if for every two distinct points $x, y\in X$ there exists an open set $U$ such that $|U\cap\{x, y\}|=1$,
\item $T_1$ if for every two distinct points $x, y\in X$ there exists an open set $U$ such that $x\in U$ and $y\notin U$,
\item $T_2$ (or \emph{Hausdorff}) if for every two distinct points $x, y\in X$ there exist disjoint open sets $U$, $V$ such that $x\in U$ and $y\in V$,
\item \emph{regular} if for every closed set $F\subseteq X$ and every point $x\in X\setminus F$ there exist disjoint open sets $U$, $V$ such that $x\in U$ and $F\subseteq V$,
\item \emph{normal} if for every two disjoint closed sets $E$, $F$ there exist disjoint open sets $U$, $V$ such that $E\subseteq U$ and $F\subseteq V$,
\item \emph{completely regular} if for every closed set $F$ and every point $x\in X\setminus F$ there exists a continuous function $f\colon X\to [0,1]$ such that $f(x)=0$ and $f(F)\subseteq \{1\}$,
\item $T_3$ if it is regular and $T_1$,
\item $T_{3\frac12}$ (or \emph{Tychonoff}) if it is completely regular and $T_1$,
\item $T_4$ if it is normal and $T_1$.
\end{itemize}
\end{defin}

\begin{pozn}
The following implications hold:
\[
T_4 \implies T_{3\frac12}\implies T_3\implies T_2\implies T_1\implies T_0,
\]
however the first implication is relatively difficult to prove (see Corollary~\ref{cor:T4isT3}), while the others are easy. None of the listed implications can be reversed; this is more difficult to show in the first three cases (see Examples~\ref{ex:separatingAxiomsEasy} and the example of the Niemytzki plane, which will be presented in exercises; the example showing that the second implication cannot be reversed is the most difficult and will be omitted here, it is written in the lecture notes).

It is easy to see that the properties $T_{3\frac12}$, $T_3$, $T_2$, $T_1$, and $T_0$ are preserved under subspaces, and that the property $T_4$ is preserved under closed subspaces. In general, normality is not preserved under subspaces; a corresponding counterexample will be mentioned later.
\end{pozn}

\begin{lemma}
Let $(X,\tau)$ be a topological space. Then the following hold:
\begin{itemize}
\item[a)] $X$ is $T_1 \iff$ every singleton subset of $X$ is closed $\iff$ every finite subset of $X$ is closed.
\item[b)] $X$ is $T_2 \iff \forall x, y\in X,\ x\neq y\;\exists U\in\mathcal U(x)\colon x\in U \text{ and }y\notin\cl U$.
\item[c)] $X$ is regular $\iff \forall x\in X\;\forall U\in\mathcal U(x)\;\exists V\in\mathcal U(x)\colon \cl V\subseteq U$.
\item[d)] $X$ is normal $\iff \forall V\in \tau\;\forall E\subseteq V \text{ closed }\; \exists U\in\tau:\;  E\subseteq U\subseteq \cl U\subseteq V$.
\end{itemize}
\end{lemma}
\dukazLehky

\begin{prop}[Uniqueness of the limit of a net]
A topological space is $T_2$ if and only if every net has at most one limit.
\end{prop}
\sDukazem

\begin{theorem}[Urysohn's Lemma]
A topological space $X$ is normal if and only if for every two disjoint closed sets $E$ and $F$ there exists a continuous function $f\colon X\to [0,1]$ such that $f(E)\subseteq \{0\}$ and $f(F)\subseteq\{1\}$.
\end{theorem}
\sDukazem

\begin{corollary}\label{cor:T4isT3}
If a topological space is $T_4$, then it is $T_{3\frac12}$.
\end{corollary}
\sDukazem

\konecPrednasky{14. 10. 2025}

\begin{defin}
We say that a topological space is \emph{zero-dimensional} if it has a neighborhood base consisting of clopen sets.
\end{defin}

\begin{examples}\label{ex:separatingAxiomsEasy}
    \begin{itemize}
        \item The Sierpiński space (i.e.\ the space $X=\{0,1\}$ with topology $\tau = \{\emptyset,X,\{0\}\}$) is $T_0$ but not $T_1$.
        \item An example of a topological space that is $T_1$ but not $T_2$ is an infinite set $X$ with the \emph{cofinite topology}, i.e.\ with topology $\tau:=\{\emptyset\}\cup \{A\subset X\colon X\setminus A\text{ is finite}\}$.
        \item An example of a topological space that is $T_2$ but not $T_3$ is the set $S:=(0,1)^2\cup\{(0,0)\}$ with topology generated by a neighborhood base defined as follows:\\
 - for $x\in (0,1)^2$ we set $\mathcal{B}(x):=\{B_\rho(x,\varepsilon)\colon \rho\text{ is the Euclidean metric and }\varepsilon>0\}$,\\
- for $x=(0,0)$ we set $\mathcal{B}(x):=\{[0,1/2)\times [0,1/n)\cap S\colon n\in\en\}$.
        \item Every zero-dimensional $T_1$ topological space is $T_{3\frac{1}{2}}$.
    \end{itemize}
\end{examples}
\sDukazem

\begin{prop}[Metrizable spaces are normal]
Metrizable spaces are $T_4$.
\end{prop}
\sDukazem

\chapter{Operations with Topological Spaces and Mappings}

\section{Projectively and Inductively Generated Topologies}

\begin{defin}
Let $X$ be a set and let $\tau, \sigma$ be two topologies on $X$. We say that $\tau$ is \emph{larger (finer, stronger)} than $\sigma$ if $\tau\supseteq\sigma$. In this case the topology $\sigma$ is called \emph{smaller (coarser, weaker)}.
\end{defin}

The discrete topology is the largest of all topologies, the indiscrete topology is the smallest.

\begin{defin}[Projective and inductive generation]
Let $X$ be a set, $(X_i, \tau_i)$, $i\in I$, a family of topological spaces, and let $f_i\colon X\to X_i$, $i\in I$, be a family of mappings.
A topology $\tau$ on the set $X$ is called \emph{projectively generated} by the mappings $f_i$ if $\tau$ is the smallest topology such that all mappings $f_i\colon (X, \tau)\to (X_i,\tau_i)$ are continuous.
If $f_i\colon X_i\to X$, then a topology $\tau$ on $X$ is called \emph{inductively generated} if $\tau$ is the largest topology such that all mappings $f_i\colon (X_i, \tau_i)\to (X,\tau)$ are continuous.
\end{defin}

\begin{theorem}[Existence and properties of a projectively generated topology]
Let $X$ be a set, $(X_i, \tau_i)$, $i\in I$, a family of topological spaces, and let $f_i\colon X\to X_i$, $i\in I$, be a family of mappings. Then there exists exactly one topology $\tau$ projectively generated by the family of mappings $f_i$, $i\in I$. This topology has the following properties:
\begin{itemize}
    \item[(P1)] A subbase of the topology $\tau$ is $\{f_i^{-1}(U)\colon U\in\tau_i\}$,
    \item[(P2)] Whenever $(Y,\sigma)$ is a topological space and $g\colon Y\to X$ is given, then the mapping $g$ is continuous if and only if for every $i\in I$ the mapping $f_i\circ g$ is continuous,
    \item[(P3)] Whenever a net $(x_\alpha)_{\alpha\in A}$ in $X$ and a point $x\in X$ are given, then $x_\alpha\to x$ in the space $(X,\tau)$ if and only if for every $i\in I$ we have $f_i(x_\alpha)\to f_i(x)$.
\end{itemize}
\end{theorem}
\sDukazem

\begin{examples}
\begin{itemize}
    \item If $Y$ is a subspace of a topological space $X$, then the topology on $Y$ is projectively generated from the topology on $X$ via the identity mapping $i:Y\to X$.
    \item A mapping $f:(X,\tau)\to (Y,\sigma)$ is an embedding if and only if it is injective and the topology $\tau$ is projectively generated by the mapping $f$.
    \item Let $X$ be a topological space. We denote by $C_p(X)$ the topological space $(C(X),\tau_p)$, where $C(X):=\{f:X\to \R\colon f \text{ is continuous}\}$ and $\tau_p$ is the topology projectively generated by the mappings $\delta_x:C(X)\to \R$, $x\in X$, defined by $\delta_x(f):=f(x)$. Then for every net of functions $(f_i)_{i\in I}$ in $C_p(X)$ and every $f\in C_p(X)$ we have $f_i\to f$ if and only if $f_i(x)\to f(x)$ for every $x\in X$. We say that $\tau_p$ is the \emph{topology of pointwise convergence}.
\end{itemize}
\end{examples}

\begin{theorem}[Existence and properties of an inductively generated topology]
Let $X$ be a set, $(X_i, \tau_i)$, $i\in I$, a family of topological spaces, and let $f_i\colon X_i\to X$, $i\in I$, be a family of mappings. Then there exists exactly one topology $\tau$ inductively generated by the family of mappings $f_i$, $i\in I$. This topology has the following properties:
\begin{itemize}
    \item[(I1)] $\tau=\{M\subset X\colon f_i^{-1}(M)\in \tau_i,\; i\in I\}$,
    \item[(I2)] Whenever $(Y,\sigma)$ is a topological space and a mapping $g\colon X\to Y$ is given, then $g$ is continuous if and only if for every $i\in I$ the mapping $g\circ f_i$ is continuous.
\end{itemize}
\end{theorem}
\dukazNaznak

\section{Product}

\begin{defin}
If $(X_i, \tau_i)$ are topological spaces, then their product is the topological space $(X,\tau)$, where $X=\prod X_i$ and the topology $\tau$ is projectively generated by the projections $\pi_i\colon X\to X_i$.
It is denoted by $\prod_{i\in I} X_i$.
A mapping $f\colon (X,\tau)\to (Y,\sigma)$ is then called an \emph{embedding} if $f$ is injective and the topology $\tau$ is projectively generated by the mapping $f$.
\end{defin}

\begin{pozn}
A base $\mathcal B$ of the product has the following form:
\[
\mathcal B=\{\pi_{i_1}^{-1}(U_1)\cap\dots\cap \pi_{i_n}^{-1}(U_n)\colon i_1, \dots, i_n\in I,\ U_{1}\in\tau_{i_1}, \dots, U_{n}\in\tau_{i_n},\ n\in\mathbb N\},
\]
and the intersection of preimages of projections can be rewritten as
\[
\pi_{i_1}^{-1}(U_1)\cap\dots\cap \pi_{i_n}^{-1}(U_n)=\left\{x\in\prod X_i\colon x({i_1})\in U_1, \dots, x({i_n})\in U_n\right\}.
\]
Convergence of nets in the product topology is straightforward: a net $(x_j)_{j\in J}$ converges to $x$ if and only if for every $i\in I$ the net $(x_j(i))_{j\in J}$ converges to $x(i)$. Thus the product topology corresponds to pointwise convergence.
For this reason, the product topology is sometimes called the topology of pointwise convergence.
The closure of a set of the form $\prod A_i$ is $\prod \overline A_i$.
\end{pozn}

\konecPrednasky{20. 10. 2025}

\begin{defin}
Given a family of mappings $f_i:X_i\to Y_i$, $i\in I$, we define the \emph{product mapping} $\prod_I f_i:\prod_I X_i\to\prod_I Y_i$ by
\[
\prod_I f_i (\{x_i\}_{i\in I})=\{f_i(x_i)\}_{i\in I}.
\]
Given a family of mappings $f_i:X\to Y_i$, $i\in I$, we define the \emph{diagonal mapping} $\Delta_I f_i:X\to\prod_I Y_i$ by
\[
\Delta_I f_i(x)=\{f_i(x)\}_{i\in I}.
\]
\end{defin}

\begin{prop}
Let $Z$, $Y_i$, $i\in I$, and $X_i$, $i\in I$, be topological spaces.
\begin{enumerate}[(i)]
	\item A mapping $f\colon Z\to \prod_I X_i$ is continuous if and only if all compositions of $f$ with the projections $\pi_i:\prod_I X_i\to X_i$ are continuous.
	\item If $f_i\colon Y_i\to X_i$, $i\in I$, are continuous mappings, then their product $\prod_I f_i\colon \prod_I Y_i \to\prod_I X_i$ is continuous.
	\item If $f_i\colon Z\to X_i$, $i\in I$, are continuous mappings, then their diagonal product $\Delta_I f_i\colon Z\to \prod_I X_i$ is continuous.
\end{enumerate}
\end{prop}
\sDukazem

\begin{corollary}
If $X$ is a topological space and $f, g\colon X \to \R$ are continuous mappings, then $f+g$, $f-g$, $f\cdot g$, $\max\{f, g\}$, $\min\{f, g\}$, and $|f|$ are continuous.
If $g$ is nonzero, then $f/g$ is also continuous.
\end{corollary}
\sDukazem

\begin{prop}[Characterization of Hausdorff spaces]
A topological space $X$ is Hausdorff if and only if $\{(x,x)\colon x\in X\}$ is closed in $X\times X$.
\end{prop}
\sDukazem

\section{Sum and Quotient}

\begin{defin}
The sum of spaces $X_i$, $i\in I$, is the space $\disjUnion_I X_i$ consisting of points $(i,x)$, where $i\in I$ and $x\in X_i$. If all sets $X_i$ are disjoint, one usually takes the points $x$ instead of the pairs $(i,x)$.

If $(X_i, \tau_i)$, $i\in I$, are topological spaces, then their \emph{topological sum} is the topological space $(\disjUnion_I X_i,\tau)$, where the topology $\tau$ is defined by
\[
\tau := \Big\{\disjUnion_I O_i\colon O_i\subset X_i\text{ are open}\Big\}.
\]
The topological sum is denoted by $\bigoplus_I X_i$.\\
(Warning: there is no uniform agreement on the notation for the topological sum; different symbols are used in the literature. The notation used above is somewhat inconvenient, since for example in Banach space theory the symbol $\bigoplus_I X_i$ denotes a Banach space which, as a topological space, is the product of the $X_i$ rather than their topological sum.)
\end{defin}

\begin{pozn}
The topology on $X:=\bigoplus_I X_i$ is inductively generated by the canonical embeddings $X_i\ni x\mapsto (i,x)\in \disjUnion X_i$.

If the $X_i$ are disjoint, then a set $U$ in the topological sum $\bigoplus_{i\in I} X_i$ is open if and only if for every $i\in I$ the set $U\cap X_i$ is open in $X_i$. The sets $X_i$ are clopen in $X$. If $f_i:X_i\to Y_i$, $i\in I$, are continuous mappings, then $\bigoplus_I f_i:\bigoplus_I X_i\to \bigoplus_I Y_i$ is continuous.
\end{pozn}

\begin{defin}
If $(X, \tau)$ is a topological space and $E\subseteq X\times X$ is an equivalence relation, then the \emph{quotient topology} on $X/E$ is inductively generated by the mapping $x\mapsto [x]_E$.

A mapping $f\colon (X,\tau)\to (Y,\sigma)$ is called a \emph{quotient mapping} if $f$ is surjective and the topology $\sigma$ is inductively generated by the mapping $f$.
\end{defin}

\begin{pozn}
A more intuitive description of the quotient topology is again direct. A set $U$ in the quotient space $Y$ is open if and only if $f^{-1}(U)$ is open in $X$.

If $f\colon X\to Y$ is a quotient mapping, then one can naturally consider an equivalence relation $\sim$ on $X$ defined by $x\sim y$ if and only if $f(x)=f(y)$. Then $X/\sim$ is naturally homeomorphic to $Y$.
\end{pozn}

\begin{examples}
\begin{itemize}
    \item If we define an equivalence relation on $\R$ by $xEy$ if and only if $\{x,y\}\in \qe$ or $\{x,y\}\subset \R\setminus\qe$, then $\R/E$ is a two-point indiscrete space. In particular, a quotient of a $T_4$ space need not even be $T_0$.
    \item If we define an equivalence relation on $\R$ by $xEy$ if and only if $x-y\in\mathbb Z$, then the quotient space $\R/E$ is homeomorphic to the circle.
\end{itemize}
\end{examples}

\begin{prop}[Characterization of a quotient mapping]
Let $(X,\tau)$ and $(Y,\sigma)$ be topological spaces and let $f\colon X\to Y$ be a mapping. Then $f$ is a quotient mapping if and only if $f$ is surjective and for every $V\subseteq Y$ we have $V\in\sigma \iff f^{-1}(V)\in\tau$.
\end{prop}
\dukazNaznak

\begin{prop}[Sufficient condition for a quotient mapping]
If a mapping $f\colon X\to Y$ is continuous, open (or closed), and surjective, then it is a quotient mapping.
\end{prop}
\sDukazem

\begin{pozn}
The open mapping theorem states that a continuous surjective linear mapping between Banach spaces is already open.
A quotient mapping between topological spaces need not be either open or closed.
\end{pozn}

\begin{pozn}
A continuous image (and hence also a quotient) of a separable space is separable.
\end{pozn}

Overview of preservation of properties by individual operations.
The symbol (+) means preservation at least under countable products or sums.

\begin{tabular}{|r|c|c|c|c|c|c|}
\hline
    & $T_0$, $T_1$, $T_2$, $T_3$, $T_{3\frac12}$ & $T_4$ & separable & countable base & countable character & metrizable \\
\hline
subspace   & + & -  & - & + & + & +  \\
\hline
(countable) sum  & + & + & -(+) & -(+) & + & +    \\
\hline
quotient  & - & - & + & - & - & -   \\
\hline
(countable) product  & + & - & -(+)  & -(+) & -(+) & -(+)   \\
\hline
\end{tabular}
(Some material was/will be covered in lectures, some was/will be covered in exercises.)

\section{Extension of Continuous Functions}

\begin{prop}
Let $X, Y$ be topological spaces and let $f, g\colon X\to Y$ be continuous mappings. If $Y$ is Hausdorff, then the set $\{x\in X\colon f(x)=g(x)\}$ is closed.
\end{prop}
\sDukazem

\begin{corollary}[Uniqueness of continuous extension]\label{jednorozsireni}
Thus, if $f\colon X\to Y$ is continuous, $Y$ is Hausdorff, and $S\subseteq X$ is dense, then the function $f\restriction S$ has a unique continuous extension to the whole space $X$.
\end{corollary}
\sDukazem

\begin{prop}
If $X$ is a topological space, $f_n\colon X\to\mathbb \R$ are continuous, and the sequence $(f_n)$ converges uniformly to $f$, then $f$ is continuous.
\end{prop}
\sDukazem

\konecPrednasky{27. 10. 2025}

\begin{theorem}[Tietze--Urysohn]
If $X$ is a normal space and $F\subseteq X$ is closed, then every continuous function $f\colon F\to\R$ can be continuously extended to the whole space $X$, i.e.\ there exists a continuous function $\tilde f\colon X\to \R$ such that $\tilde f\restriction F=f$.
\end{theorem}
\sDukazem

\chapter{Compactness}

\section{Compact Spaces}

\begin{defin}
A system of sets $\mathcal S$ is called a \emph{cover} of a space $X$ if $\bigcup\mathcal S=X$. Any subsystem $\mathcal S$ that is a cover is called a \emph{subcover}. A cover is called \emph{open} if its elements are open sets.
A topological space is called \emph{compact} if from every open cover one can select a finite subcover.
A topological space is called \emph{countably compact} if from every countable open cover one can select a finite subcover.
A topological space is called \emph{Lindelöf} if from every open cover one can select a countable subcover.

We say that a system $\mathcal F\subseteq \mathcal P(X)$ is \emph{centered} if $F_1\cap\dots\cap F_n\neq\emptyset$ for $n\in\mathbb N$ and $F_1,\dots, F_n \in\mathcal F$.
\end{defin}

\begin{defin}
Let $X$ be a topological space, $(I,\leq)$ an upward directed set, and $(x_i)_{i\in I}$ a net. A point $x\in X$ is called an \emph{accumulation point of the net} $(x_i)_{i\in I}$ if for every $U\in \mathcal{U}(x)$ and every $i\in I$ there exists $j \geq i$ such that $x_j\in U$.

Let $(J,\preceq)$ be another upward directed set. We say that a mapping $\varphi:J\to I$ is \emph{cofinal} if for every $i_0\in I$ there exists $j_0\in J$ such that $\varphi([j_0,\to])\subset [i_0,\to)$ (i.e.\ for every $j\in J$ with $j_0\preceq j$ we have $i_0\leq \varphi(j)$). If a cofinal mapping $\varphi:J\to I$ is given, then we say that $(x_{\varphi(j)})_{j\in J}$ is a \emph{subnet} of the net $(x_i)_{i\in I}$.
\end{defin}

\begin{pozn}
If a sequence $(x_n)_{n=1}^\infty$ in a space $X$ is given, then subsequences are a special case of subnets. However, there also exist subnets that are not subsequences. For example, $(x_{F(t)})_{t\in\R}$, where $F:\R\to\en$ is any mapping satisfying $\lim_{t\to \infty} F(t) = \infty$.

It is easy to see that if a net $(x_i)$ converges to a point $x$, then every subnet also converges to the point $x$.
\end{pozn}

\begin{theorem}[Characterization of compactness]
For a topological space $X$, the following conditions are equivalent.
\begin{itemize}
\item[(a)] $X$ is compact.
\item[(b)] Every centered system of closed sets has a nonempty intersection.
\item[(c)] Every net in $X$ has a convergent subnet.
\end{itemize}
\end{theorem}
\sDukazem

\begin{lemma}
Let $X$ be a topological space and $x\in X$. Then $x$ is an accumulation point of a net $(x_i)_{i\in I}$ if and only if there exists a subnet of $(x_i)_{i\in I}$ that converges to the point $x$.
\end{lemma}
\sDukazem

\begin{theorem}[Characterization of countable compactness]
For a topological space $X$, the following conditions are equivalent.
\begin{itemize}
\item[(a)] $X$ is countably compact.
\item[(b)] Every countable centered system of closed sets has a nonempty intersection.
\item[(c)] Every sequence in $X$ has a convergent subnet.
\end{itemize}
\end{theorem}
\dukazNaznak

\begin{theorem}[Characterization of the Lindelöf property]
For a topological space $X$, the following conditions are equivalent.
\begin{itemize}
\item[(a)] $X$ is Lindelöf.
\item[(b)] Every countably centered system of closed sets has a nonempty intersection.
\end{itemize}
\end{theorem}
\dukazNaznak

\begin{prop}[Preservation of properties]
Compactness, countable compactness, and the Lindelöf property are inherited by closed subspaces and continuous images.
\end{prop}
\sDukazem

\begin{corollary}[Attainment of extrema]
A continuous real-valued function on a nonempty (countably) compact space attains its maximum and minimum.
\end{corollary}
\sDukazem

\begin{examples}
\begin{itemize}
    \item If $(X,\leq)$ is linearly ordered and every (even empty) subset $A\subset X$ has a supremum and an infimum, then $X$ with the topology generated by the order is a compact space.
    \sDukazem
    \item $[0,\alpha]$ is a compact space for every ordinal $\alpha$.
    \dukazNaznak
\end{itemize}
\end{examples}
\konecPrednasky{3. 11. 2025}

\begin{lemma}[Alexander]
Let $X$ be a topological space and $\mathcal S$ its subbase. Suppose that from every cover $\mathcal U\subseteq \mathcal S$ one can select a finite subcover. Then $X$ is compact.
\end{lemma}
\sDukazem

\begin{theorem}[Tychonoff]
The product of compact topological spaces is compact.
\end{theorem}
\sDukazem

\begin{pozn}
    In ZF, Tychonoff's theorem is equivalent to the Axiom of Choice. Tychonoff's theorem restricted only to Hausdorff spaces is, in ZF, equivalent to the Boolean prime ideal theorem.
\end{pozn}

In the following statements we need compact spaces to be Hausdorff.

\begin{prop}[Compact sets are closed in Hausdorff spaces]
If $X$ is a Hausdorff topological space and $K\subseteq X$ is compact, then $K$ is closed in $X$.
\end{prop}
\sDukazem

\begin{prop}[Automatic homeomorphism]
Let $X, Y$ be Hausdorff compact topological spaces and let $f\colon X\to Y$ be continuous.
If $f$ is surjective, then $f$ is a quotient mapping.
If $f$ is bijective, then $f$ is a homeomorphism.
\end{prop}
\sDukazem

\begin{pozn}
Thus a compact Hausdorff topology on a given set is minimal among Hausdorff topologies and maximal among compact ones.
\end{pozn}

\begin{theorem}[Sufficient condition for normality]
A regular Lindelöf topological space is normal. A Hausdorff compact space is $T_4$.
\end{theorem}
\sDukazem

\begin{prop}[A continuous image of a compact space does not increase weight]
Let $X,Y$ be Hausdorff topological spaces, $X$ compact, and $f\colon X\to Y$ a continuous surjective mapping. Then $w(Y)\leq w(X)$.
\end{prop}

\section{Spaces of Continuous Functions on a Compact Space}

For two topological spaces $X, Y$ we denote by $C(X,Y)$ the set of all continuous functions from $X$ to $Y$. In the case $Y=\mathbb R$, we write simply $C(X)$. For a compact topological space $K$, the space $C(K)$ together with the supremum norm forms a Banach space whose properties are closely related to the topological properties of the compact space $K$.

\begin{prop}[Dini's criterion for uniform convergence]
Let $K$ be a compact topological space, and let $f_n\colon K\to\mathbb R$ be a sequence of continuous functions such that $f_{n+1}\geq f_n$ and that converge pointwise to a continuous function $f\colon K\to \mathbb R$. Then $(f_n)$ converges uniformly to $f$.
\end{prop}
\sDukazem

\begin{lemma}[On the square root]
There exists a sequence of polynomials that converges uniformly to the function $\sqrt{t}$ on $[0,1]$.
\end{lemma}
\sDukazem

\konecPrednasky{10. 11. 2025}

\begin{defin}
We say that a system of functions $\mathcal F$ mapping $X$ into $Y$ \emph{separates points} if for any two distinct points $x,y\in X$ there exists $f\in\mathcal F$ such that $f(x)\neq f(y)$.
\end{defin}

A ring and a lattice are algebraic notions. We will not recall their general definitions here. We only note that for $\emptyset\neq\mathcal A\subset C(K,\mathbb{F})$, where $\mathbb{F}\in\{\R,\mathbb{C}\}$, the system $\mathcal A$ is a ring if $\mathcal A$ is a subspace closed under multiplication. The system $\mathcal A$ is self-adjoint if $\ov{f}\in\mathcal A$ whenever $f\in\mathcal A$.
A system $\mathcal A\subset C(K,\R)$ is a lattice if it is closed under $\max$ and $\min$ of two (respectively finitely many) functions.

\begin{theorem}[Stone--Weierstrass Theorem]
Let $K$ be a compact topological space.
\begin{enumerate}[(a)]
\item Let $\mathcal B\subset C(K,\R)$ be a vector subspace containing the constants and separating points.
If $\mathcal B$ is a ring or a lattice, then $\mathcal B$ is dense in $C(K,\R)$.
\item Let $\mathcal B\subset C(K,\mathbb{C})$ be a vector subspace containing the constants and separating points, which is moreover self-adjoint. If $\mathcal B$ is a ring, then $\mathcal B$ is dense in $C(K,\mathbb{C})$.
\end{enumerate}
\end{theorem}
\sDukazem

\begin{corollary}
Polynomials with rational coefficients form a dense subset of the space $C([0,1])$.
\end{corollary}
\sDukazem

\begin{prop}
Let $K$ be a Hausdorff compact space. Then the following statements are equivalent.
\begin{enumerate}[(a)]
    \item $K$ has countable weight.
    \item There exist countably many functions $\mathcal A\subset C(K)$ that separate the points of $K$.
    \item $C(K)$ is separable.
\end{enumerate}
In particular, if $K$ is a metrizable compact space, then $C(K)$ is separable.
\end{prop}
\sDukazem

\begin{pozn}
Later we will see that a Hausdorff compact space has countable weight if and only if it is metrizable.
\end{pozn}

\section{Compactifications}

\begin{defin}
Let $X$ be a topological space. A pair $(j, Y)$ is called a \emph{compactification} of the space $X$ if $j\colon X\to Y$ is an embedding of $X$ onto a dense subset of a compact Hausdorff space $Y$. Often the compactification is identified with the space $Y$ itself, and the mapping $j$ is understood as an inclusion.
We say that a compactification $(j_1, Y_1)$ is \emph{larger} than a compactification $(j_2, Y_2)$ if there exists a continuous mapping $f\colon Y_1\to Y_2$ such that $f\circ j_1=j_2$.
We call compactifications $(j_1, Y_1)$ and $(j_2, Y_2)$ of the space $X$ \emph{equivalent} if there exists a homeomorphism $h\colon Y_1\to Y_2$ extending $id_X$ (more formally, extending $i_2\circ i_1^{-1}$).
\end{defin}

\begin{example}
The closed interval $Y_1=[0,1]$ is a compactification of the open interval $X=(0,1)$, where $j_1\colon (0,1)\to [0,1]$ is an embedding.
The circle $Y_2$ can naturally be regarded as a compactification of $X$. Formally, $j_2(x)=\exp(2\pi i x)$ and $Y_2$ is the unit circle in the complex plane.
The compactification $(j_1,Y_1)$ is larger than the compactification $(j_2,Y_2)$.
\end{example}

\begin{pozn}
A compactification of a Hausdorff compact space $X$ is homeomorphic to $X$.
If one compactification is larger than another and the other is larger than the first, then they are equivalent.
\end{pozn}

\begin{defin}
A topological space $X$ is called \emph{locally compact} if every point has a compact neighborhood.
\end{defin}

\begin{prop}[Alexandroff compactification]
Let $(X,\tau)$ be a Hausdorff locally compact space that is not compact and let $\infty\notin X$. Consider on the space $X\cup\{\infty\}$ the topology $\sigma$ generated by the base
\[
\B:=\tau\cup \{\{\infty\}\cup(X\setminus K)\colon K\subseteq X \text{ is compact}\}.
\]
Then $(X\cup\{\infty\},\sigma)$ is a compactification of the space $X$.
\end{prop}
\sDukazem

\konecPrednasky{24. 11. 2025}

\begin{examples}
\begin{itemize}
    \item Every compact Hausdorff space is locally compact.
    \item Euclidean spaces are locally compact.
    \item A discrete space is locally compact.
    \item The space $[0,\omega_1)$ is locally compact.
\end{itemize}
\end{examples}

\begin{lemma}
Let $X$ be a Hausdorff locally compact space. Then
\begin{enumerate}[(a)]
    \item every point $x\in X$ has a neighborhood base consisting of compact sets,
    \item if $A\subset X$ is an open or closed subspace, then $A$ is locally compact,
    \item $X$ is an open subset in each of its compactifications.
\end{enumerate}
Moreover, a Hausdorff topological space is locally compact if and only if it is homeomorphic to an open subset of some Hausdorff compact space.
\end{lemma}
\sDukazem

\begin{defin}
We say that a compactification $(j, bX)$ of a space $X$ is one-point if $|bX\setminus X| = 1$.
\end{defin}

\begin{lemma}
A Hausdorff topological space has a one-point compactification if and only if it is locally compact and not compact. Furthermore, if $X$ is a locally compact space that is not compact, then the one-point compactification of $X$ is uniquely determined up to equivalence of compactifications and the one-point compactification is the smallest compactification of $X$.
\end{lemma}
\dukazLehky

\begin{defin}
Let $X$ and $Y$ be topological spaces and let $\mathcal F$ be a system of functions mapping $X$ into $Y$. We say that $\mathcal F$ \emph{separates points and closed sets} if for every closed set $F\subseteq X$ and every $x\in X\setminus F$ there exists $f\in\mathcal F$ such that $f(x)\notin \cl{f(F)}$.
\end{defin}

\begin{lemma}[Tychonoff embedding]
Let $X$ be a topological space and let $\mathcal{F}=\{f_i\colon X\to Y_i$, $i\in I\}$ be a family of continuous mappings, and consider the diagonal mapping $\Delta\mathcal{F}:=\Delta_I f_i\colon X\to\prod_{i\in I} Y_i$.
If $\mathcal F$ separates points, then $\Delta\mathcal{F}$ is injective.
If, moreover, $\mathcal F$ separates points and closed sets, then $\Delta\mathcal{F}$ is an embedding.
\end{lemma}
\sDukazem

\begin{prop}[Tychonoff cube]
Let $X$ be a $T_{3\frac{1}{2}}$ topological space. Then there exists a set $I$ such that $X$ embeds into $[0,1]^I$.
\end{prop}
\sDukazem

\begin{corollary}
Every $T_{3\frac{1}{2}}$ topological space has some compactification.
\end{corollary}
\sDukazem

\begin{defin}
Let $X$ be a $T_{3\frac{1}{2}}$ topological space. Let $\mathcal{F}:=\C(X,[0,1])$ and let $\triangle \mathcal{F}:X\to [0,1]^{\mathcal{F}}$ be the corresponding diagonal embedding. Then the compactification $(\triangle \mathcal{F},\overline{\triangle \mathcal{F}(X)})$ (or any equivalent one) is called the \emph{Čech--Stone} compactification (or beta compactification) and is denoted by $\beta X$.
\end{defin}

\begin{theorem}[Characterization of the beta compactification]
Let $X$ be a $T_{3\frac{1}{2}}$ topological space and let $(j,Y)$ be a compactification of $X$. Then the following conditions are equivalent.
\begin{itemize}
\item[(a)] $Y$ is the Čech--Stone compactification of $X$.
\item[(b)] Every continuous function $f\colon X\to [0,1]$ can be continuously extended to the whole space $Y$, i.e.\ there exists a continuous function $\widetilde{f}\colon Y\to [0,1]$ such that $\widetilde{f}\circ j = f$.
\item[(c)] Every continuous function $f\colon X\to Z$ into a compact Hausdorff space $Z$ can be continuously extended to the whole space $Y$, i.e.\ there exists a continuous function $\widetilde{f}\colon Y\to Z$ such that $\widetilde{f}\circ j = f$.
\item[(d)] $Y$ is the largest compactification of $X$.
\end{itemize}
\end{theorem}
\sDukazem

\begin{pozn}
Every bounded continuous function on a $T_{3\frac{1}{2}}$ space $X$ can be continuously extended to a (bounded) continuous function $\beta X\to\mathbb R$. Thus it is possible to identify the Banach spaces $\ell_\infty$ and $C(\beta\mathbb N)$.

The beta compactification of simple topological spaces can be very complicated. An example is the space $\beta\mathbb N$, which can, for instance, be described as the space of all ultrafilters on $\mathbb N$ with a suitable topology. See the exercises for more details.
\end{pozn}

\chapter{Metrizability and Čech Completeness}

\section{Metrizability}

\begin{lemma}
    A topological space with a countable base is Lindelöf.
\end{lemma}
\sDukazem

\konecPrednasky{1. 12. 2025}

If $(X,\rho)$ is a metric space and $\sigma=\min\{\rho,1\}$, then $\sigma$ is again a metric on $X$, which moreover generates the same topology on $X$ as the metric $\rho$.

\begin{prop}
    A countable product of metric spaces $(X_n,\rho_n)$, $n\in\en$, is metrizable by the metric
\[
\rho(x,y):=\sum_{n=1}^\infty \frac{\min\{\rho_n(x_n,y_n),1\}}{2^n},\qquad x,y\in \prod X_n.
\]
\end{prop}
\sDukazem

\begin{theorem}[Urysohn Metrization Theorem]
Every $T_3$ topological space with a countable base is metrizable.
\end{theorem}
\sDukazem

\begin{pozn}
It follows from the proof of the previous theorem that 
every separable metrizable space has a metrizable compactification and can be embedded into the Hilbert cube $[0,1]^{\mathbb N}$.
\end{pozn}

\begin{corollary}
A compact Hausdorff space is metrizable if and only if it has countable weight.
\end{corollary}
\dukazNaznak

\begin{corollary}
A continuous image of a metrizable compact space is a metrizable compact space, provided that the image is Hausdorff.
\end{corollary}
\sDukazem

\section{Completeness in Metric Spaces}

Recall that a metric space is \emph{complete} if every Cauchy sequence is convergent.

Let $I$ be a set. Recall that on $\ell_\infty(I):=\{f:I\to\er\colon f\text{ bounded}\}$ we define a metric by
$\rho(f,g):=\|f-g\|_\infty=\sup_{i\in I} |f(i)-g(i)|$. Then $(\ell_\infty(I),\rho)$ is a complete metric space.

\begin{theorem}[Universality of $\ell_\infty(I)$] 
    Let $(M,d)$ be a metric space. Then there exists an isometric embedding $\varphi:M\to \ell_\infty(M)$.
\end{theorem}
\dukazNaznak

\begin{corollary}[On completion]
Every metric space $(X,\rho)$ has a completion, i.e.\ there exists a complete metric space $(Y,\sigma)$ and an isometric embedding $j\colon X\to Y$ such that $j(X)$ is dense in $Y$.
\end{corollary}
\dukazNaznak

\begin{prop}
    Every uniformly continuous mapping from a subspace $A$ of a metric space $(X, \rho)$ into a complete metric
space $(Y, \sigma)$ can be extended to a uniformly continuous mapping from $\overline{A}$ into $Y$.
\end{prop}
\dukazNaznak

\begin{corollary}[Uniqueness of completion]
     Let $(X,\rho)$ be a metric space, $(Y_1,\sigma_1),(Y_2,\sigma_2)$ complete metric spaces, and
$e_1\colon X\to Y_1$, $e_2\colon X\to Y_2$ two isometric embeddings such that
$\cl{e_1(X)}^{Y_1}=Y_1$ and $\cl{e_2(X)}^{Y_2}=Y_2$. Then there exists an isometric surjective mapping
$e\colon Y_1\to Y_2$ such that $e_2=e\circ e_1$.
\end{corollary}
\dukazNaznak

Finally, let us recall two well-known statements from Mathematical Analysis concerning complete metric spaces.

\begin{theorem}[Cantor]
A metric space $(X,\rho)$ is complete if and only if for every sequence $(F_n)$ of nonempty closed subsets of $X$
satisfying $F_n\subset F_{n+1}$ for every $n\in\en$ and
$\lim_{n\to\infty}\operatorname{diam} F_n = 0$, the intersection
$\bigcap_{n\in\en} F_n$ is a singleton.
\end{theorem}

\begin{theorem}[Baire Theorem]
	In a complete metric space, the intersection of countably many dense open sets is again dense.
\end{theorem}

\begin{defin}
We say that a topological space is \emph{Baire} if the intersection of countably many dense open sets is again dense.
\end{defin}

\section{Complete Metrizability}

\begin{defin}
A topological space is called \emph{completely metrizable} if there exists a complete metric generating its topology.
\end{defin}

\begin{prop}
A countable product of completely metrizable spaces is completely metrizable.
\end{prop}
\sDukazem

\begin{theorem}[Completely metrizable spaces]
Let $(X,\rho)$ be a metric space. Then the following statements are equivalent.
\begin{enumerate}[(i)]
\item $X$ is a $G_\delta$ subset of some completion of itself.
\item $X$ is a $G_\delta$ subset of every completion of itself.
\item Whenever $X$ is embedded into a complete metric space, it is embedded there as a $G_\delta$ subset.
\item The space $X$ is completely metrizable.
\end{enumerate}
\end{theorem}
\sDukazem

\konecPrednasky{8. 12. 2025}

\begin{defin}
A Tychonoff space $X$ is called \emph{Čech-complete} if it is a $G_\delta$ subset of $\beta X$.
\end{defin}

\begin{lemma}[On the remainder of compactifications]
If $(j_1, Y_1)$ and $(j_2,Y_2)$ are compactifications of a topological space $X$ and
$f\colon Y_1\to Y_2$ satisfies $f\circ j_1 = j_2$, then
$f(Y_1\setminus j_1(X))=Y_2\setminus j_2(X)$.
\end{lemma}
\sDukazem

\begin{prop}[Characterization of Čech-complete spaces]
For a Tychonoff space $X$, the following conditions are equivalent.
\begin{itemize}
\item[(a)] $X$ is Čech-complete.
\item[(b)] $X$ is a $G_\delta$ subset of every compactification.
\item[(c)] $X$ is a $G_\delta$ subset of some compactification.
\end{itemize}
\end{prop}
\sDukazem

\begin{pozn}
In the previous theorem, one cannot equivalently say ``$G_\delta$ in every compact space.'' For example,
the space $[0,1]^{\omega_1}$ is not a $G_\delta$ subset of $[0,2]^{\omega_1}$.\\
(The proof will be given in the exercises.)
\end{pozn}

\begin{theorem}[Frolík's internal characterization of Čech completeness]
Let $X$ be a Tychonoff topological space. Then the following conditions are equivalent.
\begin{enumerate}[(i)]
\item $X$ is Čech-complete.
\item There exists a sequence of open covers $\{\mathcal U_n\}$ of the space $X$ such that for every centered
system of closed sets $\mathcal F$ with the property that for every $n\in\mathbb N$ there exist
$F\in\mathcal F$ and $U\in\mathcal U_n$ such that $F\subset U$, we have $\bigcap \mathcal F\neq\emptyset$.
\end{enumerate}
\end{theorem}
\sDukazem

\begin{corollary}[Čech]
A metrizable space $X$ is completely metrizable if and only if it is Čech-complete.
\end{corollary}
\sDukazem

\konecPrednasky{15. 12. 2025}

\begin{examples}
\begin{itemize}
    \item Every locally compact Hausdorff space is Čech-complete.
    \item The space of irrational numbers is Čech-complete, while the space of rational numbers is not Čech-complete.
\end{itemize}
\end{examples}

\begin{theorem}[Baire Theorem for Čech-complete spaces]
Every Čech-complete space is Baire.
\end{theorem}
\sDukazem

\begin{prop}[Preservation of Čech completeness under operations]
Čech completeness is preserved under
\begin{enumerate}[(i)]
    \item topological sums,
    \item countable products,
    \item closed subspaces.
\end{enumerate}
\end{prop}
\sDukazem

\begin{theorem}
    The product of arbitrarily many Čech-complete spaces is Baire.
\end{theorem}
\bezDukazu

\begin{pozn}
    Čech completeness is not preserved under arbitrary products, and there therefore exist spaces that are Baire
but not Čech-complete.
\end{pozn}

\chapter{Topological Groups}

This part of the syllabus was not covered in the lectures. For an overview, one may consult the prepared lecture notes.
A more detailed treatment of this material will be given in the course General Topology~2.

\konecPrednasky{5. 1. 2026}

\end{document}