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\newcommand\sLehkymDukazem{\begin{proof}The proof was presented, it can be examined.\end{proof}}
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\title{P\v{r}edn\'a\v{s}ka - Obecná topologie 2}
\author{Marek C\'uth}

\begin{document}

\chapter{Uniform Spaces}
\section{Relation to Topological and Metric Spaces}
\begin{pozn} % or does this belong to the appendix?
For a set $X$, denote the diagonal by $\Delta(X)=\{(x,y)\in X^2\colon x=y\}$. The inverse relation to a relation $E$ is the relation $\{(x,y)\colon (y,x)\in E\}$.
For binary relations $C,D$ on a set $X$ we define
$C\circ D=\{(x,z)\in X\times X\colon (x,y)\in C, (y,z)\in D \text{ for some } y\in X\}$.
\end{pozn}

\begin{defin}
A pair $(X, \mathcal D)$ is called a \emph{uniform space} (US) if $X$ is a set and $\mathcal D\subseteq\mathcal P(X\times X)$ is nonempty and satisfies:
\begin{multicols}{2}
\begin{enumerate}[label=(U\alph*), series=uniformita]
\item $\forall D\in \mathcal D:\;\Delta(X)\subseteq D$,
\item $\forall C, D\in\mathcal D:\; C\cap D\in\mathcal D$,
\item if $D\in\mathcal D$ and $D\subseteq E$, then $E\in\mathcal D$,
\item $\forall D\in\mathcal D:\;D^{-1}\in\mathcal D$,
\item $\forall D\in\mathcal D\; \exists C\in\mathcal D:\quad C\circ C\subseteq D$.
\end{enumerate}
\end{multicols}
The system $\mathcal D$ is called a \emph{uniformity}. Elements of $\mathcal D$ are called \emph{neighborhoods of the diagonal}. 
A uniformity $\mathcal D$ is called separated if, in addition,
\begin{enumerate}[label=(U\alph*), resume*=uniformita]
    \item $\bigcap \mathcal D=\Delta(X)$. (Equivalently, for every $x, y\in X$, $x\neq y$, there exists $D\in\mathcal D$ such that $(x,y)\notin D$.)
\end{enumerate}
If $\mathcal D$ is separated, then we say that the uniform space $(X, \mathcal D)$ is $T_1$.

A system $\mathcal B\subseteq \mathcal P(X^2)$ is called a \emph{base of a uniformity} (respectively, a \emph{base of the uniformity $\mathcal D$}) if closing $\mathcal B$ under supersets yields a uniformity (respectively, the uniformity $\mathcal D$). A system $\mathcal S\subseteq\mathcal P(X^2)$ is called a \emph{subbase of a uniformity} (respectively, a \emph{subbase of the uniformity $\mathcal D$}) if closing it under finite intersections yields a base of a uniformity (respectively, a base of the uniformity $\mathcal D$).

If $(X,\mathcal D)$ and $(Y,\mathcal E)$ are uniform spaces, then we say that a mapping
$f\colon (X,\mathcal D)\to (Y,\mathcal E)$ is \emph{uniformly continuous} if $(f\times f)^{-1}(E)\in\mathcal D$ for every $E\in\mathcal E$. A mapping $f$ is called a \emph{uniform homeomorphism} if $f$ is a bijection and both $f$ and $f^{-1}$ are uniformly continuous.
\end{defin}

\begin{lemma}
A nonempty system $\mathcal B\subseteq \mathcal P(X^2)$ forms a base of some uniformity on $X$ if and only if the following conditions hold:
\begin{multicols}{2}
\begin{enumerate}[label=(\alph*)]
\item $\forall C\in\mathcal B:\;\Delta (X)\subseteq C$,
\item $\forall C, D\in\mathcal B\; \exists E\in\mathcal B:\quad E\subseteq C\cap D$,
\item $\forall C\in\mathcal B\;\exists D\in\mathcal B:\quad D\subseteq C^{-1}$,
\item $\forall D\in\mathcal B\; \exists C\in\mathcal B:\quad C\circ C\subseteq D$.
\end{enumerate}
\end{multicols}
Moreover, if $\mathcal B$ is a base of a uniformity, then it is a base of a separated uniformity if and only if $\bigcap \B=\Delta(X)$.
\end{lemma}
\dukazLehky

\begin{examples}
\begin{itemize}
\item The \emph{discrete uniformity} on a set $X$ consists of all supersets of $\Delta(X)$. 
\item If $(X,\rho)$ is a pseudometric space, define
$E_\rho(r):=\{(x,y)\in X\times X\colon \rho(x,y)<r\}$ for $r>0$. Then $\{E_\rho(r)\colon r>0\}$ is a base of a uniformity on $X$, which is a base of a separated uniformity if, in addition, $\rho$ is a metric. This uniformity is denoted by $\D_\rho$. We say that a uniformity $\D$ is \emph{metrizable} if there exists a metric $\rho$ such that $\D = \D_\rho$.

It is easy to verify that if $(X,\rho)$ and $(Y,\sigma)$ are metric spaces, then a mapping $f:(X,\rho)\to (Y,\sigma)$ is uniformly continuous if and only if it is uniformly continuous as a mapping between the uniform spaces $(X,\D_\rho)$ and $(Y,\D_\sigma)$.
\end{itemize}
\end{examples}

\begin{defin}
    If $R$ is a system of pseudometrics on a set $X$, then the \emph{uniformity generated by $R$} (denoted by $\D_R$) is the uniformity whose subbase is $\{E_\rho(r)\colon r>0,\;\rho\in R\}$.
\end{defin}

\begin{pozn}
    It is easy to see that if $R$ is a system of pseudometrics on a set $X$, then $\D_R$ is separated if and only if $R$ separates points of $X$ (i.e.\ $\forall x\neq y\; \exists \rho\in R:\; \rho(x,y)>0$). Furthermore, if $S$ is a system of pseudometrics on a set $Y$, then a mapping $f:(X,\D_R)\to (Y,\D_S)$ is uniformly continuous if and only if
    \[
    \forall \rho\in S\;\forall\varepsilon>0\; \exists \sigma\in R\;\exists \delta>0\; \forall x,y\in X:\quad \sigma(x,y)<\delta\implies \rho(f(x),f(y))<\varepsilon.
    \]
\end{pozn}

\begin{notation}
For $E\subset X\times X$ and $x\in X$ we denote $E[x]:=\{y\in X\colon (x,y)\in E\}$.
\end{notation}

\begin{prop}
    If $(X,\mathcal D)$ is a uniform space, then
\[\tau_{\mathcal D}=\{A\subseteq X\colon \forall x\in A\quad \exists D\in\mathcal D\colon D[x]\subseteq A\}\]
 is a topology on $X$. Moreover, the following hold.
 \begin{enumerate}[label=(\alph*)]
     \item If $\mathcal B$ is a base of the uniformity $\mathcal D$, then $\mathcal B(x):= \{D[x]\colon D\in\mathcal B\}$, $x\in X$, are neighborhood bases at points in $(X,\tau_\D)$.
     \item $\D$ is separated if and only if $(X,\tau_\D)$ is $T_1$.
     \item If $(Y,\E)$ is a uniform space and $f:(X,\D)\to (Y,\E)$ is uniformly continuous, then $f:(X,\tau_\D)\to (Y,\tau_\E)$ is continuous.
     \item If $\D$ is generated by a system of pseudometrics $R$, then for every net $(x_i)_{i\in I}$ and every $x\in X$ we have that $x_i\stackrel{\tau_\D}{\to}x$ if and only if $\rho(x_i,x)\to 0$ for every $\rho\in R$.
 \end{enumerate}
\end{prop}
\sLehkymDukazem

\begin{defin}
    A topological space $(X,\tau)$ is called \emph{uniformizable} if there exists a uniformity $\D$ such that $\tau = \tau_\D$.
\end{defin}

\begin{lemma}[On a pseudometric]
Let $(X,\mathcal D)$ be a uniform space and let $\{D_n\colon n\in\en\cup \{0\}\}\subset \D$ satisfy
\begin{multicols}{2}
\begin{enumerate}[label=(\roman*)]
    \item $D_0=X\times X$,
    \item $\forall n\in\en:\; D_n = (D_n)^{-1}$,
    \item $\forall n\in\en:\; D_{n+1}\circ D_{n+1}\circ D_{n+1}\subseteq D_n$.
\end{enumerate}
\end{multicols}
Then there exists a pseudometric $\rho$ on $X$ satisfying
\begin{enumerate}[label=(\alph*)]
    \item $\forall n\geq 1:\quad \{(x,y)\colon d(x,y)<2^{-n-1}\}\subseteq D_n\subseteq \{(x,y)\colon d(x,y)\leq 2^{-n}\}$,
    \item $\D_\rho\subset \D$ and $\rho\leq 1$.
\end{enumerate}
\end{lemma}
\sTezkymDukazem

\begin{corollary}\label{cor:uniformitaPseudometrikama}
    Every uniformity is generated by some system of pseudometrics. Every $T_1$ uniformity is generated by a system of pseudometrics that separates points.
\end{corollary}
\sDukazemVkontextu

\begin{theorem}
    A $T_1$ uniform space is metrizable if and only if it has a countable base.
\end{theorem}
\sLehkymDukazem

\begin{pozn}
    A countable base of $\D$ is something different from a countable base of $\tau_\D$. An example is the discrete uniformity whose base consists of a single element $\Delta(X)$, but $\tau_\D$ is the discrete topology and hence the weight of $(X,\tau_\D)$ equals $|X|$.
\end{pozn}

\begin{theorem}
    A $T_1$ topological space is uniformizable if and only if it is $T_{3\frac12}$.
\end{theorem}
\sLehkymDukazem

\section{Subspace, Sum and Product}

\begin{defin}
    \begin{itemize}
        \item Let $(X,\mathcal D)$ be a uniform space and $A\subset X$. Define $\D|_A:=\{D\cap (A\times A)\colon D\in \D\}$. Then $(A,\D|_A)$ is a \emph{subspace} of $(X,\D)$.
        \item Let $(X_i,\mathcal D_i)$ be uniform spaces. The \emph{product} of uniformities is the uniformity $\D_{\Pi_I X_i}$ on $\Pi_I X_i$ whose subbase is $\{(\pi_i\times \pi_i)^{-1}(D)\colon i\in I, D\in \D_i\}$. Then $(\Pi_I X_i, \D_{\Pi_I X_i})$ is the \emph{product of uniform spaces}.
        \item Let $(X_i,\mathcal D_i)$ be uniform spaces. On $\biguplus_I X_i := \bigcup_{i\in I} (\{i\}\times X_i)$ we define the \emph{sum} of uniformities as the uniformity
        $\biguplus_I \D_i:=\{\bigcup_{i\in I}(\{i\}\times D_i)\colon D_i\in\D_i\text{ for each }i\in I\}$. Then $(\biguplus_I X_i, \biguplus_I \D_i)$ is the \emph{sum of uniform spaces}.
    \end{itemize}
\end{defin}

\begin{pozn}
    It is easy to verify that the sum/product/subspace of uniform spaces is a well-defined uniform space. The topology generated by the uniformity $\D|_A$ is the subspace topology, and the topology generated by the uniformity $\D_{\Pi X_i}$ is the product topology.
\end{pozn}

\begin{prop}
    Let $(Z,\D)$ and $(X_i,\D_i)$ for $i\in I$ be uniform spaces.
    \begin{enumerate}[label=(\roman*)]
        \item A mapping $f:Z\to \Pi_I X_i$ is uniformly continuous if and only if the mappings $\pi_i\circ f$ are uniformly continuous for each $i\in I$.
        \item Let $f_i:X_i\to (Y_i,\E_i)$, $i\in I$, be uniformly continuous mappings. Then the mapping $\Pi_I f_i:\Pi_I X_i\to \Pi_I Y_i$ is also uniformly continuous.
        \item Let $f_i:Z\to X_i$, $i\in I$, be uniformly continuous mappings. Then the mapping $\Delta_I f_i:Z\to \Pi_I X_i$ is also uniformly continuous.
        \item If $f,g:Z\to \er$ are uniformly continuous mappings, then the mappings $f+g$, $f-g$, $\max\{f,g\}$, $\min\{f,g\}$, and $|f|$ are also uniformly continuous. Moreover, if $f,g$ are bounded functions, then the mapping $f\cdot g$ is also uniformly continuous.
    \end{enumerate}
\end{prop}
\sLehkymDukazem
\konecPrednasky{19. 2. 2026}

\section{Completeness and Total Boundedness}

\begin{defin}
    \begin{itemize}
        \item A net $(x_i)_{i\in I}$ in a uniform space $(X,\mathcal D)$ is called \emph{Cauchy} if for every $D\in\mathcal D$ there exists $i_0\in I$ such that for all $i, j\geq i_0$ we have $(x_i, x_j)\in D$.
        \item A uniform space $(X,\mathcal D)$ is called \emph{complete} if every Cauchy net converges in $(X,\tau_{\mathcal D})$.
        \item A uniform space $(X,\mathcal D)$ is called \emph{totally bounded} if for every $E\in\mathcal D$ there exists a finite set $K\subseteq X$ such that $E[K]=X$.\quad (Where $E[K]:=\bigcup_{x\in K} E[x]$.)
    \end{itemize}
\end{defin}

\begin{pozn}
    If the uniformity $\D$ is generated by a system of pseudometrics $R$, then a net $(x_i)$ is Cauchy in $(X,\D)$ if and only if
\[
\forall \rho \in R\;\forall\varepsilon\;\exists i_0\;\forall i,j\geq i_0: \quad \rho(x_i,x_j)<\varepsilon.
\]
\end{pozn}

\begin{pozn}
It is not difficult to see that in a complete metric space Cauchy nets are convergent. It then easily follows that a metric space $(X,\rho)$ is complete (resp.\ totally bounded) if and only if the uniform space $(X,\mathcal D_\rho)$ is complete (resp.\ totally bounded).
\end{pozn}

\begin{pozn}
Let $(X,\D)$ be a uniform space. It is easy to verify that convergent nets in $(X,\tau_\D)$ are Cauchy in $(X,\D)$ and that uniformly continuous mappings send Cauchy nets to Cauchy nets.
\end{pozn}

\begin{prop}
Let $(X,\D)$ be a $T_1$ uniform space. Then
\[
\text{$X$ is totally bounded $\Leftrightarrow$ every net in $X$ has a Cauchy subnet.}
\]
\end{prop}
\sTezkymDukazem

\begin{prop}
\begin{enumerate}[label=(\roman*)]
    \item If a subspace of a complete $T_1$ uniform space is complete, then it is closed.
    \item A subspace of a totally bounded uniform space (resp.\ a closed subspace of a complete uniform space) is totally bounded (resp.\ complete).
    \item The product of totally bounded (resp.\ complete) uniform spaces is totally bounded (resp.\ complete).
\end{enumerate}
\end{prop}
\sLehkymDukazem

\begin{theorem}
    Let $X$ be a $T_1$ uniform space. Then $(X,\tau_\D)$ is compact if and only if $(X,\D)$ is complete and totally bounded.
\end{theorem}
\sLehkymDukazem

\begin{prop}
    Let $X$ and $Y$ be $T_1$ uniform spaces, let $Y$ be complete, let $A\subset X$, and let $f:A\to Y$ be uniformly continuous. Then there exists a uniformly continuous mapping $F:\overline{A}\to Y$ such that $F|_A = f$.
\end{prop}
\sLehkymDukazem

\begin{defin}
Let $(X,\D)$ be a $T_1$ uniform space. Its \emph{completion} is a pair $(e,Y)$, where $Y$ is a complete $T_1$ uniform space and $e:X\to Y$ is a uniform embedding onto a dense subset (i.e.\ $\overline{e(X)} = Y$ and $e:X\to e(X)$ is a uniform homeomorphism).
\end{defin}

\begin{theorem}
    Every $T_1$ uniform space has a completion. Moreover, if $(e,Y)$ and $(e',Y')$ are completions of a $T_1$ uniform space $X$, then there exists a uniform homeomorphism $F:Y\to Y'$ such that $F\circ e = e'$.
\end{theorem}
\bezDukazu

\section{Uniformity on Compact Spaces}

\begin{theorem}
    Let $(X,\tau)$ be a compact Hausdorff space. Then there exists exactly one uniformity on $X$ that generates the topology $\tau$; a base of this unique uniformity is formed by the open neighborhoods of the diagonal $\Delta(X)$.
\end{theorem}
\sLehkymDukazem

\begin{prop}
    Let $(X,\mathcal D)$ and $(Y,\mathcal E)$ be $T_1$ uniform spaces and let $(X,\tau_D)$ be compact. Then every continuous mapping $f\colon X\to Y$ is uniformly continuous.
\end{prop}
\sLehkymDukazem

\chapter{Topological Groups}

\begin{defin}
A triple $(G, \cdot, \tau)$ is called a \emph{topological group} (TG) if $(G, \cdot)$ is a group, $(G,\tau)$ is a topological space, and the multiplication operation $\cdot\colon G\times G\to G$ (where $G\times G$ is equipped with the product topology) and the inversion operation $^{-1}\colon G\to G$ are continuous.
\end{defin}

\begin{examples}
    Examples of topological groups include the following.
    \begin{enumerate}[label=(\alph*)]
        \item Any group with the discrete topology.
        \item Any normed linear space with addition as the group operation and the topology induced by the norm. More generally, every topological vector space is a commutative topological group.
        \item $GL(n,\er)$ (the group of all real invertible $n\times n$ matrices). The group operation is given by matrix multiplication, and the topology is given by coordinatewise convergence (i.e.\ the topology inherited from the product topology on $\er^{n\times n}$).

        \konecPrednasky{26. 2. 2026}
        \konecPrednasky{5. 3. 2026}
        \item $\Iso(X)$, where $(X,\rho)$ is a metric space. The symbol $\Iso(X)$ denotes the set of all surjective isometries $f:X\to X$ with the group operation given by composition and the topology of pointwise convergence (i.e.\ the product topology inherited from $X^X$).
        \sLehkymDukazem
        \item $\Iso(V)$, where $(V,\|\cdot\|)$ is a normed linear space. Here $\Iso(V)$ denotes the set of all surjective linear isometries $f:V\to V$ with the group operation given by composition and the topology of pointwise convergence (i.e.\ the product topology inherited from $V^V$).
        \item $H(K)$, where $K$ is a compact Hausdorff space. The symbol $H(K)$ denotes the set of all surjective homeomorphisms $f:K\to K$ with the group operation given by composition and the \emph{compact-open} topology, i.e.\ the topology whose subbase is formed by the sets
        $E[L;U]:=\{f\in H(K)\colon f(L)\subset U\}$, where $L\subset K$ is compact and $U\subset K$ is open.\\
        (A proof that this is indeed a topological group is left as an exercise.)
    \end{enumerate}
\end{examples}

The neutral element of a topological group $G$ is denoted by $e_G$ (or simply $e$ if the group $G$ is clear from the context). Recall that a subgroup $N\subset G$ is a \emph{normal subgroup} (denoted $N\lhd G$) if $gNg^{-1} = N$ for every $g\in G$ (not to be confused with the unrelated notion of a normal topological space). Furthermore, recall that for $N\lhd G$ we can define the \emph{quotient group} $G/N$, i.e.\ the group whose elements are cosets of the form $xN:=\{xn\colon n\in N\}$, with the group operation defined naturally by $(xN)(yN) = (xy)N$ and $(xN)^{-1} = x^{-1}N$.

For each $g\in G$ we define the left translation $L_g:G\to G$ and the right translation $R_g:G\to G$ by $L_g(h)=gh$ and $R_g(h)=hg$ for $h\in G$. A topological space $X$ is called \emph{homogeneous} if for every $x,y\in X$ there exists a homeomorphism $f:X\to X$ such that $f(x)=y$.

\begin{lemma}
Let $G$ be a topological group. Then the following hold.
%\begin{multicols}{2}
\begin{enumerate}[label=(\alph*)]
\item The inversion mapping $^{-1}: G\to G$ is a homeomorphism.
\item For every $g\in G$, the left and right translations $L_g$ and $R_g$ are homeomorphisms of $G$.
\item $G$ is a homogeneous space.
\item $\forall U\in \mathcal{U}(e)\; \exists V\in \mathcal{U}(e):\quad V\cdot V^{-1}\subset U$.
\item If one of the sets $A,B\subset G$ is open, then $A\cdot B$ is open.
\item If $H\subset G$ is a (normal) subgroup, then $\overline{H}$ is also a (normal) subgroup.
\item If $H\subset G$ is a subgroup and $\operatorname{Int}(H)\neq \emptyset$, then $H$ is clopen.
\item The product of topological groups equipped with the product topology is a topological group.
\item A homomorphism of topological groups $f\colon G\to H$ is continuous if and only if it is continuous at the neutral element $e_G$.
\end{enumerate}
%\end{multicols}
\end{lemma}
\sLehkymDukazem

\section{Uniformities on Topological Groups}

\begin{defin}
    Let $G$ be a topological group. Then
    \begin{itemize}
        \item the \emph{right uniformity} on $G$ is the uniformity $\D_R$ whose base is given by the system $\{R_U\colon U\in\U(e)\}$, where $R_U:=\{(x,y)\colon xy^{-1}\in U\}$ for $U\in \U(e)$,
        \item the \emph{left uniformity} on $G$ is the uniformity $\D_L$ whose base is given by the system $\{L_U\colon U\in\U(e)\}$, where $L_U:=\{(x,y)\colon x^{-1}y\in U\}$ for $U\in \U(e)$.
    \end{itemize}
\end{defin}

\begin{lemma}
Let $(G,\cdot,\tau)$ be a topological group. Then $\tau = \tau_{\D_R} = \tau_{\D_L}$ (i.e.\ the topology generated by the right uniformity is the group topology). Moreover, the mappings $R_g:(G,\D_R)\to (G,\D_R)$ and $L_g:(G,\D_L)\to (G,\D_L)$ are uniform homeomorphisms for every $g\in G$.
\end{lemma}
\sLehkymDukazem

\begin{theorem}
    Every $T_1$ topological group is $T_{3\frac12}$. Moreover, a $T_1$ topological group is metrizable if and only if it has a countable character.
\end{theorem}
\sDukazemVkontextu

\begin{lemma}[On right/left invariant pseudometrics]
    Let $(G,\cdot,\tau)$ be a topological group and let $\{U_n\colon n\in\en\cup\{0\}\}\subset \U(e)$ satisfy
\begin{multicols}{2}
\begin{enumerate}[label=(\roman*)]
    \item $U_0=G$,
    \item $\forall n\in\en:\; U_n = (U_n)^{-1}$,
    \item $\forall n\in\en:\; U_{n+1}\cdot U_{n+1}\cdot U_{n+1}\subseteq U_n$.
\end{enumerate}
\end{multicols}
Then there exists a right-invariant (respectively, left-invariant) pseudometric $\rho$ on $G$ satisfying $\rho\leq 1$,
\begin{enumerate}[label=(\alph*)]
    \item $\forall n\geq 1:\quad B_\rho(e,2^{-n-1})\subseteq U_n\subseteq \overline{B}_\rho(e,2^{-n})$,
    \item $\D_\rho\subset \D_R$ (respectively, $\D_\rho\subset \D_L$).
\end{enumerate}
Moreover, if the left and right uniformities on $G$ coincide, then there exists a bi-invariant pseudometric $\rho$ with the properties above.
\end{lemma}
\sDukazemVkontextu

\begin{theorem}[Birkhoff--Kakutani]
    Every metrizable topological group is metrizable by a right-invariant (respectively, left-invariant) metric.
\end{theorem}
\sDukazemVkontextu

\begin{defin}
    We say that a topological group is SIN (Small Invariant Neighborhoods) if there exists a base $\B$ of neighborhoods of $e$ such that $gUg^{-1}\subset U$ for every $U\in\B$ and every $g\in G$.
\end{defin}

\begin{prop}
A topological group is SIN if and only if its left and right uniformities coincide.
\end{prop}
\sLehkymDukazem

\begin{theorem}
    A metrizable group is SIN if and only if it is metrizable by a bi-invariant metric.
\end{theorem}
\sDukazemVkontextu

\begin{examples}
    Typical examples of SIN topological groups are compact groups, discrete groups, and commutative groups. An example of a metrizable topological group that is not SIN is $GL(n,\R)$.
\end{examples}
\sLehkymDukazem

\section{Quotients of Topological Groups}

\begin{theorem}
    Let $G$ be a topological group and let $N\lhd G$. Consider on $G/H$ the quotient topology induced by the mapping $\pi:G\to G/H$. Then $G/H$ is a topological group and $\pi$ is an open homomorphism. Moreover, $G/H$ is $T_1$ if and only if $H$ is closed (regardless of whether $G$ itself is $T_1$).
\end{theorem}
\sTezkymDukazem

\begin{theorem}
    Let $G$ be a $T_1$ topological group and let $H\subset G$ be a locally compact subgroup. Then $H$ is closed.
\end{theorem}
\sLehkymDukazem

\section{Representations of Topological Groups}

\begin{defin}
Let $G$ be a $T_1$ topological group. We say that a function $f:G\to\R$ is right uniformly continuous if $f$ is uniformly continuous as a function from $(G,\D_R)$ to $\R$. The set of all bounded right uniformly continuous functions $f:G\to \R$ is denoted by $RUC(G)$.
\end{defin}

\konecPrednasky{12.3.2026}

\begin{lemma}
Let $G$ be a $T_1$ topological group. Then a function $f:G\to \R$ is right uniformly continuous if and only if
\[
\forall\varepsilon>0 \; \exists U\in\U(e)\; \forall u\in U\;\forall x\in G:\quad |f(ux)-f(x)|<\varepsilon.
\]
If $RUC(G)$ is equipped with the norm $\|f\|_\infty:=\sup_{x\in G} |f(x)|$, then $(RUC(G),\|\cdot\|_{\infty})$ is a Banach space. Moreover, $RUC(G)$ separates points and closed sets in $G$.  
\end{lemma}
\sLehkymDukazem

\begin{theorem}[Teleman]
    Let $G$ be a $T_1$ topological group. Then there exists a Banach space $V$ such that $G$ embeds as a topological group into $\Iso(V)$.
\end{theorem}
\sTezkymDukazem

\konecPrednasky{19.3.2026}

\chapter{Paracompact Spaces}

\begin{defin}
    If $X$ is a set and $\mathcal U$ is a cover of $X$, then a system $\mathcal V$ is called a \emph{refinement of $\mathcal U$} (denoted $\mathcal V\prec \U$) if $\mathcal V$ is a cover of $X$ and for every $V\in\mathcal V$ there exists $U\in\mathcal U$ such that $V\subseteq U$.

    Furthermore, let $X$ be a topological space and let $\mathcal S\subseteq\mathcal P(X)$. The system $\mathcal S$ is called
    \begin{itemize}
        \item \emph{locally finite} if every point of $X$ has a neighborhood intersecting only finitely many sets from $\mathcal S$,
        \item \emph{discrete} if every point of $X$ has a neighborhood intersecting at most one set from $\mathcal S$,
        \item \emph{$\sigma$-locally finite} (respectively, \emph{$\sigma$-discrete}) if it is a countable union of locally finite (respectively, discrete) systems.
    \end{itemize}
\end{defin}

\begin{pozns}
Every ($\sigma$-)discrete system is ($\sigma$-)locally finite. The system $\{(-\tfrac{1}{n}, \tfrac{1}{n})\colon n\in\mathbb N\}$ is $\sigma$-discrete but not locally finite.
\end{pozns}

\begin{fact}[Closure of a locally finite family]
If $\mathcal A$ is a locally finite family in a topological space $X$, then $\{\cl A\colon A\in\mathcal A\}$ is locally finite and
$\cl{\bigcup \mathcal A}=\bigcup\{\cl A\colon A\in\mathcal A\}$.
\end{fact}
\sDukazemVkontextu

\begin{defin}
A Hausdorff topological space $X$ is called \emph{paracompact} if every open cover of $X$ has a locally finite open refinement.
\end{defin}

\begin{examples}
All compact spaces and all discrete spaces are paracompact.\\
(Later we will prove that every metric space is also paracompact.)
\end{examples}

\begin{theorem}[Characterization of Paracompactness]\label{charakparak}
For a $T_{3}$ topological space $X$, the following conditions are equivalent.
\begin{enumerate}[label=(\alph*)]
\item $X$ is paracompact.
\item Every open cover of $X$ is refined by a $\sigma$-locally finite open cover.
\item Every open cover of $X$ is refined by a locally finite cover.
\item Every open cover of $X$ is refined by a locally finite closed cover.
\end{enumerate}
\end{theorem}
\sTezkymDukazem

\begin{corollary}
Every Lindel\"of $T_{3}$ space is paracompact.
\end{corollary}
\sDukazemVkontextu

\begin{defin}
    For a system $\mathcal S$ of subsets of a set $X$ and $x\in X$, define
    $st_{\mathcal S}(x)=\bigcup\{S\in\mathcal S\colon x\in S\}$.
    We say that a cover $\mathcal V$ \emph{star-refines} a cover $\mathcal U$
    (denoted $\mathcal{V}\prec_{st} \U$) if $\{st_{\mathcal V}(x)\colon x\in X\}$
    refines $\mathcal U$.
\end{defin}

\begin{theorem}[Characterization of Paracompactness II]
For a Hausdorff topological space $(X,\tau)$, the following conditions are equivalent.
\begin{enumerate}[label=(\alph*)]
\item Every open cover $\U$ of $X$ has an open star-refinement that is a cover.
\item There exists a uniformity $\D$ on $X$ generating the topology of $X$
(i.e.\ $\tau_\D=\tau$) such that for every open cover $\U$ of $X$ there exists
$D\in\D$ with $\{D[x]\colon x\in X\}\prec \U$.
\item $X$ is $T_{3\frac 12}$ and for every open cover $\U$ of $X$ there exists
a continuous pseudometric $\rho$ on $X$ such that
$\{B_{\rho}(x,1)\colon x\in X\}\prec \U$.
\item $X$ is $T_{3\frac 12}$ and every open cover of $X$ is refined by a
$\sigma$-discrete open cover.
\item $X$ is paracompact.
\end{enumerate}
\end{theorem}
\sTezkymDukazem

\konecPrednasky{26.3. 2026}

\begin{defin}
A Hausdorff topological space $X$ is called \emph{collectively normal} if for every discrete family $\mathcal F$ of closed sets there exists a family of pairwise disjoint open sets $\{U(F)\colon F\in\mathcal F\}$ such that $F\subset U(F)$ for every $F\in\mathcal F$.
\end{defin}

\begin{pozn}
    If the family of closed sets is finite, then it is discrete if and only if it is disjoint. In particular, every collectively normal space is normal.
\end{pozn}

\begin{prop}
Every paracompact topological space is collectively normal, and hence normal.
\end{prop}
\sLehkymDukazem

\begin{theorem}[Stone]
    Every metrizable topological space is paracompact.
\end{theorem}
\sTezkymDukazem

\begin{defin}
    Let $\mathcal G$ be an open cover of a space $X$. A family of continuous functions
    $\{f_i:X\to [0,1]\colon i\in I\}$ is called a \emph{locally finite partition of unity
    subordinate to $\mathcal G$} if the family $\{\{f_i\neq 0\}\colon i\in I\}$ is locally finite, refines $\mathcal G$, and
    \[
    \sum_{i\in I} f_i(x) = 1
    \]
    for every $x\in X$.
\end{defin}

\begin{theorem}[Partition of Unity]
In a paracompact topological space, for every open cover there exists a locally finite partition of unity subordinate to this cover.
\end{theorem}
\sLehkymDukazem

\begin{theorem}[Dugundji -- special case]
    Let $K$ be a metrizable compact space and let $L\subset K$ be a closed subset. Then there exists a linear mapping
    $E:C(L)\to C(K)$ such that $Ef|_L = f$ and $\|Ef\|_\infty\leq \|f\|_\infty$ for every $f\in C(L)$. Moreover, $Ef\geq 0$ whenever $f\geq 0$.
\end{theorem}
\sTezkymDukazem

%\begin{corollary}
 %   Let $K$ be a metrizable compact space and let $L\subset K$ be a closed subset. Then $C(L)$ is isometrically isomorphic to a complemented subspace of $C(K)$.
%\end{corollary}

\konecPrednasky{2. 4. 2026}

\begin{theorem}[Bing, Nagata, Smirnov]
For a $T_{3\frac 12}$ space $X$, the following are equivalent.
\begin{itemize}[noitemsep]
\item[(a)] $X$ is metrizable.
\item[(b)] $X$ has a $\sigma$-discrete base.
\item[(c)] $X$ has a $\sigma$-locally finite base.
\end{itemize}
\end{theorem}
\sTezkymDukazem

\chapter{Connectedness}

\begin{defin}
A topological space is called \emph{connected} if it cannot be expressed as a disjoint union of two nonempty open sets.
\end{defin}

\begin{pozn}
    There are also the notions of path connectedness and arc connectedness; however, we will not discuss these notions further in this lecture.
\end{pozn}

Note that some authors (e.g.\ Engelking) consider the empty set to be connected, while others do not.

\begin{prop}
For a topological space $X$, the following conditions are equivalent.
\begin{enumerate}[label=(\alph*)]
\item The space $X$ is connected.
\item If $X=A\cup B$ and $\cl A\cap B=\emptyset=A\cap \cl B$, then $A=\emptyset$ or $B=\emptyset$.
\item The space $X$ contains no nontrivial clopen subset.
\item Every continuous mapping $f\colon X\to \{0,1\}$ is constant (where $\{0,1\}$ is the two-point discrete space).
\end{enumerate}
\end{prop}
\sLehkymDukazem

\begin{prop}
The continuous image of a connected space is connected.
\end{prop}
\sLehkymDukazem

\begin{prop}[Union of connected sets]\label{prop:sjednoceniSouvislych}
Let $\{C_i\colon i\in I\}$ be a family of connected subsets of a space $X$ and suppose that one of the following conditions holds:
\begin{multicols}{2}
\begin{enumerate}[label=(\alph*)]
\item $\exists i_0\in I\; \forall i\in I:\quad C_i\cap C_{i_0}\neq\emptyset$;
\item $\bigcap_{i\in I} C_i\neq\emptyset$.
\end{enumerate}
\end{multicols}
Then $\bigcup C_i$ is connected.
\end{prop}
\sLehkymDukazem

\begin{corollary}
If $X$ is a topological space, $A\subseteq X$ is connected, and $A\subseteq M\subseteq \cl A$, then $M$ is connected.
\end{corollary}
\sLehkymDukazem

\begin{theorem}
Let $X$ be a $T_{3 \frac 12}$ topological space. Then $X$ is connected if and only if $\beta X$ is connected.
\end{theorem}
\sDukazemVkontextu

\begin{theorem}
Let $X_i$, $i\in I$, be nonempty topological spaces. Then $\prod_I X_i$ is connected if and only if all spaces $X_i$, $i\in I$, are connected.
\end{theorem}
\sLehkymDukazem

\begin{defin}
Let $X$ be a topological space and let $x\in X$. The \emph{connected component} of the point $x$ is the largest connected set containing $x$. It is denoted by $C_x$.
\end{defin}

\begin{pozn}
By Proposition~\ref{prop:sjednoceniSouvislych}, the connected component of every point exists. If $C_x$ and $C_y$ are two components, then either $C_x=C_y$ or $C_x\cap C_y=\emptyset$. Thus, connected components form a partition of the space $X$.
\end{pozn}

\begin{prop}
If $X_i$, $i\in I$, are topological spaces and $x = (x_i)\in \prod_I X_i$, then
\[
C_x = \prod_I C_{x_i}.
\]
(That is, the component of $x=(x_i)$ is the product of the components of the corresponding $x_i$, $i\in I$.)
\end{prop}
\sLehkymDukazem

\begin{defin}
Let $X$ be a topological space. A set $Q$ is called the \emph{quasicomponent} of the point $x$ if
\[
Q=\bigcap\{Z\colon x\in Z,\ Z \text{ is clopen}\}.
\]
It is denoted by $Q_x$.
\end{defin}

\begin{pozn}
For every $x\in X$ we have $C_x\subseteq Q_x$. Quasicomponents are closed, since they are defined as intersections of closed sets. Moreover, quasicomponents also form a partition of the space.
\end{pozn}

\begin{example}
Let $X$ be a subset of the plane consisting of the points $a=(0,0)$, $b=(0,1)$, and a countable family of line segments joining the points $(2^{-n},0)$ and $(2^{-n},1)$. Then
$C_a = \{a\}\neq  \{a,b\} = Q_a$.
\end{example}
\sLehkymDukazem

\begin{lemma}[On intersections in compact spaces]\label{prunikvkompaktu}
Let $X$ be a compact space and let $\mathcal A$ be a family of closed sets. If $\bigcap\mathcal A\subseteq U$ for some open set $U$, then there exists a finite subfamily $\mathcal F\subseteq \mathcal A$ such that $\bigcap \mathcal F\subseteq U$.
\end{lemma}
\sLehkymDukazem

\begin{theorem}\label{komponentyakvazi}
In a compact $T_2$ space, components and quasicomponents coincide.
\end{theorem}
\sLehkymDukazem

\section{Continua}

\begin{defin}
A compact, connected, nonempty $T_2$ space is called a \emph{continuum}. If it consists of a single point, it is called \emph{degenerate}.
\end{defin}

\begin{pozn}
Continuous images and arbitrary products of continua are continua.
\end{pozn}

\begin{prop}
If $\mathcal H$ is a family of continua closed under finite intersections, then $\bigcap \mathcal H$ is a continuum.\\
(In particular, the intersection of a decreasing sequence of continua is a continuum.)
\end{prop}
\sLehkymDukazem

\begin{prop}[Boundary bumping]
If $A$ is a proper closed subset of a continuum $X$, then every component of $A$ intersects the boundary of $A$.
\end{prop}
\sTezkymDukazem

\begin{theorem}[Sierpi\'nski]
Let $X$ be a continuum and let $X_n$, $n\in\N$, be pairwise disjoint closed subsets whose union is $X$. Then $X_n=\emptyset$ for all $n$ except one.
\end{theorem}
\sTezkymDukazem

\konecPrednasky{9. 4. 2026}

\begin{defin}
A continuum is called \emph{decomposable} if it can be written as the union of two proper subcontinua. Otherwise, it is called \emph{indecomposable}.
\end{defin}

\begin{example}
    There exists an indecomposable continuum in $\er^2$.
\end{example}

\konecPrednasky{16. 4. 2026}

\section{Disconnectedness}

\begin{defin}
A Hausdorff topological space $X$ is called
\begin{itemize}
\item
\emph{hereditarily disconnected} if all components are singletons;
\item
\emph{totally disconnected} if for $x\neq y$ there exists a clopen set $Z\subseteq X$ such that $x\in Z$ and $y\notin Z$;
\item
\emph{zero-dimensional} (sometimes written \emph{$0$-dim}) if it has a base consisting of clopen sets;
\item
\emph{strongly zero-dimensional} (sometimes written \emph{strongly $0$-dim}) if for every two disjoint closed sets $E,F$ there exists a clopen set $Z$ such that $E\subseteq Z\subseteq X\setminus F$.
\end{itemize}
\end{defin}

\begin{pozns}
\begin{itemize}
    \item The terminology is not completely uniform; we use the terminology from Engelking (different terminology is used in the lecture notes). The most important notion will be zero-dimensionality (where the terminology is uniform).
    \item Strong zero-dimensionality as defined here automatically implies normality, but it can also be naturally defined in a reasonable way already in Tychonoff spaces (see, for example, the lecture notes for details).
\end{itemize}
\end{pozns}

\begin{prop}
    Let $X$ be a $T_2$ topological space. Then
    \[
    \text{$X$ is strongly $0$-dim $\implies$ $X$ is $0$-dim $\implies$ $X$ is totally disconnected $\implies$ $X$ is hereditarily disconnected.}
    \]
\end{prop}
\sLehkymDukazem

\begin{examples}
    \begin{itemize}
        \item Consider $X=\er^2$ with a topology $\tau$ defined as follows: the points $\qe^2$ are isolated, and the remaining points $x$ have basic neighborhoods of the form
        $\{x\}\cup (B(x,\varepsilon)\cap \qe^2)$ for $\varepsilon>0$. Then $(X,\tau)$ is a $T_2$ space that is hereditarily disconnected but not totally disconnected.\\
        There also exists a metrizable example, but it is considerably more complicated (see the lecture notes, Example 8.46).
        \item Consider the Erd\H{o}s space, i.e.\ $E:=\ell_2\cap \qe^\omega$ with the topology inherited from $\ell_2$. Then $E$ is a metrizable totally disconnected space that is not zero-dimensional.
        \item Examples of spaces that are zero-dimensional but not normal (and hence not strongly zero-dimensional according to our definition) were already mentioned in General Topology~1 (the product of the Sorgenfrey line, or the Isbell--Mr\'owka space).
        \item In exercises we will present an example of a normal space that is zero-dimensional but not strongly zero-dimensional. There even exists a metrizable example, but it is very complicated.
    \end{itemize}
\end{examples}
\sTezkymDukazem

\begin{theorem}[Disconnectedness in compact spaces]
For a $T_2$ compact space $X$ we have:
\[
    \text{$X$ is strongly $0$-dim $\Leftrightarrow$ $X$ is $0$-dim $\Leftrightarrow$ $X$ is totally disconnected $\Leftrightarrow$ $X$ is hereditarily disconnected.}
\]
\end{theorem}
\sLehkymDukazem

\begin{theorem}[Zero-dimensionality of $\beta X$]
Let $X$ be $T_4$. Then $\beta X$ is $0$-dim if and only if $X$ is strongly $0$-dim.
\end{theorem}
\sLehkymDukazem

\begin{prop}
Let $X$ be $T_2$. Then $X$ is zero-dimensional if and only if it can be embedded into $2^I$ for some set $I$. In that case, one can choose $I = w(X)$.
\end{prop}
\sLehkymDukazem

\begin{theorem}
Every $T_2$ compact space is a continuous image of a zero-dimensional compact space of the same weight.
\end{theorem}
\sTezkymDukazem

\chapter{Topological Dimension}

\begin{defin}[Small inductive dimension: Menger, Urysohn]
For a $T_3$ space $X$ we define its small inductive dimension inductively for $n\in\en\cup\{0\}$ as follows:
\begin{itemize}[noitemsep]
\item We say that $\ind X = -1$ if and only if $X = \emptyset$.
\item $\ind X\leq n$ if for every $x\in X$ and every neighborhood $U$ of $x$ there exists an open set $V$ such that $x\in V\subseteq U$ and $\ind(\partial V)\leq n-1$.
\item $\ind X=n$ if $\ind X\leq n$ and $\ind X\leq n-1$ does not hold.
\item $\ind X=\infty$ if $\ind X\leq n$ does not hold for any $n\in\N$.
\end{itemize}
We call $\ind X$ the \emph{small inductive dimension} of the space $X$.
\end{defin}

\begin{pozns}
Let $X$ be a $T_3$ space. Then:
    \begin{itemize}[noitemsep]
        \item $\ind X\leq 0$ if and only if $X$ is zero-dimensional;
        \item if $M\subset X$, then $\ind M\leq \ind X$;
        \item $\ind [0,1] = 1$.
    \end{itemize}
\end{pozns}

\begin{defin}[Large inductive dimension: Brouwer, \v{C}ech]
For a $T_4$ space $X$ we define its large inductive dimension inductively for $n\in\en\cup\{0\}$ as follows:
\begin{itemize}[noitemsep]
\item We say that $\Ind X = -1$ if and only if $X = \emptyset$.
\item $\Ind X\leq n$ if for every closed set $E$ and every open set $U\supseteq E$ there exists an open set $V$ such that $E\subseteq V\subseteq U$ and $\Ind(\partial V)\leq n-1$.
\item $\Ind X=n$ if $\Ind X\leq n$ and $\Ind X\leq n-1$ does not hold.
\item $\Ind X=\infty$ if $\Ind X\leq n$ does not hold for any $n\in\en$.
\end{itemize}
We call $\Ind X$ the \emph{large inductive dimension} of the space $X$.
\end{defin}

\begin{pozns}
Let $X$ be a $T_4$ space. Then:
    \begin{itemize}[noitemsep]
        \item if $M\subset X$ is closed, then $\Ind M\leq \Ind X$;
        \item $\Ind X\leq 0$ if and only if $X$ is strongly zero-dimensional;
        \item $\ind X\leq \Ind X$;
        \item $\Ind [0,1] = 1$.
    \end{itemize}
\end{pozns}

\begin{defin}
We say that a family $\mathcal A$ of subsets of a set $X$ has order $n$ if $n$ is the largest natural number for which there exist distinct elements $A_1,\dots,A_{n+1}\in\mathcal A$ such that $\bigcap A_i\neq \emptyset$.
\end{defin}

\begin{defin}[Covering dimension: \v{C}ech, Lebesgue]
For a $T_4$ space $X$ we define its covering dimension inductively for $n\in\en\cup\{0\}$ as follows:
\begin{itemize}[noitemsep]
\item $\dim\emptyset = -1$.
\item $\dim X\leq n$ if every finite open cover of $X$ is refined by a finite open cover of order at most $n$.
\item $\dim X=n$ if $\dim X\leq n$ and $\dim X\leq n-1$ does not hold.
\item $\dim X=\infty$ if $\dim X\leq n$ does not hold for any $n$.
\end{itemize}
We call $\dim X$ the \emph{covering dimension} of the space $X$.
\end{defin}

\begin{pozns}
Let $X$ be a $T_4$ space. Then:
    \begin{itemize}[noitemsep]
        \item if $M\subset X$ is closed, then $\dim M\leq \dim X$;
        \item $\dim [0,1] = 1$.
    \end{itemize}
\end{pozns}

\konecPrednasky{23. 4. 2026}

\begin{prop}
In a $T_4$ topological space $X$, we have $\dim X\leq 0$ if and only if $X$ is strongly $0$-dimensional.
\end{prop}
\sLehkymDukazem

\begin{defin}
Let $X$ be a set and let $\mathcal S\subseteq \mathcal P(X)$ be a cover of $X$. An indexed family $\{T_S\colon S\in\mathcal S\}$ is called a \emph{shrinking} of the family $\mathcal S$ if it is a cover and $T_S\subseteq S$ for each $S\in\mathcal S$.
\end{defin}

\begin{lemma}[On shrinking]
    Let $X$ be a $T_4$ space and let $\{G_1,\ldots,G_n\}$ be an open cover of $X$. Then there exists an open cover $\{H_1,\ldots,H_n\}$ of $X$ such that $\overline{H_i}\subset G_i$ for $i\in\{1,\ldots,n\}$.\\
    (That is, every finite open cover has a closed shrinking whose interiors also form a cover.)
\end{lemma}
\sLehkymDukazem

\konecPrednasky{30. 4. 2026}

\begin{lemma}[On swelling]
    Let $X$ be a $T_4$ space, let $\{F_1,\ldots,F_n\}$ be a finite family of closed subsets of $X$ of order at most $n$, and let $\{U_1,\ldots,U_n\}$ be open sets such that $F_i\subset U_i$ for $i=1,\ldots,n$. Then there exists a family $\{V_1,\ldots,V_n\}$ of open subsets of $X$ such that $\{\overline{V_1},\ldots,\overline{V_n}\}$ has order at most $n$ and
    \[
    F_i\subset V_i\subset \overline{V_i}\subset U_i
    \]
    for each $i=1,\ldots,n$.
\end{lemma}
\sLehkymDukazem

\begin{theorem}[Characterization of covering dimension]
For a $T_4$ space $X$, the following conditions are equivalent.
\begin{enumerate}[label=(\alph*)]
\item $\dim X \leq n$.
\item Every finite open cover of $X$ has an open shrinking of order at most $n$.
\item Every finite open cover of $X$ has a closed shrinking of order at most $n$.
\item Every finite open cover of $X$ is refined by a finite closed cover of order at most $n$.
\end{enumerate}
\end{theorem}
\sLehkymDukazem

\begin{theorem}[Sum theorem for the dimension $\dim$]\label{souctovavetadim}
If a $T_4$ space $X$ is the union of countably many closed subspaces $F_i$ and $\dim F_i\leq n$, then $\dim X\leq n$.
\end{theorem}
\sTezkymDukazem

\begin{theorem}
    If $X$ is $T_4$, then $\dim X\leq \Ind X$.
\end{theorem}
\sLehkymDukazem

\section{Topological Dimension in Metrizable Spaces}

\begin{theorem}
    If $X$ is a metrizable space, then $\dim X = \Ind X$.
\end{theorem}
\begin{proof}The proof was presented only for the special case of compact $X$, this special case can be examined.
\end{proof}

\konecPrednasky{7. 5. 2026}

\begin{lemma}
    Let $X$ be a metrizable space and let $Z\subset X$ be strongly $0$-dimensional. Then for every closed set $F\subset X$ and every open set $U\subset X$ with $F\subset U$ there exists an open set $V\subset X$ such that
    \[
    F\subset V\subset \overline{V}\subset U
    \]
    and $Z\cap \partial V = \emptyset$.
\end{lemma}

\begin{theorem}
    Let $X$ be a metrizable separable space. Then
    \[
    \ind X = \dim X = \Ind X.
    \]
\end{theorem}

\begin{theorem}
    Let $X$ be a metrizable space and let $n\in \en\cup\{0\}$. Then the following statements are equivalent.
    \begin{enumerate}[label=(\alph*)]
        \item $\Ind X\leq n$,
        \item $X = Y \cup Z$, where $\Ind Y\leq n-1$ and $\Ind Z\leq 0$.
    \end{enumerate}    
\end{theorem}

\begin{corollary}[On separation]
Let $X$ be a metrizable space and let $n\in \en\cup\{0\}$. If $\Ind X\leq n$, then for every sequence of $(n+1)$ pairs of closed disjoint sets
$(F_1,H_1),\ldots,(F_{n+1}, H_{n+1})$, there exist open sets $U_i$, $i=1,\dots, n+1$, such that
\[
F_i\subseteq U_i\subseteq\cl {U_i}\subseteq X\setminus H_i
\quad\text{and}\quad
\bigcap_{i=1}^{n+1} \partial U_i=\emptyset.
\]
\end{corollary}

\begin{theorem}
    Let $X$ and $Y$ be nonempty metrizable spaces. Then
    \[
    \Ind (X\times Y)\leq \Ind X + \Ind Y.
    \]
\end{theorem}

\section{Dimension and Euclidean Spaces}

\begin{theorem}[Brouwer's fixed point theorem]
Every continuous mapping $f\colon [0,1]^n\to [0,1]^n$ has a fixed point, i.e.\ there exists $x\in [0,1]^n$ such that $f(x)=x$.
\end{theorem}

\begin{theorem} 
For every $n\in\en$ we have $\dim [0,1]^n=\dim\R^n=n$.
\end{theorem}

\begin{corollary}
    If $n,m\in\en$ and $n\neq m$, then $\er^n$ is not homeomorphic to $\er^m$.
\end{corollary}

\end{document}