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A. Lelek, ZBIORY (Sets), PZWS Warszawa 1966, pages
9095.
The first hereditarily indecomposable
continuum has been constructed by B. Knaster in 1922. This
continuum K, discovered by Knaster, is just named the pseudoarc or Knaster's continuum. The pseudoarc K
is obtained as the common part of a decreasing sequence of
continua in the plane, every of which is the union of
finitely many discs, forming a chain of sets.
We
will consider chains
of discs D_{i} in the plane, having the property that any
two consecutive disc D_{i} and D_{i+1} have interior
points in common, i.e., that their intersection  nonempty
by assumption  is not contained in the boundary of any of
them (Figure A). Discs D_{i} are called links of the
chain . The union of all links of the chain
is a continuum. Observe that if we delete from the chain
a link which is not an end link, i.e., a link D_{i}
such that 1 < i < k, then the union of the remaining links
is a not connected set, having two components.
Let two
chains of discs and be given. We say that
the chain refines the chain if each
link D'_{s} of the chain is contained in the
interior of at least one link D_{i} of the chain ,
i.e., if for every
there exists a link
such that
and that the
disc D'_{s} has no common points with the boundary of the
disc D_{i} (Figure B).
We say that the chain
is crooked in the chain if it refines and if, for every pair of links D_{i} and D_{j} of the
chain such that
and for every pair of links D'_{s} and D'_{v} of the chain
, intersecting the links D_{i} and D_{j}
respectively, i.e.,
there are, in the chain , links D'_{t} and D'_{u}, lying between
the link D'_{s} and D'_{v} in the same order, i.e.,
such that

(2) 
In other words, a part of the chain which is
contained between the links D'_{s} and D'_{v} has to form a
fold in the chain between the links D_{i} and D_{j}
(Figure C). The chain , to go from the link D_{i}
to the link D_{j} has first to come to the link D_{j1},
next go back to the link D_{i+1}, and only after this it
can reach the link D_{j}. Such a fold must exist for every
two links D_{i} and D_{j} having no adjoining (i.e.
neighbor) link in common.
Now, let there be given
in the plane two points a and b, and an infinite
sequence
of chains of discs, satisfying, for every
, the
following conditions:
 1^{o}
 the point a belongs to the first, and the point b to the
last link of the chain ;
 2^{o}
 the diameter of every disc in the chain is less
than ;
 3^{o}
 the chain
is crooked in the chain .
Thus, for example, the chain is crooked in the chain
, and the chain is crooked in the chain (on
Figure D only a part of the chain is presented). The foldings are
more and more condensed and they overlapped themselves.
Denote by K_{n} the union of all links of the chain D_{n}. From condition
3^{o} it follows in particular that the each next chain refines the previous
one, and therefore the continua K_{n} form a decreasing sequence
The Pseudoarc K is defined as the common part of all continua
K_{n}, i.e.,
By virtue of condition 1^{o} points a and b belong to the continuum K,
thus it is nondegenerate. We shall prove that the continuum K is
hereditarily indecomposable.
Let be an arbitrary continuum. To prove that the continuum
K is indecomposable it is enough to show that every continuum distinct from K' is nowhere dense in K'. Consider an
arbitrary point and number
.
Since , hence there is a point
which
therefore is at a distance at least from each point ,
where is a fixed number, independent from y. In fact, if not, then
some points of the set would be arbitrarily close to the point q, and so
the point q would belong to the closure of the set Y, which is
impossible, because the set Y is closed, and it does not contain the point
q. Let us take a natural number n so large that the inequalities are
satisfied
Denote by D_{i} and D_{j} (with ) the links of the chain
whose union contains the points p and q. Since , hence
, (here means the metric in the plane) and thus
it follows that between the links D_{i} and D_{j} there are at least two
links of the chain . Indeed, in the opposite case the links D_{i}
and D_{j} would have a neighbor common link, and choosing suitable points
p' and q' in the intersections of this link with the links D_{i} and
D_{j} we would have
by condition 2^{o}. Thus we can assume that the indices i and j satisfy
the inequality (1). Further, we can assume also that the point p belongs
to one, and the point q to the other of the links D_{i} and D_{j}.
Denote by D'_{s} and D'_{v} the links of the chain
which
contain points p and q, respectively. Additionally we assume that
. In the opposite case, i.e. if , the further part of
the proof runs in the same way (with changing of the roles of the links
D_{i} and D_{j} as well as D_{i+1} and D_{j1}).
So,
whence we infer, by virtue of condition 3^{o} and the definition of folding
of chains, that there are two links D'_{t} and D'_{u} in the chain
which lie between the links D'_{s} and D'_{v} in the same order and
which satisfy the inclusions (2). The union of the links of the chain
distinct from the link D'_{u} is a not connected set (see
above) containing the point p in one component, and the point q in the
other. Since the continuum K' joins the points p and q and is
contained in the union K_{n+1} of all links of the chain
,
hence K' must pass thru the link D'_{u}. Thus there is a point
(Figure E). We shall prove that the point x does not belong to the
continuum Y. Indeed, if it would be , then by the same reason as
previously for the continuum K' the continuum Y would have to pass thru
the link D'_{t}. Then there would exist a point
which would be in the link D_{j1} according to (2). But
the neighbor links D_{j} and D_{j1} have their diameters
less than by condition 2^{o}, and therefore we
would have
which is impossible because .
So, the point
x belongs to the set
and to the link
D_{i+1} simultaneously. The neighbor links D_{i} and
D_{i+1} have diameters less than , whence it
follows that
Since was an arbitrary positive number, hence, in
this way we have proved in that there are points of the set
which lie arbitrarily closely to the point
p. The point p was an arbitrary point of the continuum
Y. Thus we have proved that the continuum Y is nowhere
dense in K', and the proof of hereditary indecomposability
of the continuum K is finished.
An example of a chain crooked in a chain with 7 links is on
Figure F.
Figure (
A
)
a chain with links, construction of the
Pseudoarc
Figure (
B
)
refining a chain, construction of the
Pseudoarc
Figure (
C
)
crookedness used in the construction of the
Pseudoarc
Figure (
D
)
crookedness used in the construction of the
Pseudoarc
Figure (
E
)
crookedness used in the construction of the
Pseudoarc
Figure (
F
)
an example of a chain crooked in a chain with 7 liks, construction of the
Pseudoarc
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Next: Dyadic Selenoid
Up: New examples
Previous: Whyburn's Curve
Janusz J. Charatonik, Pawel Krupski and Pavel Pyrih
20010221