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## Pseudo-arc

A. Lelek, ZBIORY (Sets), PZWS Warszawa 1966, pages 90-95.

The first hereditarily indecomposable continuum has been constructed by B. Knaster in 1922. This continuum K, discovered by Knaster, is just named the pseudo-arc or Knaster's continuum. The pseudo-arc 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 Di in the plane, having the property that any two consecutive disc Di and Di+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 Di 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 Di 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 Di 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 Di (Figure B).

We say that the chain is crooked in the chain if it refines and if, for every pair of links Di and Dj of the chain such that

 i + 2 < j, (1)

and for every pair of links D's and D'v of the chain , intersecting the links Di and Dj 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 Di and Dj (Figure C). The chain , to go from the link Di to the link Dj has first to come to the link Dj-1, next go back to the link Di+1, and only after this it can reach the link Dj. Such a fold must exist for every two links Di and Dj 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:
1o
the point a belongs to the first, and the point b to the last link of the chain ;
2o
the diameter of every disc in the chain is less than ;
3o
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 Kn the union of all links of the chain Dn. From condition 3o it follows in particular that the each next chain refines the previous one, and therefore the continua Kn form a decreasing sequence

The Pseudo-arc K is defined as the common part of all continua Kn, i.e.,

By virtue of condition 1o 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 Di and Dj (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 Di and Dj there are at least two links of the chain . Indeed, in the opposite case the links Di and Dj would have a neighbor common link, and choosing suitable points p' and q' in the intersections of this link with the links Di and Dj we would have

by condition 2o. 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 Di and Dj.

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 Di and Dj as well as Di+1 and Dj-1).

So,

whence we infer, by virtue of condition 3o 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 Kn+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 Dj-1 according to (2). But the neighbor links Dj and Dj-1 have their diameters less than by condition 2o, and therefore we would have

which is impossible because .

So, the point x belongs to the set and to the link Di+1 simultaneously. The neighbor links Di and Di+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 Pseudo-arc

Figure ( B ) refining a chain, construction of the Pseudo-arc

Figure ( C ) crookedness used in the construction of the Pseudo-arc

Figure ( D ) crookedness used in the construction of the Pseudo-arc

Figure ( E ) crookedness used in the construction of the Pseudo-arc

Figure ( F ) an example of a chain crooked in a chain with 7 liks, construction of the Pseudo-arc

Source files: a.cdr . a.eps . a.gif . a.txt . b.cdr . b.eps . b.gif . b.txt . c.cdr . c.eps . c.gif . c.txt . d.cdr . d.eps . d.gif . d.txt . e.cdr . e.eps . e.gif . e.txt . f.cdr . f.eps . f.gif . f.txt . latex.tex . title.txt .

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Next: Dyadic Selenoid Up: New examples Previous: Whyburn's Curve
Janusz J. Charatonik, Pawel Krupski and Pavel Pyrih
2001-02-21