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Let C be a connected and completely metrizable subset of the Euclidean plane. Can C fail to be locally connected at each of its points?

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Is the Knaster-Kuratowski fan completely metrizable? Gerhard "Ask Me About System Design" Paseman, 2011.08.09 –  Gerhard Paseman Aug 10 '11 at 0:24

4 Answers 4

The pseudo-arc is another planar example. In fact, isn't Victor's example a pseudo-arc?

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Bill, pseudo-arc was my first thought, but since it's homeomorphic to its nondegenerate subcontinua, doesn't this mean that its closed metric neighborhoods are connected and, consequently, that it is locally connected at each point? –  Victor Protsak Aug 23 '10 at 22:04
No, because e.g. a locally connected continuum is arcwise connected. (Continuum is important here; there are connected and locally connected metric spaces that do not contain any arc.) There are strong uniqueness properties of the pseudo-arc. I don't remember what they are but suspect that they imply that your example is a pseudo-arc. –  Bill Johnson Aug 23 '10 at 22:29
Thank you, that does explain it. The fallacy in my objection was a silly circular reasoning: the closed neighborhood would be homeomorphic to the whole pseudo-arc iff it is a subcontinuum, i.e. connected (which is what we are trying to determine). Meanwhile, I found a much more elementary example and updated my answer. –  Victor Protsak Aug 23 '10 at 22:51
By the way, I don't think that a continuous image of a solenoid can be a pseudo-arc: while a solenoid isn't locally connected, it does contain a closed arc through every point, which would have to contract to a point, since a pseudo-arc doesn't contain a nondegenerate pathwise connected subcontinuum. –  Victor Protsak Aug 24 '10 at 0:14
See e.g. Henon attractor for a picture. en.wikipedia.org/wiki/Henon_attractor –  Ian Agol Aug 24 '10 at 4:28

Edit Here is an easier example. Let $K$ be the middle third Cantor set and

$$B=\{(t,ty): 0\leq t\leq 1, y\in K\}$$

be the cone over $K,$ the Cantor branch and $C=B\cup\varphi(B),$ where $\phi$ is the central symmetry about the midpoint $(1/2,0)$ of the bottom twig $[0,1]\times\{0\}.$ Thus $C$ is obtained by gluing two Cantor branches rotated by $\pi$ relative to each other along their bottom twigs. This space $C$ is compact and pathwise connected, but it is not locally connected at each point. Indeed, a small neighborhood of every point $P$ will contain parts of Cantor twigs not passing through $P.$

Yes. A solenoid is a homogeneous continuum (=compact connected metric space) embeddable in $\mathbb{R}^3$ that is not locally connected at any point, in fact, a small neighborhood of each point looks like the Cantor set crossed with an interval. Its generic projection to $\mathbb{R}^2$ is compact, connected, and not locally connected at each of its points. [Edit I am not longer confident that the last claim is true.]

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Another description of the same space: connect the origin with every point of the Cantor set on the vertical segment between $(1,0)$ and $(1,1)$ and the point $(1,0)$ with every point of the Cantor set on the vertical segment between the origin and $(0,-1)$. –  Victor Protsak Aug 24 '10 at 0:08
Nice and very simple, Victor. –  Bill Johnson Aug 24 '10 at 1:00
Thanks, Victor, for a very neat example which is easy to visualize and is actually a compact continuum. –  Garabed Gulbenkian Aug 24 '10 at 18:29

Take the subset $A=\left(\{1\}\times[0,1]\right)\cup\left(\bigcup_{n\in\mathbb{N}}[0,1]\times\{\frac{1}{n}\}\right)\cup\left([0,1]\times \{0\}\right)$. In this way describe, it might seem awful, but if you draw it, it's a very simple set.

This set is closed, and so completely metrizable. It is also connected (three segments are enough to connect any two points, so it is even pathwise connected). But at the point $(0,0)$ it has no neighbourhood base of connected subsets.

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Any indecomposable continuum has the property you desire.

(A continuum is indecomposable if it cannot be written as a union of two proper subcontinua.)

One way to see this is that any indecomposable continuum has uncountably many composants, all of which are mutually disjoint, and all of which are dense in $x$. (Here the composant of a point $x$ in $X$ is the union of all proper subcontinua of $X$ that contain $x$.)

Here is a more direct proof: Suppose that $C$ is a proper subcontinuum of $X$ that is a neighborhood of $x$. Then every connected component of $V := X\setminus C$ contains a point of $C$ in its boundary. (This is known as the 'boundary bumping theorem'.)

If $\overline{V}$ is connected, then $\overline{V}$ and $C$ are proper subcontinua of $X$ whose union is $X$.

If $V$ is disconnected, decompose $V$ into two relatively closed disjoint subsets $A$ and $B$; then $A\cup C$ and $B\cup C$ are the desired subcontinua.

A simple example of an indecomposable continuum is given by the Knaster bucket-handle, see

http://commons.wikimedia.org/wiki/File%3aThe_Knaster_%22bucket-handle%22_continuum.svg .

The solenoid, mentioned in another answer, is another indecomposable continuum. You can also get such examples from "Lakes of Wada" continua. Of course the double Cantor brush given by Victor is not indecomposable (and in fact hereditarily decomposable).

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