# Can a connected planar compactum minus a point be totally disconnected?

What the title said. In a slightly more leisurely fashion:-

Let $X$ be a compact, connected subset of $\mathbb{R}^2$ with more than one point, and let $x\in X$. Can $X\smallsetminus\{x\}$ be totally disconnected?

Note that the Knaster-Kuratowski fan shows that, in the absence of the compactness hypothesis, the answer can be 'yes'.

To give credit where it's due, this question was inspired by one that I was asked by Barry Simon.

• I'm almost embarrassed to say this, but perhaps you want to add the condition that X has more than one point. – Tom Leinster Feb 27 '10 at 3:15
• Do you have an example of non-planar one? – Anton Petrunin Feb 27 '10 at 4:35
• Is there an example of a compact connected set such that no two points can be joined by a path? – gowers Feb 27 '10 at 12:36
• that can be inbedden in the plane, and it doesn't seem connected to me. – faridrb Feb 27 '10 at 18:27
• Now I'm confused about what you mean by the 1-point compactification of Q. The complement of a neighborhood of infinity in the usual 1-point compactification must be compact, so I don't see a disconnection. It's not the induced topology on the rationals plus infinity from the 1-point compactification of the reals. However, it's also not Hasudorff, and therefore can't be embedded in the plane. – Douglas Zare Feb 27 '10 at 20:45

Being planar has nothing to do with the problem. Suppose a totally disconnects $X$ and choose $b$ different from $a$. By passing to a sub continuum, assume that no proper sub continuum contains both $a$ and $b$. Take non empty disjoint open sets $U$ and $V$ whose union is $X\sim a$. WLOG $b$ is in $U$, and observe that $U\cup \{a\}$ is closed and connected.

• Bill, I'm sorry to look foolish: What is a continuum here? Is it a set of a certain cardinality, an image of an interval, a connected planar compactum, or something else? (I am no topologist, so I assume that the term is standard; I just haven't encountered it.) The “pass to a minimal sub-continuum” step worries me a little. – LSpice Feb 27 '10 at 20:12
• Oops, never mind, I see from above that it is a compact, connected metric space. – LSpice Feb 27 '10 at 20:19
• The intersection of a nested family of continua is again a continuum. You are right to ask about this--it is the only place that compactness is needed. – Bill Johnson Feb 27 '10 at 20:35
• I like this. So, Bill, just to be clear, you're invoking Zorn's Lemma to construct your minimal continuum? – HJRW Feb 27 '10 at 20:46
• Yes, although by separability you can lower the logical strength a bit if you care about that. – Bill Johnson Feb 27 '10 at 20:49

Let denote by $U_n\subset \mathbb R^2$ a sequence of open bounded neigborhoods of $X$, so that $$U_{n+1}\subset U_n\ \ \text{and}\ \ \bigcap_n U_n=X.$$ We can assume that all $U_n$ are connceted and therefore path-connected. Coose a point $p\in X$ distict from $x$ and consider a sequence of paths $\gamma_n$ in $U_n$ from $p$ to $x$. Fix $\epsilon>0$ such that $\epsilon<|p-x|$. For each path choose the smalest value $t_n\in[0,1]$ so that $|\gamma_n(t_n)-x|=\epsilon$. The image $Z_n=\gamma([0,t_n])$ is connected compact set. Let $Z$ be a Hausdorff limit of a subsequence of $Z_n$. Note that $Z$ is a compact connected subset of $X$. Clearly, $Z\not\ni x$ and it contains at least two points; a contradiction

• P.S. From Kuratowski embedding, any compact metric space is isometric to a subset of $L^\infty$. Thus the same argument works for any continuum (=compact connected metric space). – Anton Petrunin Feb 27 '10 at 20:30
• Oh, I see we "crossed paths", Anton. Your proof also works for a general continuum by making the Hilbert cube the ambient space. It also shows that any minimal sub continuum joining two points can be written as a Hausdorff limit of (even piecewise linear) arcs in some suitable larger space. – Bill Johnson Feb 27 '10 at 20:44
• OK; another crossing of paths... – Bill Johnson Feb 27 '10 at 20:44
• Thanks for the excellent answer, Anton. I marked Bill's answer as accepted, for its extra simplicity. If I had two ticks to give, I'd give them! – HJRW Feb 27 '10 at 20:53

Two great answers have already been given, and I don't claim to add much, but here is something anyway.

A totally disconnected locally compact Hausdorff space has a basis of clopen sets, according to Proposition 3.1.7 of Arhangel'skii and Tkachenko, for example. A closed set in $X-\{a\}$ need not be closed in $X$, but if $X$ is a metric space then the clopen subsets of $X-\{a\}$ at positive distance to $a$ will be clopen in $X$. Thus if $X$ is a compact metric space with more than one point and $X-\{a\}$ is totally disconnected, then $X$ is not connected.

I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.

Proposition. Let $$X$$ be a connected metric space that contains more than one point. Let $$A\subset X$$ be totally disconnected and locally compact with respect to the subspace topology. Then $$A$$ is nowhere dense.

Proof. First, let us show $$int A =\varnothing$$. Assume that $$U$$ is an open set in $$X$$ such that $$\overline{U}$$ is compact and contained in $$A$$, and $$x\in U$$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $$V\subset U$$ of $$x$$ that is clopen in $$A$$. Then, there are closed set $$F$$ in $$X$$ and open set $$W$$ in $$X$$ such that $$V=A\cap F=A\cap W$$. Since $$V\subset U\subset A$$, $$V=U\cap W$$ is open in $$X$$. Since $$V\subset \overline{U}\subset A$$, $$V=\overline{U}\cap F$$ is closed in $$X$$. Hence, $$V$$ is nonempty and clopen in $$X$$ which contradicts its connectedness.

Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $$A=\overline{A}\cap U$$, for some open $$U\subset X$$. Assume $$int \overline{A}\ne\varnothing$$. Then, there is $$x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$$, from where $$x\in int \overline{A} \cap U\subset A$$, and so $$x\in int A$$. Contradiction with the previous step.

Corollary. If $$X$$ is a connected complete metric space that contains more than one point, it cannot be covered by a countable collection of totally disconnected locally compact subsets.

In particular, we cannot remove a closed totally disconnected set (e.g. a single point) from a continuum to make it totally disconnected.