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.

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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. –  jef 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.

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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. –  L Spice Feb 27 '10 at 20:12
Oops, never mind, I see from above that it is a compact, connected metric space. –  L Spice 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 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

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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
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.