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26
7
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What the title said. In a slightly more leisurely fashion:-
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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21
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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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7
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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 |
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20
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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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