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Let C be an infinite, separable and connected metric space. If C becomes totally disconnected when one of its points p is removed, does every closed ball of C with positive radius and center p always contain an infinite connected subset?

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I have a question: if C becomes totally disconnected when the point p is removed, does it necessarily mean that every connected component of C has to include p? Thx for clarifying. –  Carlo Von Schnitzel Aug 18 '10 at 23:15
    
@alephomega: I believe the answer is yes. For if $B\subset C$ and contains more then one point, but $p\notin B$, then when removing $p$ we still have $B\subset C\setminus\{p\}$, and $B$ is still connected, which is a contradiction to the totally disconnectedness of $C\setminus \{p\}$. –  Dror Atariah Aug 19 '10 at 8:07

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I don't know if this answers your question precisely but here's an interesting example.

Let $K$ be the Knaster-Kuratowski fan, also called the Cantor's teepee. The space $K$ is defined as follows: let $C$ be the Cantor Set. Let $Q$ be the set of endpoints of the deleted middle-third intervals. Then $Q \subset C$. Let $P=C \backslash Q$. We also let $p=(1/2,1/2) \in \mathbb R^2$. Now $\forall x \in C$, we let $L_x$ be the line joining $p$ and $x$. Now our space $K$ is the union over all $x \in C$ of the lines $L_x=\ {(x,y):y\in \mathbb Q \ }$ if $x\in Q$ and $L_x=\ {(x,y):y\in Irr \ }$ if $x \in P$. The space $K$ is connected but $K\backslash {p}$ is totally disconnected. Also $K$ is punctiform, that is it contains no compact connected T2 subspace. Let $f$ be a continuous function from $K$ to $K$. Let $U$ be some closed connected open set about a point $x$, $x \neq p$, then there exists a closed connected open set $V$ about $p$, which can be written down as the set of all $(x,y)$ with $1/2 - \epsilon < y < 1/2$ such that $f(V) \subset U$ since $f$ is continuous. Then $V$ is necessarily homeomorphic to the space $K$

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