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Recently I've started studying dimension (inductive) theory and there's a logic behind the statement below I don't understand, so I need help. If there already exists a duplicate to my question, I apologize in advance, but I haven't found any. I'm only hoping answer isn't trivial.

Let $\left(X,\mathcal{T}\right)$ be separable, metrizable topological space.

Statement Let $p \in X$ and $C$ nonempty, compact subset of $X$ not containing $p$. If $p$ and $c$ are separated in $X$ for every $c \in C$, then $p$ and $C$ are also separated in $X$.

Problem I don't understand why converse isn't true and why is $C$ being compact important.

Definition I'm using is:

Disjoint sets $A,B \subset X$ are said to be separated in $X$ if there exists a clopen set $U \subset X$ such that $A \subset U$ and $B \subset X/U $.

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Where did you learn that the converse does not hold? – Harald Hanche-Olsen Apr 23 '14 at 21:10
@HaraldHanche-Olsen Author of the book I'm using (Nadler - Dimension Theory: An Indroduction with Exercises) does not include converse implication in that statement, which is actually stated as lemma. What I'm thinking is that he could have left out converse implication because it is trivial (and that was my first thought after reading it), but still it doesn't seem right to do so, since it doesn't mention (after or before) anything about converse implication. I doubted because statement is not written in its complete form. – aquarius Apr 23 '14 at 21:44
Ah. Well, you shouldn't read too much into an omission like that. The converse is indeed trivial. Especially when it is stated as a lemma: Lemmas tend to be tightly focused on the intended usage, so adding the converse would just add noise if it is not required. – Harald Hanche-Olsen Apr 24 '14 at 8:32
Oh, I see your point. Thank you very much for your advice and help. – aquarius Apr 24 '14 at 10:01
up vote 1 down vote accepted

Importance of compactness: Let $p=0$, $C=\{1,1/2,1/3,\ldots\}$, $X=\{p\}\cup C$.

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