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I have shown that if $X$ is a $T_1$ space, $A\subset X$ is connected and $|A|>1,$ then $$A\subset A'.$$

I was wondering if this is also true for an arbitrary topological space $X$?

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  • $\begingroup$ What's $A'$? $\left.\right.$ $\endgroup$
    – 5xum
    May 9, 2017 at 8:35
  • $\begingroup$ The derived set of $A$ $\endgroup$
    – samantha
    May 9, 2017 at 8:41

2 Answers 2

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No, the result does not hold for an arbitrary topological space. Consider Sierpiński space: $A = \{a,b\}$, $\tau=\{\emptyset,\{a\},\{a,b\} \}$. Then $A' = \{b\}$, which is not a superset of A. And $(A,\tau)$ is connected (do you know why?).

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  • $\begingroup$ Happy to help. This topology and discrete topology are really useful for easy counterexamples. $\endgroup$
    – ElChorro
    May 9, 2017 at 9:09
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Suppose $X$ is $T_1$ and $A$, $|A| > 2$ is connected. Let $a \in A$, if $a \notin A'$ this means there is an open set $O$ of $X$ such that $O \cap A = \{a\}$. But this means that $B = \{a\}$ is open in $A$ and closed in $A$ by $X$ being $T_1$, so $A =B \cup A \setminus B$ is a disconnection of $A$, as $|A| > 2$ implies the other part is nonempty too, contradiction. So $A \subseteq A'$.

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  • $\begingroup$ Thanks for your prove Henno. It will be useful for further readers :) $\endgroup$
    – ElChorro
    May 9, 2017 at 9:21
  • $\begingroup$ @ElChorro I misread the question. He was looking for a non-$T_1$ counterexample which you gave. I thought he was looking for the proof. I'll keep it here as a reference. $\endgroup$ May 9, 2017 at 9:23

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