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I want to prove the following:

Let $(X, \mathcal{T})$ be a topological space. Let $A \subset X$ be a subset. Then $\mathrm{int}A$ is the largest open subset of $A$.

Our definition is $\mathrm{int}A = \{x \in A : \exists \text{ neighbourhood } N \text{ of } x \text{ such that } N \subset A\}$

I'm not sure if my proof is correct and would appreciate a check of it:

We have already previously shown that $\mathrm{int}A$ is open (this was a previous proposition). Let $U \subset A$ be open. Let $x \in U$. Then $\exists$ neighbourhood $N$ of $x$ such that $N \subset U$ (since $U$ is open). But $N \subset U \subset A$, i.e. $x \in \mathrm{int}A$, and we have shown that $U \subset \mathrm{int}A$.

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    $\begingroup$ Yes, this is fine. $\endgroup$ May 12, 2016 at 2:10
  • $\begingroup$ Thanks for the verification - it was just that it seemed too short to me, and since it was set as an exercise it planted seeds of self doubt. $\endgroup$ May 12, 2016 at 2:12
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    $\begingroup$ In this case it really is that easy, and your writeup is very clear. $\endgroup$ May 12, 2016 at 2:17
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    $\begingroup$ My pleasure! $\,$ $\endgroup$ May 12, 2016 at 2:21
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    $\begingroup$ Note also that the interior of $A$ is the union of all open subsets of $A$, and that the closure of $A$ is the intersection of all closed supersets of $A$. $\endgroup$
    – user21820
    May 12, 2016 at 4:17

1 Answer 1

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For the sake of removing this question from the unanswered queue: Brian M. Scott has confirmed that this proof is correct.

user21820 also makes the point that the interior of $A$ is the union of all open subsets of $A$, and that the closure of A is the intersection of all closed supersets of $A$.

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