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I have been working on this question. Let A be an open set. Does Int(A-closure)= A?

Here is my answer:

Let my space be [0,1] ∪ {2} and my topology be the subspace topology.

Let A = (0,1) ∪ {2}.

Then, A-closure = [0,1] ∪ {2}

But then, int(A-closure) = [0,1] ∪ {2} = A-closure

Hence int(A-closure) does not equal A.

Kara
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  • Your example does not work -- if your topological space is $X=[0,1]\cup{2}$ with the subset topology, then $2$ is certainly in the interior of $X$ itself, since $X$ contains a neighborhood of $2$, namely $X$. – hmakholm left over Monica May 26 '16 at 16:25
  • So then is {2} not an isolated point? – Kara May 26 '16 at 16:28
  • x @Kara: $2$ is an isolated point yes -- and an isolated point of the topological space is always interior in any subset it is a member of. – hmakholm left over Monica May 26 '16 at 16:33
  • The interior of A is defined as the union of all open sets contained in A. So {2} is an open set in A, so it is in the interior. I see it now.

    So A-closure = [0,1] union {2}. Then int(A-closure) = (0,1) union {2}, which equals A.

     – Kara May 26 '16 at 18:11
  • x @Kara, actually, since $\bar A$ is the entire topological space in this case, $\bar A$ is both open and closed, and it its own interior. So your counterexample is a counterexample, just not in the way you imagined. – hmakholm left over Monica May 26 '16 at 18:38
  • So wait...........are you saying that Int(A-closure) = [0,1] union {2}? The interior of A is defined as the union of all open sets contained in A, so the whole topological space IS an open set. Right, ok, I see it now. – Kara May 26 '16 at 18:42

1 Answers1

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Consider $A=(0,1)\cup (1,2)$ which is open w.r.t. the usual topology on $\mathbb{R}$. Then, $\bar{A}=[0,2]$, so int$\bar{A}=(0,2)\not= A$

ervx
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