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I'm trying to understand the following claim in a proof that $\text{Bd}(A) \cap \text{Int}(A) = \emptyset$, as well as a claim in the proof of $\overline{A} = \text{Int}(A) \cup \text{Bd}(A)$. I will bold the misunderstandings below.

First misunderstanding: If $x \in \text{Int}(A)$, then $x \in U$ for some open set $U \subset A$. Then, $U$ is a neighbourhood of $x$ disjoint from $X-A$, so $x$ is not in the closure of $X-A$.

Second misunderstanding: It is obvious that $\text{Int}(A) \cup \text{Bd}(A) \subset \overline{A}$. (Why is this obvious?)

If anyone can help clarify, that would be greatly appreciated.

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    $\begingroup$ One thing which is always useful in these problems is to state the precise definitions of terms - what are the definitions of the boundary, interior and closure of a set? What do you mean by an open set? $\endgroup$ – Mark Bennet Sep 6 '18 at 22:03
  • $\begingroup$ Common alternative notations for the boundary of $A$ are $\partial A$... (\partial A...) and Fr($A$)... (for Frontier). $\endgroup$ – DanielWainfleet Sep 7 '18 at 3:22
  • $\begingroup$ If $U$ is open and disjoint from a set $S$ then $U$ is disjoint from $\bar S$... See my answer. $\endgroup$ – DanielWainfleet Sep 7 '18 at 4:16
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(1). We have $Int (A)\subset A\subset \overline A.$ And we have $Bd(A)=\overline A\cap \overline {X\backslash A}\subset \overline A.$

So the union $Int(A)\cup Bd(A)$ is also a subset of $\overline A.$

(2a). The closure $\overline Y$ of a set $Y$ is defined as the common intersection of all closed sets that have $Y$ as a subset. So the complement, in $X,$ of $\overline Y$, is the union of all open sets that are disjoint from $Y.$

(2b). So any open subset of $X$ that is disjoint from $Y=X-A $ is also disjoint from $\overline Y=\overline {X-A}.$ In your Q, we have open $U$ disjoint from $X-A,$ so $U$ is disjoint from $\overline {X-A}.$ And $p\in U.$

Remark. From the last sentence of (2a) we have $X-\overline Y=Int (X-Y).$ In particular, if $Y=X-A, $ then $X-\overline {X-A}=Int (X-(X-A))=Int(A)$.

This tells us that $\overline A- Int(A)=$ $\overline A - (X-\overline {X-A})=$ $=\overline A \cap \overline {X-A}$ $=Bd(A).$ That is worth repeating: $$\overline A-Int(A)=Bd(A).$$

So, since $\overline A\supset A\supset Int(A)$, we have $\overline A=(\overline A -Int A)\cup Int(A)=Bd(A)\cup Int (A).$

This also tells us that we don't need (1) to prove that $\overline A=Int(A)\cup Bd(A).$

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  • $\begingroup$ The general fact that an open set that is disjoint from $Y$ is also disjoint from $\overline Y$ is used all the time. $\endgroup$ – DanielWainfleet Sep 9 '18 at 17:17
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  1. Use theorem x in cl K iff
    for all open U nhood x, U $\cap$ K is not empty.

  2. Because bd A = cl A - int A.
    where cl A = $\overline A.$

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  • $\begingroup$ For the proof $bound(A)=\overline{ A}\setminus int(A)$ $\endgroup$ – Math geek Feb 24 at 11:36
  • $\begingroup$ Let $x\in bound(A)=\overline{A}\cap\overline{X\setminus A}$, So $x\in \overline{A}$ and $x\in \overline{X\setminus A}\implies $every neighebrhood of $x$ intersects $X\setminus A$. So, there is no neighbourhood of $x$ lies entirely in $A$. So, $x$ is not a interior point. $\endgroup$ – Math geek Feb 24 at 11:40
  • $\begingroup$ for reverse inclusion $x\in \overline A \setminus inerior(A)$. Every neighbourhood of $x$ inersects $X\setminus A$. Hence $x\in boundary (A)$ $\endgroup$ – Math geek Feb 24 at 11:42
  • $\begingroup$ Am I correct @WilliamElliot $\endgroup$ – Math geek Feb 24 at 11:43
  • $\begingroup$ @Mathgeek. Exercise. Prove cl X\A = X - int A. $\endgroup$ – William Elliot Feb 24 at 23:13

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