# Proving that $A\subset B \implies \hat A \subset \hat B$

Proving that $A\subset B \implies \hat A \subset \hat B$ where, $\hat X$ implies closure of $X$. I want to prove this strictly using contradiction. So,I started out assuming $\exists x \in \hat A \text{ and } x \notin \hat B$

Since, $A \subset B, x\in (\hat A-A)$ (else it would be in B automatically).

Now, I need to somehow prove that any point in $\hat A - A$ either is a part of B or a part of $\hat B - B$.

Any leads? I don't want solution. This is homework.

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What definition of closure are you using? –  Trevor Wilson Sep 10 '12 at 23:03
@TrevorWilson, The closure of A would be the smallest closed set such that A would be a subset of closure(A). –  Inquest Sep 10 '12 at 23:06
You're thinking too hard. –  robjohn Sep 10 '12 at 23:20

If you must use contradiction, consider that if $a\in\hat{A}$ but $a\not\in\hat{B}$, then there is an open neighborhood $U$ containing $a$ so that $U\cap\hat{B}=\varnothing\Rightarrow U\cap A=\varnothing$.

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This is what I used: If $x \in (\hat A - A)$, x is a limit point of A. Thus, every punctured neighborhood of x has a non-empty intersection with A. Since $A \subset B$, every neighborhood of x has a non-empty intersection with B. Thus, x is also a limit point of B. Thus, $x \in \hat B$ which is a contradiction. Is this correct? –  Inquest Sep 11 '12 at 14:20
@Inquest: The proof is fine. However, this is a direct proof, thinly veiled as a proof by contradiction. What you have done is say "assume not Y", then you prove Y without using that assumption, then you say "contradiction". To shorten the proof, leave out the "assume not Y" and the "contradiction" and just leave the direct proof. –  robjohn Sep 11 '12 at 16:40
I kinda noticed that but I am on a holy war to prove everything with contradiction. :P My notebook is filled with Let us assume that <> is not true. –  Inquest Sep 11 '12 at 18:58
@Inquest: Following what I had started leads to a less artificial seeming contradiction. Finish off by saying that since $a\in\hat{A}$, for each open $U$ containing $a$, we have $U\cap A\not=\varnothing$; contradiction. –  robjohn Sep 11 '12 at 19:46

You have already given your definition of closure. So you know that $A \subseteq B \subseteq \hat{B}$ yes? This is by definition of $\hat{B}$ containing $B$ (if you drop the word "closed" in the definition and just look at set containment). Then $\hat{B}$ is a closed set containing $A$ yes? And then $\hat{A}$ by definition is.....

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$A\subset B\subset X$. $\bar{B}$ is a closed set containing $A$, hence...

$\bar{A}\subset\bar{B}$.

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The OP has explicitly stated not to give a full solution. –  user38268 Sep 10 '12 at 23:11
@BenjaLim: I have been trying to find a hint that is not a full answer. It is not easy. –  robjohn Sep 10 '12 at 23:18