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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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    You're thinking too hard.2012-09-10

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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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    @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.2012-09-11
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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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    @BenjaLim: I have been trying to find a hint that is not a full answer. It is not easy.2012-09-10