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So we are given Proving that if $A \subseteq B \subseteq \mathbb{R}^n$ then $\overline{A} \subseteq \overline{B}$. Here is my attempt at the problem:

Since $B \subseteq \mathbb{R}^n$, by a theorem in the book I'm using we have that $B \subseteq \overline{B}$. To show that $\overline{A} \subseteq \overline{B}$ we can take an element of $\overline{A}$ and find it in $\overline{B}$.By definition $\overline{A}$ $=$ $\cap$ {T:T $\supseteq$ A, T is closed}. So take some element $x \in \overline{A}$. Thus we have that $x \in T\supseteq A$. But isn't this the same as $A \subseteq T$?

If not would it just follow that $x \in T\supseteq A \subseteq B = T\supseteq B$. Thus by definition we would have that $x \in \overline{B}$. But this doesn't seem right, is there a different way to go about it, or is this in general wrong?

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To show $\overline{A}\subseteq\overline{B}$, we take $$x\in\overline{A}=\bigcap_{T\supseteq A\\ T\,\text{closed}} T$$

Since $A\subseteq B\subseteq\overline{B}$ closed, we have that $x\in \overline{A}$ then $x\in\overline{B}$. As $\overline{B}$ is one of the sets where we take the intersection to get $\overline{A}$.

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  • $\begingroup$ why do we have that $x \in \overline{A}$? $\endgroup$ – Joey Oct 17 at 22:33
  • $\begingroup$ @Joey That is the assumption. To show that $X\subseteq Y$ we have to show that for every $x\in X$ we have $x\in Y$. So this is what we get from the definition of the subset-inclusion. It is the starting point of the proof. $\endgroup$ – Cornman Oct 17 at 22:34
  • $\begingroup$ right right I see that. But how do we get that $\overline{B}$ is one of the sets where we take the intersection to get $\overline{A}$ $\endgroup$ – Joey Oct 17 at 22:42
  • $\begingroup$ Because, we know that $A\subseteq B$. Since $B\subseteq \overline{B}$, we have that also $A\subseteq\overline{B}$. So $\overline{B}$ (which is a closed set) contains $A$ and therefor is one of the sets which we intersect to get $\overline{A}$. Or phrased differently. $\overline{B}$ is a closed set that contains $A$. But the closure of $A$ (which is $\overline{A}$) is the smallest(!) closed set that contains $A$ (by definition), so it must be $\overline{A}\subseteq\overline{B}$. $\endgroup$ – Cornman Oct 17 at 23:00

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