# Proving that if $A \subseteq B \subseteq \mathbb{R}^n$ then $\overline{A} \subseteq \overline{B}$

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?

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 $$\overline{A}$$.
• why do we have that $x \in \overline{A}$? – Joey Oct 17 at 22:33
• @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. – Cornman Oct 17 at 22:34
• 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 $\overline{A}$ – Joey Oct 17 at 22:42
• 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 $\overline{A}\subseteq\overline{B}$. – Cornman Oct 17 at 23:00