# Show that any set $A,B\subseteq\mathbb{R}^n,\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$

Show that any set $$A,B\subseteq\mathbb{R}^n,\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$$

Def. (open ball)

$$B(a;r)=\{x∈R^n:|x−a|.

Def. (closure)

$$\overline{S}=\{{x∈R^n:∀ε>0,B(x;ε)∩S≠\varnothing}\}$$.

Proof.

Let $$A,B\subseteq\mathbb{R}^n$$

We start from

$$\overline{A\cap B}=\{x∈R^n:∀ε>0,B(x;ε)∩A\cap B≠\varnothing\}$$

$$=\{x∈R^n:∀ε>0,\{b∈R^n:|b−x|<\varepsilon\}∩A\cap B≠\varnothing\}$$

Since $$\exists \phi_1,\phi_2, s.t. A=\{x:\phi_1(x)\},B=\{x:\phi_2(x)\}$$

Then we have

$$=\{x∈R^n:∀ε>0,\exists b∈R^n,s.t.|b−x|<\varepsilon\wedge \phi_1(x)\wedge \phi_2(x)\}$$

Since $$\exists x,\phi_1(x)\wedge\phi_2(x)\Rightarrow \exists x,y,\phi_1(x)\wedge\phi_2(y)$$, but not converse, so we can only use $$\subseteq$$ here:

(I'm not sure about this, why I can't use $$=$$ here)

$$\subseteq\{x∈R^n:∀ε>0,\exists b∈R^n,s.t.|b−x|<\varepsilon\wedge \phi_1(x)\}$$

$$\cap\{y∈R^n:∀ε>0,\exists b∈R^n,s.t.|b−y|<\varepsilon\wedge \phi_2(y)\}$$

$$=\{x∈R^n:∀ε>0,B(x;\varepsilon)\cap A\neq\varnothing\}$$

$$\cap\{y∈R^n:∀ε>0,B(y;\varepsilon)\cap B\neq\varnothing\}$$

$$=\overline{A}\cap\overline{B}$$

Therefore

$$\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}\tag*{\square}$$

$$\dots$$ Is my proof correct ? Any suggestions would be appreciated.

Also please tell me if there is a better method to prove it.

Thanks for your help.

• I do not follow "$\exists \phi_1 \mathrm{s.t.} A = \{x:\phi_1(x)\}$" - on what space is the $\phi_1$ defined? Is the $\phi_1$ supposed to be continuous? I don't see why we can assume this of $A$ (or the same about $\varphi_2$ and $B$). – Math1000 Oct 15 '19 at 2:51

## 2 Answers

A quick proof: Note that the closure of a set $$E$$ is the smallest closed set containing $$E$$.

Since $$A\cap B\subset \overline{A}\cap\overline B$$ and $$\overline{A}\cap\overline B$$ is closed, so $$\overline{A\cap B}\subset \overline{A}\cap\overline B.$$

First show that for any two sets $$A$$ and $$B$$ in $$\Bbb R^n$$, if $$A\subseteq B$$ then $$\overline{A}\subseteq\overline{B}$$.

From this, it follows that $$\overline{A\cap B}\subseteq\overline{A}$$ (since $$A\cap B\subseteq A$$) and that $$\overline{A\cap B}\subseteq\overline{B}$$. So, $$\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$$

To prove the first, assume $$A\subseteq B$$ and let $$\varepsilon>0$$. Set $$x\in\overline{A}$$. Then we have that $$B_\varepsilon(x)\cap A\neq \varnothing$$, let's say there is some $$y\in B_\varepsilon(x)\cap A$$.

In particular $$y\in A$$, and since $$A\subseteq B$$, then $$y\in B$$. This shows that $$y\in B_\varepsilon(x)\cap B$$, that is, $$B_\varepsilon(x)\cap B \neq \varnothing$$ showing that $$x\in\overline{B}$$ as we want to prove.