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

Def. (open ball)

$B(a;r)=\{x∈R^n:|x−a|<r\}$.

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.

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  • $\begingroup$ 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 $B$). $\endgroup$ – Math1000 Oct 15 at 2:51
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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 $$\overline{A\cap B}\subset \overline{A}\cap\overline B.$$

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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.

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