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I am trying to proof $\overline{A \cup B}= \overline{A} \cup \overline{B} $, where $\overline{A}$ is the topological closure of A. I did some thinking and I came up with this. My Question is: Is my proof correct?

First I am trying to proof $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$

Let $x \in \overline{A \cup B}=(\cap_{A \cup B \subseteq P, P closed}P)$ For every P that statisfies the condition below the cap $x$ is in P. Furthermore $A \subseteq \overline{A}$ and $B \subseteq \overline{B}$, as result we get $A \cup B \subseteq \overline{A} \cup \overline{B}$. As union of two closed sets $\overline{A} \cup \overline{B}$ is obviously closed. So the condition for P is statisfied and as consequence $x \in \overline{A} \cup \overline{B}$

The Second direction is: $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$

Let $x \in \overline{A} \cup \overline{B}$, without loss of generality let $x \in \overline{A}=(\cap_{A \subseteq P, P closed}P)$

This means x is in every P that statisfies A $\subseteq$ P and P is closed.

$\overline{A \cup B}:=(\cap_{A \cup B \subseteq H, H closed}H)$

trivially $A \subseteq A \cup B$, so if $A \cup B$ is a subset of a set $H$ that is closed, so is $A$, which means that all $H$ satisfy the conditions for $P$. This means $x \in H$ for all $H$ that satisfies $A \cup B \subseteq H$, $H$ closed. So as a result $x \in \overline{A \cup B}$

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John.W
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2 Answers2

2

What you did looks correct.

However, it can be greatly simplified if you know (or prove) that if a subset $A$ is included in a closed set $C$, then so is $\overline{A}$.

As an example, to prove $$\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$$ you could just say:

As $A,B \subseteq \overline{A} \cup \overline{B}$, we have $A \cup B \subseteq \overline{A} \cup \overline{B}$. And as $\overline{A} \cup \overline{B}$ is closed as a (finite) union of closed subsets, we get $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$.

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1

I am not sure I fully understand the proof for the direction $\overline{A\cup B}\subset \overline{A}\cup \overline{B}$.

I suggest the following more streamlined method to prove that direction.

  1. Show that taking closures preserves closed sets: $\overline{X}=X$ whenever $X$ is closed.
  2. Show that taking closures preserves inclusions: $X\subset Y \leadsto \overline{X} \subset \overline{Y}$. Now apply to $A\cup B \subset \overline{A}\cup \overline{B}$.

The other direction looks fine, but now easily follows from 2) applied to $A\subset A\cup B$ and $B\subset A\cup B$.

  • I would then get, $\overline{A \cup B} \subseteq \overline{\overline{A} \cup \overline{B} }=\overline{A} \cup \overline{B}$. Nice one, thanks! – John.W Sep 01 '21 at 16:23