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$\overline{A}$ denotes the closure of the set $A$.

I can show that $\overline{A}\cup\overline{B}\subset \overline{A\cup B}$, but I don't know how to go about the reverse direction?

Andrés E. Caicedo
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Sid Caroline
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    Try using the following property of closed set: the closure of a set $E$ is the unique closed set $\overline{E}$ containing $E$ with the property that if $F$ is any other closed set containing $E$, then $\overline{E} \subseteq F$. – D_S Aug 03 '17 at 23:55

5 Answers5

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$\overline{A} \cup \overline{B}$ is a closed set containing $A \cup B$. Hence $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$.

Trevor Gunn
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Consider a point $x$ in $\overline{A\cup B} $. Then for any nbhd $U$ of $x$, $U\cap (A\cup B)\neq\emptyset $. So for any nbhd $U$ of $x, U\cap A\neq \emptyset$ or $U\cap B\neq \emptyset$. Hence $x\in\overline A$ or $x\in\overline B$ and we are done.

Janitha357
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  • You need to explain why in this case, $\forall (U \in N_x), (U \cap A \ne \emptyset \vee U \cap B \ne \emptyset) \rightarrow (\forall U \in N_x, U \cap A \ne \emptyset) \vee (\forall U \in N_x, U \cap B \ne \emptyset)$, where $N_x$ is the collection of neighborhoods of $x$. – Daniel Schepler Aug 04 '17 at 01:04
  • $\exists U_1\in N_x:U_1\cap A=\emptyset$ and $\exists U_2\in N_x:U_2\cap B=\emptyset\implies V\cap A=V\cap B=\emptyset,$ where $V=U_1\cap U_2\in N_x$. – Janitha357 Aug 04 '17 at 07:19
  • Exactly - I'd say that needs to be incorporated into the main answer. – Daniel Schepler Aug 04 '17 at 17:12
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Here's an argument for first-countable spaces (e.g. $\mathbb{R}^n$):

It suffices to show that a limit point of $A \cup B$ is either a limit point of $A$ or a limit point of $B$. If $x$ is a limit point of $A \cup B$, then this means we can find a sequence $\{x_k\}_{k=1}^\infty$ that converges to $x$ where $x_k \in A$ or $x_k \in B$ for each $k$. This sequence admits a subsequence $\{x_{k_m}\}_{m=1}^\infty$ of points contained entirely in $A$ or entirely in $B$. Since subsequences of convergent sequences also converge to the same point, we therefore have a sequence of points in either $A$ or $B$ that converges to $x$. Hence, $x \in \overline{A}$ or $x \in \overline{B} \implies x \in \overline{A} \cup \overline{B} \implies \overline{A} \cup \overline{B} \subset \overline{A \cup B}$.

Kaj Hansen
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  • Note that this proof only works for first countable spaces. For more general spaces, you would need to pass to nets - and then there are complications depending on whether the directed set has a largest element (in which case the limit of the net is trivial) or not. – Daniel Schepler Aug 04 '17 at 00:00
  • Ah! I didn't see the "general topology" tag, and I assumed this was a real analysis question for the $\mathbb{R}^n$ setting. Will delete. – Kaj Hansen Aug 04 '17 at 00:02
  • I don't think it's necessary to delete it, the proof for first countable spaces based on convergence of sequences is still enlightening. – Daniel Schepler Aug 04 '17 at 00:04
  • Cool, I added the relevant caveat and ended up just filling out the proof since it's not the most general setting. Thanks for pointing out the error. – Kaj Hansen Aug 04 '17 at 00:10
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hint

$$A\subset A\cup B \implies $$ $$\overline{A} \subset \overline{A\cup B} $$

by the same, $$\overline {B}\subset \overline {A\cup B} $$

done.

hamam_Abdallah
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Let $p\in \overline{A\cup B}$. Then $p\in A\cup B$ or $p$ is an accumulation point of $A\cup B$. We proceed by cases.

Assume first that $p\in A\cup B$, then $p\in A$ or $p\in B$. If $p\in A$, then $p\in\overline{A}$ so $p\in\overline{A}\cup\overline{B}$. If $p\in B$, then $p\in\overline{B}$, so $p\in\overline{A}\cup\overline{B}$. Therefore, $p\in\overline{A}\cup\overline{B}$.

Now, assume $p$ is an accumulation point of $A\cup B$, but $p\not\in A\cup B$. Assume, for contradiction, that $p$ is not an accumulation point of $A$ and not an accumulation point of $B$. Then, there are open sets $S_1$ and $S_2$ containing $p$ such that $S_1\cap A=\emptyset$ and $S_2\cap B=\emptyset$ (these are empty sets because the intersection can only be $p$ and $p\not\in A\cup B$).

Let $S=S_1\cap S_2$. This is an open set containing $p$ and $S\cap A=\emptyset=S\cap B$. Therefore, $S\cap (A\cup B)=\{p\}$, so $p$ is not an accumulation point of $A\cup B$, a contradiction.

Now, we know that $(\overline{A\bigcup B})\subset \overline{A}\bigcup \overline{B}$. Let $p\in \overline{A}\bigcup \overline{B}$, then $p$ is an accumulation point of $A$ or $p$ is an accumulation point of $B$, or both. Then, $p$ is certainly in $(\overline{A\bigcup B})$ which implies $\overline{A}\bigcup \overline{B} \subset (\overline{A\bigcup B})$.

Wolfy
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