Show that

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

$\overline A=A\cup A'$ where $A'$ are the limit points

My attempt:

Since $\overline A$ is closed and $\overline B$, the union of two closed sets is closed.

Hence $\overline A\cup \overline B$ can be rewritten as $\overline{A\cup B}$

My Second attempt:

$\overline{A\cup B}=(A\cup B)\cup (A\cup B)'=(A\cup B)\cup (A'\cup B')=A\cup A' \cup B \cup B' = \overline{A}\cup \overline{B}$

  • $\begingroup$ Why does the last line follow from the previous one? $\endgroup$ May 28, 2014 at 7:18
  • $\begingroup$ Actually, I do not understand the previous to last line. You write "Since $\alpha$, $\beta$". The usual interpretation of this sentence is that $\beta$ is a consequence of $\alpha$. But $\beta$ is "the union of two closed sets is closed". How is this a consequence of $\bar A$ and $\bar B$ being closed? $\endgroup$ May 28, 2014 at 7:28
  • $\begingroup$ Please look at my second attempt. $\endgroup$
    – iordanis
    May 28, 2014 at 7:41
  • 3
    $\begingroup$ How do you justify $(A \cup B)' = (A' \cup B')$? $\endgroup$
    – IAmNoOne
    May 28, 2014 at 7:44

5 Answers 5


Your second attempt only needs one more fact: $(A \cup B)' = A' \cup B'$.

To see this: pick $x \in A'$. Then every neighbourhood $U$ of $x$ intersects $A$ in at least one point different from $x$. This point is also in $A \cup B$, so then every neighbourhood of $x$ intersects a point of $A \cup B$ not equal to $x$, so $x \in (A\cup B)'$. Similarly, if $x \in B'$, $x \in (A \cup B)'$, so we have $A' \cup B' \subset (A \cup B)'$.

On the other hand, if $x \in (A \cup B)'$, we need to show $x$ is in $A' \cup B'$. So assume not, $x \notin A'$, which means there is a neighbourhood $U$ of $x$ that intersects $A$ in an empty set or just $\{x\}$. Also $x \notin B'$, so there is a neighbourhood $V$ of $x$ such that $V$ intersects $B$ either in the empty set or $\{x\}$. Now, $U \cap V$ is a neighbourhood of $x$, and $(U \cap V) \cap (A \cup B) = (U \cap V \cap A) \cup (U \cap B \cap V) \subset (U \cap A) \cup (V \cap B)$, which is also either the empty set or $\{x\}$, from the properties of $U$ and $V$. But this contradicts that $x \in (A \cup B)'$. So our assumption was incorrect: $x$ must be in $A' \cup B'$, as required.

  • $\begingroup$ Sir, please elaborate how $(U \cap V \cap A) \cup (U \cap B \cap V) \subset (U \cap A) \cup (V \cap B)$ $\endgroup$ Jul 30, 2019 at 14:05
  • 1
    $\begingroup$ @AkashPatalwanshi the left hand set of the union is a subset of the lhs on the right and ditto for the rhs. $\endgroup$ Jul 30, 2019 at 14:06

You have only proved that $\overline A\cup\overline B$ is closed. You must prove two things more:

  1. $A\cup B\subset\overline A\cup\overline B$. This is very easy.
  2. Every closed set that contains $A\cup B$ also contains $\overline A\cup\overline B$. This is not so easy.

Some hints for the second part: Consider a closed set $F$ that contains $A\cup B$. Suppose now that there is some $x\in\overline A\cup\overline B$ such that $x\notin F$. Then $x$ can be in $\overline A$ or in $\overline B$.
If it is in $\overline A$, for example, consider an open neighbourhood $U$ of $x$. Now, since $x\notin F$, consider an open neighbourhood $V$ of $x$ such that $F\cap V=\emptyset$. What is $U\cap V$? Is empty $U\cap V\cap A$? Can you arrive at a contradiction?
The other possibility is that $x\in\overline B$. Can you make the same reasoning?


Note that closures preserve set inclusion, i.e. if $A \subset B$ then $\overline A \subset \overline B$.

We want to show $\overline{A\cup B} = \overline{A} \cup \overline{B}$. $A \subset A\cup B$ and $B \subset A\cup B$ so $\overline{A} \subset \overline{A\cup B}$ and $\overline{B} \subset \overline{A\cup B}$. So, $\overline{A} \cup \overline{B} \subset \overline{A\cup B}$. On the other hand, $A \subset \overline{A}, B\subset \overline{B}$, so $A\cup B \subset \overline{A} \cup \overline{B}$. The latter is closed, so $\overline{A\cup B} \subset \overline{A} \cup \overline{B}$.

We conclude $\overline{A\cup B} = \overline{A} \cup \overline{B}$. $\ \ \blacksquare$


You wrote that the union of two sets is closed, which proves that $\overline A\cup\overline B$ is closed. It does not prove that it equals $\overline{A\cup B}$.

The problem with your second attempt is that you wrote that $$(A\cup B)'=A'\cup B'.$$ Do you have proof that this is so?

  • $\begingroup$ look at my second attempt. $\endgroup$
    – iordanis
    May 28, 2014 at 7:42
  • $\begingroup$ @μακακας I did and I edited my answer. $\endgroup$
    – 5xum
    May 28, 2014 at 7:45
  • $\begingroup$ Here is a related proof. math.stackexchange.com/questions/550920/… $\endgroup$
    – iordanis
    May 28, 2014 at 7:47
  • $\begingroup$ I mean it is true if you google it. But yet it makes my proof incomplete. How can I better reason? $\endgroup$
    – iordanis
    May 28, 2014 at 7:48
  • 2
    $\begingroup$ "It's true if you google it" is not an answer your professor would want to hear. In homeworks, you are usually only allowed to use things that you can prove. After all, $\overline{A\cup B} = \overline A\cup \overline B$ is also true if you google it. $\endgroup$
    – 5xum
    May 28, 2014 at 7:50

A standard way of showing two sets $S,T$ are equal to each other is by considering an element $s$ of $S$ and showing that $s$ belongs to $T$, and, conversely, by showing that every element $t$ in $T$ is contained in $S$.

You can use the fact here that the closure $\overline {A \cup B}$ equals $A\cup B $ together with the union of the respective limit points. Now, consider an element $a$ in $\overline {A \cup B}$ and show it must belong to $\overline A \cup \overline B$.

Let me get you started: let $x$ be in $\overline {A \cup B}$ , then $x$ is either in $A$, in $B$ , or in the closure of the union (notice that $A\subseteq \overline A$). If x is in $A$, then $x$ is in $A\cup B \subseteq \overline {A\cup B}$; similar if $x$ is in $B$. Now consider $y$ in $\overline {A \cup B}-A\cup B$...


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