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I have the feeling I'm missing something obvious, but here it goes...

I'm trying to prove that for a subset $A$ of a topological space $X$, $\overline{\overline{A}}=\overline{A}$. The inclusion $\overline{\overline{A}} \subseteq \overline{A}$ I can do, but I'm not seeing the other direction.

Say we let $x \in \overline{A}$. Then every open set $O$ containing $x$ contains a point of $A$. Now if $x \in \overline{\overline{A}}$, then every open set containing $x$ contains a point of $\overline{A}$ distinct from $x$. My problem is: couldn't $\{x,a\}$ potentially be an open set containing $x$ and containing a point of $A$, but containing no other point in $\overline{A}$?

(Also, does anyone know a trick to make \bar{\bar{A}} look right? The second bar is skewed to the left and doesn't look very good.)

Alex Petzke
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    You can try and prove two separate things that will let you conclude that: $(1)$ $\operatorname{cl} A$ is closed. $(2)$ If $A$ is closed, then $\operatorname{cl} A = A$. The fact that ${\operatorname{cl} ({\operatorname{cl} A})}=\operatorname{cl} A$ then follows at once. – Pedro Jul 20 '12 at 02:10
  • Isn't $\bar{A} \subseteq \bar{\bar{A}}$ true just because every set is a subset of its closure? – Promethèus Aug 08 '22 at 15:17

5 Answers5

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How do you define the closure of a set $A$? If it's not already your definition, it might be a useful thing to prove that the closure of a set is precisely the intersection of all closed sets containing it.

Kris
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The condition you want to check is \[ x \in \bar A \quad \Leftrightarrow \quad \text{for each open set $U$ containing $x$, $U \cap A \neq \emptyset$} \] This definition implies, among other things, that $A \subset \bar A$. Indeed, with the notation above we always have $x \in U \cap A$. Is it clear why this implies the remaining inclusion in your problem?

If you instead require that $U \cap (A - \{x\}) \neq \emptyset$ then you have defined the set $A'$ of limit points of $A$. We have $\bar A = A \cup A'$. Simple examples such as $A = \{0\}$ inside of $\mathbb R$, for which $\bar A = A$ but $A' = \emptyset$, can be helpful in keeping this straight.

Dylan Moreland
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  • Most topological properties, by the way, have a lot of equivalent forms. Learning the different definitions of the closure listen on Wikipedia can really save you work — Kris' hint, for example, would give maybe the quickest proof possible of the inclusion that you already did. – Dylan Moreland Jul 20 '12 at 02:13
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Suppose $x\in\overline{\overline{A}}$. Let $U$ be an open set containing $x$; we want to show that $U\cap A\neq\varnothing$. We know that $U\cap\overline{A}\neq\varnothing$, so there exists $y\in \overline{A}$ such that $y\in U$. But since $U$ is an open set that contains $y$ and $y\in\overline{A}$, then...

Arturo Magidin
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  • then $\overline{A} = A$? – Guerlando OCs May 30 '16 at 00:02
  • Is $x\in\bar{\bar{A}}?$ – Verónica Rmz. Apr 05 '22 at 03:11
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    @V.R.M.: Isn't that the first sentence of the post? – Arturo Magidin Apr 05 '22 at 03:13
  • yes it is; ok sorry for asking about your proof. – Verónica Rmz. Apr 05 '22 at 03:17
  • I think 2 or 1 more bars are needed on the rest of the A's. – Verónica Rmz. Apr 05 '22 at 03:37
  • @V.R.M. Although it's been almost 10 years since I wrote the above, I believe that you are incorrect. We are trying to show that the closure of the closure is the closure; the only nontrivial inclusion is to show that any $x$ in $\overline{\overline{A}}$ is in $\overline{A}$; that is we want to show that if $U$ is an open set containing $x$, then $x\cap A$ is nonempty, exactly as I wrote. Because $x$ is in the closure of $\overline{A}$, we know that any open set containing $x$ will intersect $\overline{A}$, exactly as I wrote. So, no. There are no bars missing. – Arturo Magidin Apr 05 '22 at 03:40
  • @ArturoMagidin sorry, again :) (don't be upset :) my mistake.. I don't know what I was thinking – Verónica Rmz. Apr 05 '22 at 05:43
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This depends on what is admissible. The closure of a closed set is itself. So just show the closure is closed.

Btw, remark on Kris's answer: The fact that closure of closure is closure is the same thing as saying either of these

  1. the intersection of all closed sets containing $A$ is equal to the intersection of all closed sets containing $\overline{A}$

  2. the intersection of all closed sets containing $A$ is equal to the intersection of all closed sets containing (the intersection of all closed sets containing $A$)

BCLC
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A set is closed iff every convergent sequence in the set has its limit in that set. The closure operation adds a set's limit points to the set. A closed set already contains its limit points, so repeating the operation doesn't change the answer (idempotent).

For Example:

The sequence 1/n; 0 < n < inf goes like this..

1, 1/2, 1/3, 1/4, ... , 1/465674564, ...

It's limit is 0, which is not part of the sequence, so closure would add the 0.

NOTE - The above example is just meant to illustrate how closure works. As far as I know, this sequence is not a subset of any metric space.

Tetrafold
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