I want to show that given $(X, \mathcal{T})$, we define $\overline A = \{x \in X| \forall U \in \mathcal{T}, x \in U \implies U \cap A \neq \varnothing\}$ (definition of closure from Munkres), then

Show that $\overline A = \bigcap\{C \subseteq X | C \text{ is closed }, A \subseteq C\}$

I find this really hard to tackle because some unnaturalness in that $\overline A$ is specified with respect to open sets, but then it is alternatively defined as intersection of closed sets..how to juggle between open and closed?

Several other posts also doesn't help...

  1. The closure of A is the smalled closed set containing A is proved in terms of accumulation points and limit points which I do not define

  2. Proving that the closure of a subset is the intersection of the closed subsets containing it is defined wrt of metric spaces

I am stuck on both inclusions and need some help


$(\overline A \subseteq \bigcap\{C \subseteq X | C \text{ is closed }, A \subseteq C\})$

  • Let $x \in \overline A$, then $\forall U \in \tau, x \in U \implies A \cap U \neq \varnothing$. We want to show that $x \in \bigcap\{C \subseteq X | C \text{ is closed }, A \subseteq C\}$

  • So we know that $x$ is contained in some $U' \in \mathcal{T}$ that has non-empty intersection with $A$, $x$ not necessarily in $A$.

  • Let $C_1$ be a closed set containing $A$, then $U' \cap C_1 \neq \varnothing$. Let $C_2$ be a closed set containing $A$, then $U' \cap C_2 \neq \varnothing$. Assuming $C_1 \subseteq C_2$, then $U' \cap C_1 \cap C_2 \neq \varnothing$.

  • Continue this way, $U' \cap \bigcap\limits_{\alpha \in I} C_\alpha \neq \varnothing$, where $\bigcap\limits_{\alpha \in I} C_\alpha$ is the intersection of all closed sets containing $A$

  • We know already that $x \in U'$, but from above how can we see that $x \in \bigcap\limits_{\alpha \in I} C_\alpha$? From figure below, it seems that $x$ will not be in $\bigcap\limits_{\alpha \in I} C_\alpha$

enter image description here

$( \bigcap\{C \subseteq X | C \text{ is closed }, A \subseteq C\} \subseteq \overline A)$

  • Let $x \in \bigcap\{C \subseteq X | C \text{ is closed }, A \subseteq C\}$, we want to show that $x \in \overline A$. It suffices to show that $\forall U \in \mathcal{T}, x \in U \implies U \in A \neq \varnothing$.

  • Since $x \in \bigcap \{C\}$, then there exists some closed set $C' \subseteq X$ such that $x \in C'$. Let $U \in \mathcal{T}$ containing $x$, then we will show that $U \cap A \neq \varnothing$

  • We know that $x \in C' \cap U$, then $x \in \bigcap{C} \cap U$. At this point however I still don't know whether $U \cap A \neq \varnothing$. Couldn't we have a case in figure below where $x \in \bigcap \{C\}$ and $x \in U$, but $U \cap A = \varnothing$?

enter image description here

  • $\begingroup$ Hint: For the second inclusion you only need to know that $\overline{A}$ is closed. $\endgroup$ – user60589 Jun 9 '16 at 16:53
  • $\begingroup$ For the other inclusion assume $x \notin \overline{A}$. I.e. you have an open $U$ containing $x$ which has trivial intersection with $A$. Keep in mind that the complement of an open set is closed. $\endgroup$ – user60589 Jun 9 '16 at 16:59
  • $\begingroup$ A limit point is just a member of the closure that is not in the interior. $\endgroup$ – Jacob Wakem Jul 12 '16 at 2:50

You’re getting bogged down in the details of the definitions and thereby making it much harder than it really is.

For the first inclusion, start, as you did, with an arbitrary $x\in\operatorname{cl}A$. Let $C$ be any closed set such that $A\subseteq C$. Suppose that $x\notin C$: then $x\in X\setminus C$, and $X\setminus C$ is open, so $(X\setminus C)\cap A\ne\varnothing$. But on the other hand we know that $A\subseteq C$, so $A\cap(X\setminus C)=\varnothing$. This contradiction shows that $x\in C$, and since $C$ was an arbitrary closed set containing $A$, we conclude that

$$\operatorname{cl}A\subseteq\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}\;.$$

For the opposite inclusion just observe that $\operatorname{cl}A$ is one of the closed sets containing $A$, so if $x\in\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}$, then automatically $x\in\operatorname{cl}A$. It follows that

$$\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}\subseteq\operatorname{cl}A$$

and hence that

$$\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}=\operatorname{cl}A\;.$$

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  • $\begingroup$ wow perfect solution $\endgroup$ – Carlos - the Mongoose - Danger Jun 9 '16 at 18:14
  • $\begingroup$ @Brian M. Scott, can you please explain the part "$(X\setminus C)\cap A\ne\varnothing$"? you claim it's not $\varnothing$ because based on the definition of closure of $A$ , $x \in A$? $\endgroup$ – user715522 Jan 25 at 8:18

Not sure if this will help:


$A'$ is the set of all accumulation or limit points.

$\overline{A} = A \cup A'$ - this is known as the closure of $A$.

$\bar{A}$ is closed.

Proof - Suppose $p$ is not in $\bar{A}$. Since $p$ is not in $\overline{A}$, it is not in $A$ nor is it a limit point of $A$. Therefore there must be some neighborhood $N$ of $p$ that does not intersect $A$ at all.

Can $N$ contain any limit points of $A$? No. If it contained one, $a$. Then by definition of limit point $N$ must contain another point of $A$. But $N$ contains no points of $A$, so this is ridiculous. Thus $N$ must be disjoint from both $A$ and its set of limit points, so $N \subseteq \overline{A}^c$, as desired. Thus the compliment of the closure is open, so the closure is closed.

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  • $\begingroup$ thanks but I am just honestly not taught what a limit point is $\endgroup$ – Carlos - the Mongoose - Danger Jun 9 '16 at 18:14
  • $\begingroup$ would you like me to expand upon my answer to include information on what a limit point is? $\endgroup$ – Wolfy Jun 9 '16 at 18:15
  • $\begingroup$ Oh no that's okay, I have Munkres as reference and I think I am the only person who studies topology without knowing what limit point is. I think other people can easily understand your proof. My prof do not wish for us to know what limit point is he thinks it is detriment to learning topology... $\endgroup$ – Carlos - the Mongoose - Danger Jun 9 '16 at 18:17
  • $\begingroup$ I see okay well best of luck $\endgroup$ – Wolfy Jun 9 '16 at 18:17
  • $\begingroup$ A limit point is just a member of the closure that is not in the interior. $\endgroup$ – Jacob Wakem Jul 12 '16 at 3:11

All the points far from A form an open set (because it is a union of some open set for each of its points, definitionally). Moreover it is the largest open set without elements in A (for consider a neighborhood of a close point...) . Thus its complement, the closure, is the smallest closed set containing A.

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Let $x$ in $\bar{A}$. Note that $\bar{A}$ is closed, so $x$ is in a closed set. Suppose for some closed $C$ containing $A$ that $x$ is not in $C$. That means $x$ is in $C$'s complement, which is open. But that contradicts the fact that $x$ was in a closed set, so $x \in C$. Since $x$ was arbitrary in $\bar{A}$, and $C$ was an arbitrary member of all $C$ containing $A$, we have proved $\bar{A} \subset \mathscr{C}$, where the fancy "C" is the set of all closed sets containing $A$. Keep in mind that if a set exists in every one of a collection of sets, then it must also be in their intersection.

Now, for equality we need to show $\bar{A} \supset \bigcap{C}$. Let $x$ in $\bigcap C$. Suppose $x$ not in $\bar{A}$, then $x$ is in $\bar{A}^c$ which is open. But we just said $x$ is in an intersection of closed sets, so it must be in a closed set. Contradiction again.

The one gap in this proof is the fact that these sets could have been $\mathbb{R}^d$. But in that case equality is automatic.

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