Show that the closure of $A$ is the intersection of all closed sets containing $A$, topology proof needed 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...


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
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

Attempt:

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




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 $\bigcap\limits_{\alpha \in I}
   C_\alpha$




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


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 $U \cap A = \varnothing$?
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

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