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 $U \cap A = \varnothing$?

 A: Not sure if this will help:
Definitions:
$A'$ 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.
A: 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
$$\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}=\operatorname{cl}A\;.$$
A: 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. 
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.
