Cfrgtkky

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

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, $U' cap bigcaplimits_{alpha in I} C_alpha
  neq varnothing$, where $bigcaplimits_{alpha in I} C_alpha$ is
  the intersection of all closed sets containing $A$




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

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

• 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 $xinoperatorname{cl}A$. Let $C$ be any closed set such that $Asubseteq C$. Suppose that $xnotin C$: then $xin Xsetminus C$, and $Xsetminus C$ is open, so $(Xsetminus C)cap Anevarnothing$. But on the other hand we know that $Asubseteq C$, so $Acap(Xsetminus C)=varnothing$. This contradiction shows that $xin C$, and since $C$ was an arbitrary closed set containing $A$, we conclude that

$$operatorname{cl}Asubseteqbigcap{Csubseteq X:Asubseteq Ctext{ and }Ctext{ is closed}};.$$

For the opposite inclusion just observe that $operatorname{cl}A$ is one of the closed sets containing $A$, so if $xinbigcap{Csubseteq X:Asubseteq Ctext{ and }Ctext{ is closed}}$, then automatically $xinoperatorname{cl}A$. It follows that

$$bigcap{Csubseteq X:Asubseteq Ctext{ and }Ctext{ is closed}}subseteqoperatorname{cl}A$$

and hence that

$$bigcap{Csubseteq X:Asubseteq Ctext{ and }Ctext{ is closed}}=operatorname{cl}A;.$$

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.

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.

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.