Closure A is equal to intersection of closed sets, that contain A(check the proof/idea)

$X = (X,T)$ is topological space and $A \subset X$. I want to prove, that $\bar A = \cap C_i$ where $C_i$ - are closed sets, which contains A.

$(\Rightarrow)$ We will show, that $\bar A \subseteq \cap C_i$. Let $x \in \bar A$. Let's consider two cases:

a) If $x \in \bar A$ and $C_i$ closed sets, that contains $A \Rightarrow \cap C_i$ will contain $A \Rightarrow\cap C_i$ will contain $x$ (by definition of intersection)

b)If $x$ is a limit point of $A$: By definition closed set contains all it's limit points. If $C$ contains set $A$ and $C$ contains a l.p of $C\Rightarrow$ it contains a l.p of $A$. If every set $C$ contains limit point of $A$ - then intersection contains l.p. of $A$ as well.

Hence, $\bar A \subseteq \cap C_i$

$(\Leftarrow)$ we will show, that $A \supseteq \cap C_i$

Suppose $x \in \cap C_i$:

a) Because every $C_i$ contains $A$, then $\cap C_i$ contains $A\Rightarrow x \in A$.

b)

I have problem with the reverse direction. Give me some hints please

It's of course more the idea how the proof should like. If the idea is correct, I would like to ask you how should i formalize my arguments.

If it's not correct, what shoud I correct?

• You should take into account the fact that $\bar{A}$ contains $A$, so it is a member of your family $C_i$. Also in your proof you are trying to prove that $A = \cap C_i$ instead of $\bar{A} = \cap C_i$. – Francesco Carzaniga Oct 12 '17 at 7:05
• @FrancescoCarzaniga I splited my proof on part a) and b) ,because $\bar A = A \cup A^{'}$. Why is it not correct way of proof? – Daniel Yefimov Oct 12 '17 at 7:09
• Because at the beginning you state "We will show, that $A \subseteq \cup C_i$". Then you should make it clear that you are considering $\bar{A}$ as $A \cup A'$. In general I think you could make it much simpler without considering that. – Francesco Carzaniga Oct 12 '17 at 7:14
• @FrancescoCarzaniga Ah, yes, I made a typo. But I agree, I can make it simplier. Thank you:) – Daniel Yefimov Oct 12 '17 at 7:15

and K is closed, closure A $\subseteq$ closure K = K.
K $\subseteq$ closure A. Thus closure A = K.