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Prove closure of intersections is a subset of intersection of closures:

Proof:

Given a topological space $X$ and a family of open sets $A_i (i∈I)$, where $I$ is an index set, then from the definition of closure follows: $$A_i⊆\overline{A_i}\tag{∀i∈I}$$

so we have: $$\bigcap_{i∈I}A_i⊆\bigcap_{i∈I}\overline A_i$$

since for every $i∈I$ ,$ A_i$ is closed so their intersection is also a closed set containing $\bigcap_{i∈I}A_i$, on the other hand $\overline{\bigcap_{i∈I}A_i}$ is a closed set which is in the intersection of all closed sets containing $\bigcap_{i∈I}A_i$, so $$\color{blue}{\overline{\bigcap_{i∈I}A_i}⊆\bigcap_{i∈I}\overline A_i}$$

The problem I have is that in my opinion the index set $I$ must be a finite set because a topological space is not closed over taking an intersection of an infinite number of sets , and hence taking the intersection of $\overline{A_i}$'s considering the index set $I$ to be an infinite set is not guaranteed to be in our topological space $X$.

Of course I'm talking about a topological space defined via open sets.

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  • $\begingroup$ The intersection of an infinite number of subsets of a topological space is in the space (because it’s just a normal set), the problem comes when you take an infinite intersection of open sets, then, the resulting set is not neceserally open. $\endgroup$ Commented Jan 26, 2020 at 15:44
  • $\begingroup$ @AmadeusMaldonado, well your definition is totally true, but it's true if we are working with a topological space defined via closed sets, because then taking the intersection of an infinite numbers of sets is again in that topological space, but in a topological space defined via open sets it's different. $\endgroup$
    – user715522
    Commented Jan 26, 2020 at 15:51
  • $\begingroup$ Do you mean ‘definition via open sets’ the same way defined here en.m.wikipedia.org/wiki/Topological_space or is it something else? $\endgroup$ Commented Jan 26, 2020 at 15:58
  • $\begingroup$ @AmadeusMaldonado, yes exactly $\endgroup$
    – user715522
    Commented Jan 26, 2020 at 16:01
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    $\begingroup$ intersections and unions are just set theory not topology and are always defined. Unions of open are open and intersections of closed sets are closed (that is topology). $\endgroup$ Commented Jan 26, 2020 at 16:21

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Of course, any infinite intersection of sets within $X$ exists, the index set does not matter for the definition, as long as $I \neq \emptyset$:

$$x \in \bigcap_{i \in I} C_i \iff \forall i \in I: x \in C_i$$

The set in itself is well-defined, just like the infinite union is, and any intersection of closed sets is closed (this follows from the fact that any union of open sets open and complementation reverses intersection and union (de Morgan's laws). So the argument is completely valid as given.

You can also use the definition via adherence (or limit) points: let $x \in \overline{\bigcap_{i \in I} C_i}$, we want to show $x \in \bigcap_{i \in I} \overline{C_i}$ and to this end take any arbitrary $i \in I$, and let $O$ be any open set containing $x$. Then $O \cap (\bigcap_{i \in I} C_i) \neq \emptyset$ so pick $z \in \bigcap_{i \in I} C_i$ which also lies in $O$. In particular, $z \in O_i \cap O$ for our arbitrary $i$, and as $O$ was arbitrary, $x \in \overline{C_i}$. So we are done, and $x \in \bigcap_{i \in I} \overline{C_i}$ and the inclusion has been shown.

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  • $\begingroup$ thank you for your consideration but please look at this link , proofwiki.org/wiki/… in the link since the author defines a topological space via closed sets hence it has been mentioned that the index set $I$ is a finite set, so duo to my definition of a topological space vie open sets I use a finite index set for intersecting the family of sets $A_i$, (indeed the only difference between mine and the proof in the link is that we both use different definitions but based on our definition we observe the closeness of our topological space) $\endgroup$
    – user715522
    Commented Jan 26, 2020 at 16:30
  • $\begingroup$ @user715522 that is about finite unions, which corresponds to finite intersections of open sets again. Totally different situation. $\endgroup$ Commented Jan 26, 2020 at 16:34
  • $\begingroup$ so if I use a finite index set to intersect a family of open sets is it ok? $\endgroup$
    – user715522
    Commented Jan 26, 2020 at 16:54
  • $\begingroup$ @user715522 any intersection of open sets is well-defined, only finitely many ones is garantueed to be open again. $\endgroup$ Commented Jan 26, 2020 at 17:02

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