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I started reading Topology and Groupoids by Ronald Brown. The context for the following is the real line, neighborhoods and interiors are (at this point) defined using open balls around points on the real line using euclidean distance $|a-b|$.

The following exercise appears:

Let $(A_n)$ be sequence of subsets of $\mathbb{R}$. Does the following hold?

$Int(\underset{n}{\bigcap} A_n) = \underset{n}{\bigcap} IntA_n$

I can satisfy myself using the sequence $A_n = [0, 1+1/n[$

… that the set on the left evaluates to $]0,1[$

… whereas the set on the right evaluates to $]0, 1]$

Incidentally this holds also for the sequence $B_n = [0, 1+1/n]$. I'd wager that this holds in the general case, i.e. that the interior of the intersection of an infinite number of sets is a subset of the intersection of the interiors of said sets.

However I have no intuition why this is so (assuming of course it is indeed true).

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It is true that $$\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)\subset \bigcap_{i\in I}\text{int}(A_i),$$ and the inclusion can be proper if the index set is infinite. Keeping our attention only on metric spaces, which are Hausdorff, if metric space $(X,d)$ has a non-isolated point $x$, the sets $A_i=\{y\in X:d(x,y)<1/i\}$ are individually open with $x\in \text{int}(A_i)=A_i$, but $\cap_{i=1}^\infty A_i=\{x\}$. As, by hypothesis, $x$ is not isolated, $\{x\}$ is not open, and therefore does not contain $x$ (or anything else) in its interior. Basically the same example can be found in any $T_1$ space with a non-isolated point.

This inclusion $$\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)\subset \bigcap_{i\in I}\text{int}(A_i)$$ is true in any topological space.

Since you are working in the real line, let's start there with that as an example.

If $A\subset \mathbb{R}$, then a point $x\in\mathbb{R}$ satisfies $x\in \text{int}(A)$ if and only if there exists $\delta>0$ such that $(x-\delta,x+\delta)\subset A$.

Therefore $x\in \text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)$ if and only if there exists $\delta>0$ such that $$(x-\delta,x+\delta)\subset \bigcap_{i\in I}A_i,$$ which happens if and only if there exists $\delta>0$ such that for all $i\in I$, $$(x-\delta,x+\delta)\subset A_i.$$

On the other hand, $x\in \bigcap_{i\in I}\text{int}(A_i)$ means that for each $i\in I$, $x\in \text{int}(A_i)$, which means there exists $\delta_i>0$ such that $(x-\delta_i,x+\delta_i)\subset A_i$. But in this case, the different $i$ can have different $\delta_i$.

So membership in $\bigcap_{i\in I}\text{int}(A_i)$ means that for each $i\in I$ there exists $\delta_i>0$ such that $(x-\delta_i,x+\delta_i)\subset A_i$, while membership in $\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)$ means there is one $\delta>0$ which will accomplish this same thing for all $i$ at the same time.

To get down to it, it is the distinction between "For all $i\in I$ there exists $\delta$ such that $\ldots$" and "There exists $\delta$ such that for all $i\in I$ $\ldots$".

If $I$ is finite, we can take $\delta=\min_{i\in I}\delta_i$, because a finite set of positive numbers has a minimum, and these two notions are equivalent. You already came up with examples to show the failure if $I$ is infinite with $A_i=[0,1+1/i[$. So for $x=1$, the best $\delta_i$ for $A_i$ is $\delta_i=1/i$, and there is not one $\delta>0$ which will work simultaneously for all $i$.

In general metric spaces, say $(X,d)$, this is true if we replace intervals $(x-\delta,x+\delta)$ with $$B_\delta(x)=\{y\in X:d(x,y)<\delta\}.$$ These are intervals in $\mathbb{R}$, disks in $\mathbb{R}^2$ (with Euclidean distance, although there are other interesting distances in $\mathbb{R}^2$), etc.

It's still true in general topological spaces. Membership in $\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)$ means there exists an open set $U$ such that for all $i\in I$, $$x\in U\subset A_i.$$ So a single open set $U$ works simultaneously for all $i$. But $x\in \bigcap_{i\in I}\text{int}(A_i)$ means that for all $i\in I$, there exists an open set $U_i$ such that $$x\in U_i\subset A_i.$$ This is exactly the same idea, that "For all $i\in I$, there exists $U\in \tau$ such that $\ldots$" is weaker than "There exists $U\in \tau$ such that for all $i\in I$, $\ldots$". Here, $\tau$ is our topology (set of open sets).

A few interesting things that we can get out of this:

We always have $$\overline{A}^c=\text{int}(A^c),\tag{$I$}$$ where the superscript $c$ denotes the set complement and $\overline{A}$ denotes the closure of a set. To see why $I$ holds, for $x\in X$, either every open set containing $x$ intersects $A$ (which is equivalent to $x\in \overline{A}$) or there exists an open set containing $x$ which doesn't intersect $A$ (which is equivalent to $x\in \text{int}(A^c)$).

Since $(A^c)^c=A$, $I$ is equivalent to $$\overline{A^c}=\text{int}(A)^c.$$

We also have DeMorgan's laws for arbitrary unions/intersections: $$\Bigl(\bigcap_{i\in I}A_i\Bigr)^c = \bigcup_{i\in I}A_i^c$$ $$\Bigl(\bigcup_{i\in I}A_i\Bigr)^c = \bigcap_{i\in I}A_i^c.$$

Also, if $A\subset B$, then $A^c\supset B^c$.

What can we do with these observations? Let $(B_i)_{i\in I}$ be a collection of subsets (imagine subsets of $\mathbb{R}$, if for concreteness). Let $A_i=B_i^c$ for all $I$. We know $$\Bigl[\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)\Bigr]^c \supset \Bigl[\bigcap_{i\in I}\text{int}(A_i)\Bigr]^c.$$ Applying $\overline{A^c}=\text{int}(A)^c$ with $A=\bigcap_{i\in I}A_i$, we get $$\overline{\Bigl(\bigcap_{i\in I}A_i\Bigr)^c }\supset \Bigl[\bigcap_{i\in I}\text{int}(A_i)\Bigr]^c.$$

Using DeMorgan's laws, we get $$\overline{\bigcup_{i\in I}A_i^c}=\overline{\Bigl(\bigcap_{i\in I}A_i\Bigr)^c }\supset \Bigl[\bigcap_{i\in I}\text{int}(A_i)\Bigr]^c = \bigcup_{i\in I}\text{int}(A_i)^c.$$ Using $\overline{A^c}=\text{int}(A)^c$ again, and the fact that $A_i^c=B_i$, we get $$\overline{\bigcup_{i\in I}B_i}=\overline{\bigcup_{i\in I}A_i^c}\supset \bigcup_{i\in I}\text{int}(A_i)^c = \bigcup_{i\in I}\overline{A_i^c}=\bigcup_{i\in I}\overline{B_i}.$$

So now we also know something analogous about closures: $$\overline{\bigcup_{i\in I}B_i}\supset\bigcup_{i\in I}\overline{B_i}.$$

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    $\begingroup$ wonderful intuition! $\endgroup$ Feb 23 at 16:50
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    $\begingroup$ Nice answer! I think "and the inclusion can be proper if the index set is infinite. This is true in any topological space" is a bit misleading though. I can't tell if that was your intention or not but it seems to suggest that for any topological space $X$, there is a family of sets making that inclusion proper, but that's not true (consider eg discrete spaces or indiscrete spaces). $\endgroup$ Feb 23 at 19:59
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    $\begingroup$ Thank you for that comment! It was unclear. I attempted to clean it up. $\endgroup$
    – user469053
    Feb 23 at 20:15
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If $x \in int(\cap_n A_n)$, then $x$ is an interior point of $\cap_n A_n$. There exists $r>0$ such that $]x-r;x+r[\subset \cap_n A_n$. $$\forall n : ]x-r;x+r[\subset A_n $$ $\forall n : x$ is an interior point of $A_n$. $$\forall n: x\in int(A_n)$$ Hence $x\in \cap_n{int(A_n)}$. That's why "this holds in the general case, i.e. that the interior of the intersection of an infinite number of sets is a subset of the intersection of the interiors of said sets. "

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    $\begingroup$ No, this does not follow. What you have is that for every n x belongs to A_n. This doesn't automatically mean that x belongs to the interior of A_n. $\endgroup$ Feb 23 at 15:39
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    $\begingroup$ read again and understand. What i have written is that for every $n$, $x$ is an interior point of $A_n$. I have never said in my proof "for every n x belongs to A_n". $\endgroup$
    – arofenitra
    Feb 24 at 4:22
  • $\begingroup$ You are right; I stand corrected. $\endgroup$ Feb 26 at 11:48

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