Intersection of Interiors contains Interior of Intersection

Question

I'm teaching my self topology using a book I found. This is the third part of a 4 part question. links to other parts: one, two.

I'm trying to prove the following problem from a book I found:

Let $X$ be a topological space and let $\mathscr{A}$ be a collection of subset of $X$. Prove
$int( \bigcap \limits_{A \in \mathscr{A}} A)\subseteq \bigcap \limits_{A \in \mathscr{A}} int(A)$

I want to know if my proof is valid, and before i prove it I will given an example of $int( \bigcap \limits_{A \in \mathscr{A}} A)\subset \bigcap \limits_{A \in \mathscr{A}} int(A)$.

Example of $int( \bigcap \limits_{A \in \mathscr{A}} A)\subset \bigcap \limits_{A \in \mathscr{A}} int(A)$.

Let: $\mathscr A = [-1, \frac{1}{n}]; $ where $ n \in \mathbb N$. So then, $int( \bigcap \limits_{A \in \mathscr{A}} A) = (-1,0)$ and $\bigcap \limits_{A \in \mathscr{A}} int(A)= (-1,0]$.

Proof of: $int( \bigcap \limits_{A \in \mathscr{A}} A)\subseteq \bigcap \limits_{A \in \mathscr{A}} int(A)$

To prove this, I will use 3 corollaries, that will not be proven here.

(1) $ \forall S \in \mathscr S : \bigcap \limits_{S \in \mathscr{S}} S \subseteq S$; The intersection of a collection of sets is a subset of any of the individual sets in the collection.

(2) $U \subseteq V \implies int(U) \subseteq int(V)$ : IF U is a subset of V, THEN the interior of U is a subset of the interior of V.
(3)$\forall Y \in \mathscr Y : X \subseteq Y \implies X \subseteq \bigcap \limits_{Y \in \mathscr{Y}} Y$: IF X is a subset of every set in a collection of sets THEN X is also in the intersection of every set in the collection.

Proof: Let, $\mathscr A$ be any collection of sets in a topological space, and let $A$ be any individual set in the collection.

$\bigcap \limits_{A \in \mathscr A} A \subseteq A $ for all $A \in \mathscr A$; (1, $S=A$)

$int(\bigcap \limits_{A \in \mathscr A} A )\subseteq int(A)$ for all $A \in \mathscr A$; (2, $U= \bigcap \limits_{A \in \mathscr A} A $ and $V = A $)

$int( \bigcap \limits_{A \in \mathscr{A}} A)\subseteq \bigcap \limits_{A \in \mathscr{A}} int(A)$ (3,$X=int(\bigcap \limits_{A \in \mathscr A} A )$ and $Y=int(A)$)

QED

Is this proof valid? Am I using the $\forall$ symbol correctly?

Your proof is valid, though you should better write $$\forall A\in\mathcal A: B\subseteq A$$ or $$B\subseteq A \text{ for all } A\in\cal A$$ instead of $$\color{red}{B\subseteq\forall A}$$ — Stefan Hamcke, Jul 13 '16 at 20:17
@Stefan Hamcke I just changed the "for all", is it in the correct syntax now? — Michael Maliszesky, Jul 13 '16 at 20:32
By the way, if the collection is finite, then the intersection of the interiors is actually equal to the interior of the intersection. — Stefan Hamcke, Jul 13 '16 at 20:38

score 1 · Answer 1 · answered Feb 23 '24 at 14:05

Your proof is valid. The following one is an alternative.

Let $B:=\bigcap\limits_{A\in\mathscr A}A$. If a point $x$ belongs to $\operatorname{int}(B)$, i.e. if $B$ is a neighborhood of $x$, then every superset of $B$ also is, hence for every $A\in\mathscr A$, $A$ is a neighborhood of $x$, i.e. $x\in\operatorname{int}(A)$. Therefore, $x\in\bigcap\limits_{A\in\mathscr A}\operatorname{int}(A)$.

Intersection of Interiors contains Interior of Intersection

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