I am not sure how to finish the proof of showing that the intersection of finitely many open subsets is open.

Question

I am trying to prove the following:

Let $(X,d)$ be a metric space. An intersection of finitely many open subsets of $X$ is open.

The following is my attempt:

Let $x \in \bigcap_{i=1}^n U_i$ be arbitrary, where $U_i$'s are open subsets in $X$, for each $i=1,2,3,\ldots,n-1,n.$

Since $x \in \bigcap_{i=1}^n U_i$, $x \in U_i$ for all $i=1,2,3, \ldots,n-1,n.$

Because all $U_i$'s are open, for each $i,$ there exists $\epsilon_i>0$ such that $B_{\epsilon_i}(x) \subseteq U_i.$

Let $\epsilon = \min\{\epsilon_1, \epsilon_2, \epsilon_3, \ldots , \epsilon_{n-1}, \epsilon_n\}.$

Then $B_\epsilon(x)\subseteq B_{\epsilon_i}(x)$ for all $i=1,2,3, \ldots,n-1,n.$

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Question:

I am not quite sure of how to finish the proof, by showing that $B_\epsilon(x) \subseteq \bigcap_{i=1}^n U_i.$

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My reasoning for the ending of the proof:

Intuitively, it makes sense that $B_\epsilon(x) \subseteq \bigcap_{i=1}^n U_i$, since the open ball centred at $x$, with radius $\epsilon$ contains all points that are only in all of the sets $B_{\epsilon_1}(x),\,B_{\epsilon_2}(x),\,B_{\epsilon_3}(x),\ldots,\,B_{\epsilon_{n-1}}(x)$ and $\,B_{\epsilon_n}(x).$

Would it be correct in writing \begin{equation*} B_{\epsilon}(x) = \bigcap_{i=1}^n B_{\epsilon_i}(x)? \end{equation*}

Then would it be the case that $$B_{\epsilon}(x) =\bigcap_{i=1}^n B_{\epsilon_i}(x) \subseteq \bigcap_{i=1}^nU_i,$$

therefore $\bigcap_{i=1}^nU_i$ is open?

Answer 1

The proof consists in exhibiting an open ball that is included in the intersection (around every point).

For every set you know that you can find a suitable ball. Then the ball with minimum radius is the intersection of all these balls.

Hence the minimal ball is included in every ball, hence in every set, thus in their intersection.

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The proof will not work with infinitely many sets, because then the minimum radius may fail to be positive.

Answer 2

$B_\epsilon(x)\subseteq B_{\epsilon_i}(x)\ $$\ $ for all $i=1,2,3, \ldots,n-1,n.$
$B_{\epsilon_i}(x) \subseteq U_i\ $ for all $i=1,2,3, \ldots,n-1,n.$
$B_\epsilon(x)\subseteq U_i\ $ for all $i=1,2,3, \ldots,n-1,n.$

Answer 3

For the last portion, the only weakness is to prove that all the points in a ball is a subset of the points of a larger radius, when the center is the same point $p$. And also, the intersection of a finite number of such balls is the smallest one.

Though this step can be considered trivial, we can't deduce the intersection of the balls by their openness, since it is yet to be proved.

For two balls, if that doesn't hold, there exits a point $a\in B_{\epsilon_1}(p)$, but $a \notin B_{\epsilon_2}(p)$, where $\epsilon_2 > \epsilon_1$. We have $\epsilon_1 > \|a-p\| \ge \epsilon_2$, which contradicts the assumption. Then it can be applied to finitely many.

It seems the manipulation of open balls is more fundamental than openness or closeness.