check out this proof-finite intersection of open set is open [closed]

original lecture notes

I am confusing about the last comment. I am happy with $B_{r_1}(x)\subset A_1$ and $B_{r_2}(x)\subset A_2$. But even if this is the case, what guarantee the ball centered at $x$ with $r=\min(r_1,r_2)$ will be fully contained in $A_1\cap A_2$. This neighbourhood seems not necessary to "lies in" $A_1 \cap A_2$

Solution 1:

This proof is an incorrect proof. The mathematical idea of the correct proof is present, but the proof technically only shows that either $A \cap B$ is empty, or there is some $x \in A \cap B$, and some $r > 0$, such that $B_r(x) \subseteq A \cap B$.

In order to prove the statement “$A \cap B$ is open”, we must look at the definition of “open”.

The definition of the statement “$A \cap B$ is open” is: for all $x \in A \cap B$, there exists $r > 0$ such that $B_r(x) \subseteq A \cap B$.

In order to prove a statement of the form $\forall x \in A \cap B (P(x))$, we consider an arbitrary $x \in A \cap B$ and prove $P(x)$.

So consider an arbitrary $x \in A \cap B$. Take some $r_1, r_2 > 0$ such that $B_{r_1}(x) \subseteq A$ and $B_{r_2}(x) \subseteq B$; these exist by the fact that $A$ and $B$ are open and that $x$ is an element of both $A$ and $B$.

Now let $r = \min (r_1, r2)$. We see that since $r \leq r_1$, it follows that for any $y$, if $d(y, x) < r$ then $d(x, y) < r_2$, and therefore $B_r(x) \subseteq B_{r_1}(x)$ (and similarly for $r_2$). Therefore, we have $B_r(x) \subseteq B_{r_1}(x) \cap B_{r_2}(x) \subseteq A \cap B$. Finally, note that since both $r_1$ and $r_2$ are positive, $r$ must also be positive.

Thus, we can conclude that $A \cap B$ is open.

Note that this method requires no discussion of whether $A \cap B$ is empty. It’s common for beginners to try to prove a statement of the form $\forall x \in S (P(x))$ by doing case analysis on whether $S$ is empty, but this is always unnecessary and a distraction from the mathematical idea of the proof.