In a metric space, we have that the intersection of countably many open sets is open:
Let $A_1, A_2, \ldots, A_n$ be open sets and $A = \bigcap_{i=1}^n A_i$. For $x \in A$, $x \in A_i$ for all $i$, and since each $A_i$ is open, $\exists r_i > 0$ with $B_{r_i}(x) \subseteq A_i$. Set $r = \min\{r_1, r_2, \ldots, r_n\}$, then $B_r(x) \subseteq A$. Since $x$ was arbitrary, $A$ is open.
I am trying to understand why this proof, does not hold for uncountably infinite sets and I believe the problem lies in this step: $r = \min\{r_1, r_2, \ldots, r_n\}$
I think this because:
For any collection of elements of any space that I am considering the intersection of, there is a associated $r_i$ with each element.
Then since we are considering a metric space$\{r_1, r_2, \ldots, r_i, \ldots\} \subset \mathbb{R}$. Then it is not necessarily true that there exists a minimum $r_i$. As say $\{r_1, r_2, \ldots, r_i, \ldots\}$ was the set $(0,1)$.
I am aware that I can consider the many counterexamples for the the uncountable union of open sets, but I want to identify, which will help me examine when the future proofs I do hold, where the proof given falters when I try and make a more general statement.
If someone could point out if the reasoning I provided is the correct way to think about things, that would be great!
Best Answer
The issue is not "countable vs uncountable" but rather "finite vs infinite".
It is not true in general that a countable intersection of open sets is open, and the problem is exactly what you mention that an infinite set of positive radii may have infimum zero. For example in $\mathbb R$, $$ \bigcap_n\Big(-\frac1n,\frac1n\Big)=\{0\}, $$ which is not open.