3

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!

palt34
  • 121
  • 6

1 Answers1

8

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.

Martin Argerami
  • 205,756
  • 1
    I think it is faster to find a duplicate of this FAQ than to answer it. – Jyrki Lahtonen Mar 29 '24 at 12:46
  • $r = \min{r_1, r_2, \ldots, r_n, \ldots}$ may be equal to $(0,1)$ or similar sets which have an infimum of $0$. This means that there is only one common element of all the sets. (Is this part right or can I just infer that there are $\geq 1$ points around which I can't form a open set?)

    This means that the intersection (may) form a singleton as in the counterexample. In a metric space which has the standard metric the singleton is not open (in fact closed here). This means that the infinite intersection of open sets are not always open in metric spaces generally. Is my reasoning correct?

     – palt34 Mar 29 '24 at 14:05
  • 1
    Yes, as long as the metric space is not discrete. If the singletons are open, then every set is open. But if there is an accumulation point $a$, then ${a }=\bigcap_nB_{1/n}(a)$. – Martin Argerami Mar 29 '24 at 14:22
  • Does it matter if I take the minimum or the infimum of $r = \min{r_1, r_2, \ldots, r_n}$ in the original proof as they mean the same in the context of finite sets? In our discussion you specify infimum in the context of infinite sets, is that because in more cases the minimum would not be defined so it is more natural to mention infimums? (i.e. is there a reason for this?)
    • – palt34 Mar 29 '24 at 22:08
  • Say we considered a finite collection then $r = \min{r_1, r_2, \ldots, r_n}$ could not result in$r=0$. Since we know that each set in the finite collection is open which means for any x in it we can find a$ r>0$ such that an open ball of that size is in the set, so all the $r_i$s are $>0$ so $r>0$?
    • – palt34 Mar 29 '24 at 22:08
  • 1
    For a finite set, the infimum is the same as the minimum, so one would usually call it the minimum. For an infinite set a minimum might not exist, so it's not that it is "more natural" to talk about infimum, it is that the infimum exists (we are talking about sets of positive numbers so they are bounded below). – Martin Argerami Mar 29 '24 at 22:45