Co-Countable topology cases

Question

Let X be the reals. The $\tau$ $=$ $\{$ $\varnothing$ $\}$ $\cup$ $\{$ $A$ $:$ $A^c$ is countable $\}$

What I want to show is that the union of an arbitrary collection of open sets is open. So I broke it into 3 cases:

(1) $U_i = \varnothing$ for every $I\in I$

(2)$U_i$ is non-empty for all I$\in I$

(3)$U_i$ is non empty for some $I\in I$

The professor told me that I could disregard the second case, as the third implies the second. I'm not sure why that is the case, because

$\exists x : P(x)$ doe not necessarily imply $\forall x\in X: P(x)$

may someone please clarify? I feel like this is related to the drinkers paradox, but I'm not sure.

Answer 1

You are getting they other way around, while it is true that $\exists x : P(x)$ does not necessarily imply $\forall x\in X: P(x)$ it is certainly true that $\forall x : P(x)$ does imply $\exists x\in X: P(x)$

That is to say: 2) is the case where all of the $U_i$ are nonempty, this is clearly a particular case where some $U_i$ being non-empty. (Indeed, if all of them are non-empty, the definitely SOME are non-empty)

Answer 2

Suppose that you have proven your third case. What this would mean to me is that you have shown that if you have an arbitrary collection of open sets $U_i$, some of which are non-empty, then their union is open.

If some of them are non-empty (and you don't know which), and you've proven that their union is open, then you could happily assume that they're all non-empty, right? You would have proven that even when you don't know which are/are not non-empty, the union is open, so knowing that they're all non-empty shouldn't cause an issue.

Indeed, your assertion that showing that a statement is true for some $x$ does not imply that it is true for all $x$ is true, but I think you're getting confused. If you prove the third case, you are indeed showing that in any case, the union of non-empty/empty sets is open. This would be showing that the statement is true for all $x$. If you showed case $2$ instead, you could certainly not assume that statement $3$ is true, because you've only shown it true when all sets are non-empty. Does this help?