I am trying to show the co-countable topology $\tau$ is a topology on a set $X$.

The first point to show $\emptyset,X\in\tau$ is not difficult.

Now let $S_i\in\tau$. From the definition of co-countable, $S_i=\emptyset$ or $X\backslash S_i$ is countable.

I struggle to prove the second point $\cup S_i\in\tau$:

If $S_i=\emptyset$, then $\cup S_i=\cup\emptyset=\emptyset\in\tau$. The part I got stuck is what if $S_i\neq\emptyset$? So far I have not used $X\backslash S$?

The thing that troubles me a lot is that we are not told beforehand whether $X$ is countable or uncountable. Is it necesary to know?

How are we going to work on the condition being countable?

The third point $\cap S_i\in\tau$ should be similar but I am also not very sure about how can we make use of $X\backslash S_i$?

Thanks for any help on this matter.

  • $\begingroup$ What is the complement of $\bigcup_i S_i$? $\endgroup$ – Arthur Sep 22 '16 at 13:26
  • $\begingroup$ @Arthur By using De Morgan we get $\cap_i X\backslash S_i$ right? $\endgroup$ – user338393 Sep 22 '16 at 13:27
  • $\begingroup$ Appropriate notation not only demonstrates you understand the definitions. It also aids in communicating your points to others. $\endgroup$ – Matthew Leingang Sep 22 '16 at 13:29
  • $\begingroup$ @MatthewLeingang yes you're right thanks, I have made some edit. $\endgroup$ – user338393 Sep 22 '16 at 13:31
  • $\begingroup$ That's exactly what you get. How many elements are in that intersection, compared to, say, $X\setminus S_1$? (Assuming $S_1\neq \emptyset$.) $\endgroup$ – Arthur Sep 22 '16 at 13:32

Suppose $S_{i} \in \tau$. First prove that

$$ X \setminus (\bigcup S_{i}) = \bigcap (X \setminus S_{i})$$

Now there are two cases. Either all of the $S_{i}$ are the empty set, or at least one is not the empty set. The first case is straightforward. For the second case, ask yourself what the intersection of two countable sets looks like.

  • $\begingroup$ The intersection of two countable set is countable right? Our goal is to show $\cup S_i\in\tau$ isn't it? How can knowing $\cap(X\backslash S_i)$ is countable help us show $\cup S_i\in\tau$? $\endgroup$ – user338393 Sep 22 '16 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.