No problem, glad we could help! And welcome to the site! :)

[judelaw]

Let X be any infinite set. The countable closed topology is defined to be the topology having as its closed sets X and all countable subsets of X. Prove that this is indeed a topology on X.

Any help would be greatly appreciated. Thanks!

 

[micromass]

Hi judelaw!

What have you already tried to solve this? For being a topology, you need to satisfy three axioms, which ones? And which ones are troubling you?

 

[judelaw]

haha, like all of them?

I'm not really sure what countable subsets of X means. I sort of understand the definition of countable (exists a bijection between it and the set of natural numbers?), but I don't really know how to relate that to this problem.

 

[disregardthat]

First of all, what are the open sets in this (soon to be proved) topology?

1) Is the empty set an open set? Is X an open set?
2) Is a union of open sets an open set?
3) Is a finite intersection of open sets an open set?

or equivalently:

1) is the empty set and X closed sets?
2) Is a finite union of closed sets a closed set?
4) Is an intersection of closed sets a closed set?

Being countable means to be either finite or in bijection with the natural numbers. Any subset of a countable set is countable.

 

[micromass]

Well a subset of X is indeed countable if there exists a bijection between the subset and the naturals, or if the subset is finite.

Maybe some examples will help you: in [itex]\mathbb{R}[/itex], we have the following countable subsets: [itex]\mathbb{N}[/itex], [itex]\mathbb{Q}[/itex], {1,3,5,7,9,11,...}. While [0,1] or [itex]\mathbb{R}^+[/itex] are not countable.

So, what are the axioms that the closed sets must satisfy? Well, the first one is that the empty set and X are both closed. Can you show this?

 

[judelaw]

Ah okay. I got it! I misunderstood "countable subsets. "

Thank you micromass and disregardthat!

And I just found a new website to frequent.