# Topology, counter examples needed.

## Homework Statement

I need two counter examples, that show the following two theorems don't/B] hold:
Let X be a topological space.

1. If from the closeness of any subset A in X follows compactness of A, then X is compact.
2. If from the compactness of a subset A in X follows closeness of A, then X is housdorff.

I proved the *opposite* theormes which do hold, but I cannot seem to find counter examples.

That means, I need to find a non-compact space, in which every closed subset is compact, and a non-housdorff space, in which every compact subset in closed.

## Homework Equations

Are there any equations in Topology?

## The Attempt at a Solution

I just tried to take for "1" spaces I know that are not compact, but then couldn't find a space in which every closed subset is a compact one...
In "2" I thought of non-hausdorff spaces I know, but couldn't directly see whether indeed every compact set is closed.

Thanks,
Tomer.

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HallsofIvy
Homework Helper
You aren't going to be able to find a counter example to (1) because it is true. Every space is closed in itself. If it is true that every closed subset of A is compact, then A itself is compact because A is a closed subset of itself.

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Thanks a lot for the reply.
I see you're point :-) That is strange - in the task we were to prove the next theorem (part a):
If X is compact, it follows from the closeness of A that A is compact.

Then we need to give counter examples to show why the opposite doesn't hold. I translated the "opposite claim" correctly, right?

But I definitely agree with what you just said :-)

For (2), you might want to try the cocountable topology.

Thanks, I will! :-)