I have a topology question which is:
Give an example of a topological (non-Hausdorff) space X and a a non-closed compact subspace.
I've been thinking about it for a while but I'm not really getting anywhere. I've also realised that apart from metric spaces I don't really have a large pool of topological spaces to think about (and a metric sapce won't do here-because then it would be hausdorff and any compact set of a metric space is closed)
Is there certain topological spaces that I should know about (i.e. some standard and non-standard examples?)
Thanks very much for any help
Best Answer
Any topology with finitely many open sets must have the property that all sets are compact, simply because any open cover is already finite.
Here is a much less trivial example:
Let $(\mathbb R,\tau)$ be the real numbers with the co-finite topology, namely a set is closed if and only if it is finite; and a set is open if and only if its complement is finite.
Consider the natural numbers as a subset of the real line, this is an infinite set, but it is clearly not co-finite so it is neither open nor closed. Suppose that $\{U_i\mid i\in I\}$ is an open cover of $\mathbb N$. There is some $i_0\in I$ such that $0\in U_{i_0}$, and since $U_{i_0}$ is open it means that it contains everything except finitely many points, in particular it must contain all the natural numbers, except maybe finitely many of them. For every $n\in\mathbb N\setminus U_i$ we can find some $U_{i_n}$. We found, therefore, a finite subcover of this open cover, and so $\mathbb N$ is compact.
Exercise: Prove that in fact every set of real numbers is compact in this topology.