Finite Complement Topology, Compactness, and Closedness

310 Views Asked by At

Munkres says of the necessity of the Hausdorff Condition in the Proposition "Every compact supsapce of a Hausdorff space is closed":

One needs the Hausdorff condition in the hypothesis. Consider, the finite complement topology on $\mathbb{R}$. The only proper subsets of $\mathbb{R}$ that are closed in this topology are finite sets. But every subset of $\mathbb{R}$ is compact in this topology.

I'm having trouble seeing this. In particular, how does one build sets in this topology that are anything but $\mathbb{R}-\{x_0,...x_n\}$? In order to get a finite set, I need to intersect uncountably many sets like this -- which I can't do in a topology. I can see that every subset of $\mathbb{R}$ will be compact -- but not every subset is attainable in this topology. So this statement confuses me.

I guess what I'm looking for are examples of both statements.

1

There are 1 best solutions below

2
On BEST ANSWER

There is some confusion in what you wrote. If I have a topology $\tau$ on $\mathbb R$ and I want to defined a certain subset of $\mathbb R$, then $\tau$ doesn't matter. It is only a matter of defining the set. And, for instance, $[-1,1]$ is compact subset of $\mathbb R$ with respect to finite complement topology, in spite of the fact that it is not closed.