I'm learning about compactness in topological spaces and I wonder if there is an example of a compact topological space $X$ and an open cover $\{U_\alpha\}$ of $X$ for which we can't show explictly a finite subcover. I thought about it for a while but I couldn't come up with anything.
Asked
Active
Viewed 122 times
8
-
In what sense of "can't" do you want such an example? – nomen Jan 05 '14 at 08:19
-
5I don't know the answer, but you might want to try something like an uncountable product of compact spaces using the product topology. Tychonoff's theorem says that this product is compact, but the theorem requires the Axiom of Choice. So in this case it might be hard to actually say explicitly what the finite subcover is. Just a thought. – bryanj Jan 05 '14 at 08:32
-
Consider the closure of the space of all differentialble functions $f:[0,1]\rightarrow\mathbb R$ such that $|f'(x)|<1$ for all $x$, with the open cover consisting of all sets $U_\varepsilon(f)$ for some small $\varepsilon$ where $U_\varepsilon(f)={g:|g(x)-f(x)|<\varepsilon }.$ We know that this set is compact by the zarzuela-Ascoli theorem. Maybe you can find a finite subcover, but I can't. – Plutoro Jun 21 '15 at 22:18
-
@bryanj. A very good thought indeed. AC is needed to show that $\prod_{i\in I}A_i$ is not empty whenever $I\ne \phi$ and no $A_i=\phi.$.... But we do not need AC to show that $[0,1]^S$ is not empty for every $S\ne \phi.$ But an "explicit" finite sub-cover of the cover by members of the canonical open base would be equivalent to an "explicit" proof of AC. – DanielWainfleet Jan 30 '17 at 22:40