I think i have the proof, however I am worried about some assertions.
Pf: Let where is compact
Let be an open cover of
is also an open cover for
Since is compact , a finite subcover
is a finite subcover for
I am not sure if I can just say that an open cover of K exists. It makes sense, to me anyway, that I could cover any set with a bunch of open sets. Still, I am not sure if an open cover of K exists.
And the union of the finite subcovers remains finite?
Thanks for helping.
Sure, that's fine. Maybe add Plato's comments if you want to be even more explicit, but either way it works.
Two things...I am not sure if I can just say that an open cover of K exists. It makes sense, to me anyway, that I could cover any set with a bunch of open sets. Still, I am not sure if an open cover of K exists.
First of all, an open cover of any set ALWAYS exists because the whole space is open by definition. For instance, suppose is a space and . Then is an open cover of .
Second, even if an open cover didn't exist, we don't need one to show that a set is compact. In fact, that would GUARANTEE that it was compact!
Recall that a set A is compact if and only if it satisfies the following:
For all Y, if Y is an open cover of A, then there is a finite subcover.
In other words, a set A is non-compact if and only if it satisfies the following:
Not for all Y, if Y is an open cover of A, then there is a finite subcover.
This is true if and only if
There exists Y such that Y is an open cover of A, and it has no finite subcover.
If there is no open cover of A, then it cannot satisfy this property and hence is compact.
Yup. Any finite union of finite sets is again finite.And the union of the finite subcovers remains finite?