Proving Compactness of K ∩ F Using Convergent Sequences

[t3128]

Homework Statement

Show that if K is compact and F is closed, then K n F is compact.

Homework Equations

A subset K of R is compact if every sequence in K has a subsequence that converges to a limit that is also in K.

The Attempt at a Solution

I know that closed sets can be characterized in terms of convergent sequences. Am I suppose to use that to prove the question?I really have no idea how to do this question.

 

[ystael]

Yes. Think of a sequence in $$K \cap F$$; then think about what it means that the sequence is both in $$K$$ and in $$F$$.

 

[t3128]

Sorry I didn't quite understand that, would you please be able to explain it a bit more?

 

[Office_Shredder]

You want to determine if $$K\cap F$$ is compact using a statement about sequences. So if you have a general sequence in $$K\cap F$$, you can learn one thing about it by knowing the sequence is in K, and another thing about it by knowing the sequence is in F. Combine those two things and see if you get what you need to show that $$K\cap F$$ is compact

 

[t3128]

Ok, this is what I have got:

Let x_n be in KnF.
=> x_n is in K. =>We know that K is compact, so every sequence in K has a subsequence that converges to a limit that is also in K.
=> x_n is in F. => By definition, if x_n -> c , then c is in F. By B-W theorem, it must have a convergent subsequence which converges to the same limit c.
So x_n is compact.

Am I getting closer?

 

[ystael]

You have the right set of ideas, but they're put together in a sequence that doesn't make sense.

The start is right. Let $$(x_n)$$ be a sequence in $$K \cap F$$. Since $$K$$ is compact, you can find a subsequence $$(x_{n_k})$$ of $$(x_n)$$ which converges to some point $$\overline{x} \in K$$.

Now the next sentence needs to begin "Since $$F$$ is closed and the subsequence $$(x_{n_k})$$ is a sequence in $$F$$..."

And the third sentence should end "therefore $$K \cap F$$ is compact."

Try filling that in.