# Thread: Topological Compactness and Compactness of a Set

1. ## Topological Compactness and Compactness of a Set

I need help in each of the problems. See attached picture.

Thanks

2. Originally Posted by meks08999 I need help in each of the problems. See attached picture.

Thanks
9. Let $\displaystyle \{U_i\}_{i\in I}$ be an open cover of $\displaystyle F$ and let $\displaystyle U_k\in\{U_i\}_{i\in I}$ be the set that contains $\displaystyle x_0$. Since $\displaystyle x_0$ is a cluster point, $\displaystyle U_k$ also contains all but finitely many $\displaystyle x_n$. Assume that it covers all but $\displaystyle N$ points in $\displaystyle \{x_n\}$.Thus there are at most $\displaystyle N$ more open sets in $\displaystyle \{U_i\}_{i\in I}$ (in addition to $\displaystyle U_k$) that completely cover $\displaystyle \{x_n\}$, and so you have proved that every open cover has a finite subcover, and $\displaystyle F$ is compact.

I'm not sure about 10. I know in a compact space every sequence has a convergent subsequence, but I don't know how to find two of them.

3. For 10 you have to show that there are at least two cluster points in the sequence. Assume that there is $\displaystyle \alpha$ only one cluster point

1. First, show that there is $\displaystyle \varepsilon$ such that there are infiniely many $\displaystyle (x_n)$'s outside of $\displaystyle B(\alpha,\varepsilon)$. If such an $\displaystyle \varepsilon$ does not exist we would have that the sequence converge to $\displaystyle \alpha$

2. For the subsequence formed by the $\displaystyle (x_n)$'s outside of $\displaystyle B(\alpha,\varepsilon)$, observe that it is in the compact set $\displaystyle K$. Hence there exists a cluster point which can't be $\displaystyle \alpha$.

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