Check that a set is closed, and that another is compact

[bobby2k]

I have a problem with this excercise. Ironically I think I can manage the part that is supposed to be hardest, here is the problem:

Let $(V,||\cdot||)$, be a normed vector-space.

a), Show that if A is a closed subset of V, and C is a compact subset of V, then $A+C=\{a+c| a \in A, c \in C\}$ is closed.

b) Show that if both A and C are compact then A+C is compact.

I think b was ok,so I tried that first: if we have a sequence from A+C: $\{a_n+c_n\}$, then $\{a_n\}$, has a subsequence converging to a, if we look the original sequence but only the indexes from the subsequence converging to a, then of these indexes of the sequence $\{c_n\}$, must have a subsequence converging to c, and since we then have convergence of a subsequence to a+c, we are done?I struggle more with a).

I thought that I could show that if ther is a sequence from A+C converging to a point, then this point must be in A+C. So I start with the sequence $\{a_n+c_n\}$, which I assume converges to b, I must show that b is in A+C.
I get that since C is compact there must be a subsequence so that $\{c_{n_k}\}$, converges to an element c in C. Then using the same indexes $\{a_{n_k}\}$ must converge to b-c. But how do I proceed to show that b is in A+C?

 

[micromass]

Will the sequence $(a_{n_k} + c_{n_k})$ converge?

 

[bobby2k]

Will the sequence $(a_{n_k} + c_{n_k})$ converge?

Yeah, that is a subsequence of the original sequence. And I started with the original sequence converging, and every subsequence of a converging sequence converges.

 

[micromass]

So you know that $(a_{n_k})$ converges to $b-c$. What can you conclude from $A$ being closed?

 

[bobby2k]

So you know that $(a_{n_k})$ converges to $b-c$. What can you conclude from $A$ being closed?

Yeah I understand, b-c is in A, so (b-c)+c is in A+C.

Thanks!

 

[micromass]

Yeah I understand, b-c is in A, so (b-c)+c is in A+C.

Thanks!

That's it!

 

[bobby2k]

Thanks, vector-spaces are a little tricker than metric-spaces.