Rudin Exercise 4.25(a) reads:
If $K$ is compact and $C$ is closed in $\mathbb{R}^k$, prove that $K + C$ is closed.
The hints in the problem suggest a proof by proving that the complement of $K + C$ is open, a path which I was able to follow into a successful proof. However, I want to prove it "directly", by showing that any limit point of $K + C$ must be within $K + C$, however I run into the following problem:
My Attempt:
Suppose $z$ is a limit point of $K + C$. Then there is a sequence $\{z_n\} \to z$ in $K + C$. Since each $z_i$ is an element of $K + C$, we can write $z_i = k_i + c_i$ for sequences $\{k_i\}$, $\{c_i\}$ in $K$ and $C$ respectively. Now, we simply must show that $\{k_i\} \to k \in K$ and $\{c_i\} \to c \in C$ to be done.
However, I noticed that $\{k_n\}$ and $\{c_n\}$ do not necessarily converge when their sum does. As an example, in $\mathbb{R}$, take $k_i = (-1)^i$ and $c_i = (-1)^{i+1}$. Then neither $\{k_n\}$ nor $\{c_n\}$ converge, but their sum does.
Are there any suggestions on how to get around this problem and complete this more "direct" proof?
Thanks!
Best Answer
As $K$ is compact, there is a subsequence $k_{i_j}$ such that coverges to some $k \in K$. In particular, the subsequence $z_{i_j} \rightarrow z$. Thus, $c_{i_j}= z_{i_j} -k_{i_j} \rightarrow z - k \in C$, since $C $ is closed. Then, $$ z= k + (z-k) \in K + C. $$