If $C_1\subset \mathbb{R}^d$ is closed and $C_2\subset \mathbb{R}^d$ is compact then $C_1+C_2$ is closed Pardon me for the simple question but I can't figure it out.
Can someone give me the hint?
Well, I was thinking of using $C_1+C_2=\cup_{x\in C_1} (x+C_2)=\cup_{x\in C_2} (x+C_1)$ but this doesn't give me an answer.
 A: For closedness we can use the sequence criterion: $A \subseteq R^d$ is closed iff for every sequence $a_n \in A$ that converges to $a$ in $\Bbb R^d$, we have that $a \in A$ too.
So let $(x_n)$ converge to $x \in \Bbb R^d$, where all $x_n \in C_1 + C_2$. We want to show that $x \in C_1 + C_2$ as well.
We can write $x_n = a_n + b_n$ where $a_n \in C_1$ and $b_n \in C_2$.
We know that $(b_n)$, being a sequence in $C_2$, has a convergent subsequence so that we have $n_1 < n_2 < n_3 < \ldots$ and $b \in C_2$ so that $$b_{n_k} \to b \text{ as } k \to \infty $$
But then $$x_{n_k} = (a_{n_k} + b_{n_k}) \to x$$ as subsequences of convergent sequences have the same limit.
So $$a_{n_k} = x_{n_k} - b_{n_k} \to x - b$$ as the vector operations are continuous on $\Bbb R^d$ and now, as $C_1$ is closed, $$x-b \in C_1$$ so that $$x=(x-b) + b \in C_1 + C_2$$ as required and we've shown $C_1 + C_2$ is closed.
Do note that your formula works for infinite unions and so if either $C_1$ or $C_2$ are open, so is $C_1 + C_2$.
