If $f_1 \in C_c (\mathbb{R}^N)$, Why is the set $\overline{B(0,1)} + supp(f_1)$ compact?

Question

I'm studying Brezis book: Functional Analysis, Sobolev spaces and partial differential equations.

Theorem 4.22 states that if $f \in L^p(\mathbb{R}^N)$, with $1 \leq p < \infty$ and if $\rho_n$ is a sequence of mollifiers then $\rho_n * f \to f$ in $L^p(\mathbb{R}^N)$ (Where $\rho_n * f$ denotes the convolution between $\rho_n$ and $f$).

There is a part of the proof that uses a function $f_1 \in C_c(\mathbb{R}^N)$ and then the book claims that the set $$ \overline{B(0,1)} + supp(f_1)$$

is compact

(here $B(0,1)$ is the unit ball and if $A$ and $B$ are two subsets of $\mathbb{R}^N$, $A+B$ is defined as the set $\{ a+b : a \in A, \ b \in B \}$).

Why is that? I know that $\overline{B(0,1)}$ and $supp(f_1)$ are both compact sets, but I also know that in general the sum of compact sets may not be compact. I'm trying to show that $\overline{B(0,1)} + supp(f_1)$ is closed and bounded but it does not seem clear to me.

Answer 1

The sum of compact sets is in general compact. To see this, note that for a topological vector space $X$, addition is a continuous map $X\times X\to X$, and that if $K$ and $C$ are compact subsets of $X$, then so is $K\times C$. Finally, the continuous image of a compact set- in this case, $K+C$- is compact.