Additive function in $\mathbb{R}^n$ is continuous, and related subspaces compact

44 Views Asked by At

I want to show that the function:

$A: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n, (x, y) \mapsto x + y$

is continuous.

Also, why is it that if $K, L$ are compact subspaces of $\mathbb{R}^n$, $K + L = \{x + y | x \in K, y \in L\}$ is also compact?

Thanks in advance. Intuitively it sounds right. For proving it, maybe one could use the tube lemma here?

1

There are 1 best solutions below

0
On BEST ANSWER

Your phrasing of the problem is wrong, assuming that you wanted to say $A: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ and it is defined as the addition of the vector space $\mathbb{R}^n$, then we will show that for any sequence $a_k = (x_k,y_k) \in \mathbb{R}^n\times \mathbb{R}^n$ converging to $a=(x,y)$ we will show that $A(a_k) $ converges to $A(a)$ which will establish the continuity. Now $a_k $ converges to $(x,y)$ means $x_k$ and $y_k$ converges to $x$ and $y$ respectively. So $A(a_k)=x_k+y_k$ converges to $x+y$ since sum of two convergent sequences is convergent and sandwich theorem will tell us that the limit is the desired one.

For the second part note that $K \times L$ is compact in $\mathbb{R}^n\times \mathbb{R}^n$ by Tychonoff's theorem. And $K+L= A(K\times L)$. Now since continuous image of a compact set is compact, the required set is compact.