Translation of Open Set in Normed Vector Space is Open

Theorem

Let $\struct {X, \norm \cdot}$ be a normed vector space.

Let $U \subseteq X$ be an open set.

Let $x \in X$.

Then:

$U + x$ is open.

Proof

Let:

$v \in U + x$

then:

$v = u + x$

for some $u \in U$.

So:

$v - x \in U$

Since $U$ is open, there exists $\epsilon > 0$ such that whenever $v' \in X$ and:

$\norm {\paren {v - x} - \paren {v' - x} } < \epsilon$

we have $v' - x \in U$.

That is:

$v' \in U + x$

Note that:

$\norm {\paren {v - x} - \paren {v' - x} } = \norm {v - v'}$

So, whenever $v \in U + x$ and $v' \in X$ is such that:

$\norm {v - v'} < \epsilon$

we have:

$v' \in U + x$

Since $v$ was arbitrary:

$U + x$ is open.

$\blacksquare$