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 $u$ was arbitrary:


 * $U + x$ is open.