Let $E$ be a topological linear space, if $U$ is a neighbourhood of $0$ why is $U+x$ a neighbourhood of $x$?

Question

Let $E$ be a topological linear space, if $U$ is a neighborhood of $0$ why is $U+x$ a neighborhood of $x$?

With linear space I mean a vector space over the real or complex numbers.

I know the definition of topological linear space, but I don't understand that property.

Answer 1

What's the definition of neighborhood (to use US spelling : )? $V$ is a neighborhood of $x$ iff it contains some open ball $B(x, \delta)$ around $x$ — that is, $B(x, \delta) \subseteq V$.

We know that for some $\delta > 0, B(0, \delta) \subseteq U$, so $x + B(0, \delta) \subseteq x + U$. However, $$ B(x,\delta) \subseteq x + B(0, \delta) \tag {*} $$ Proof: if $y \in B(x,\delta)$, then $\lvert y - x \rvert = \lvert (y- x) - 0\rvert < \delta$, so $(y-x) \in B(0,\delta)$, so $y = x + (y-x) \in B(x,\delta)$. (Actually, all these steps are reversible, so (*) can be an equality; but that's more than we need.)

Note, this proof applies to normed/metrizable spaces only.

Answer 2

Translations are homeomorphisms, hence $x+U$ is open. Since $U$ is a neighbourhood of $0$, then $x+0=x \in U$. Therefore, $x+U$ is an open set containing $0$.