Classification of Open Neighborhoods in Topological Vector Space

Theorem

Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$.

Let $U \subseteq X$.

Let $x \in X$.

Then $U$ is an open neighborhood of $x$ if and only if $U - x$ is an open neighborhood of ${\mathbf 0}_X$.

Equivalently, the open neighborhoods of $x$ are precisely the sets of the form $V + x$ where $V$ is an open neighborhood of ${\mathbf 0}_X$.

Proof

Necessary Condition

Suppose that $U$ is an open neighborhood of $x$.

Then ${\mathbf 0}_X \in U - x$ and from Translation of Open Set in Topological Vector Space is Open, $U - x$ is open.

So $U - x$ is an open neighborhood of ${\mathbf 0}_X$.

$\Box$

Sufficient Condition

Suppose that $U - x$ is an open neighborhood of ${\mathbf 0}_X$.

Note that $U = \paren {U - x} + x$.

So, we have $x \in U$ and from Translation of Open Set in Topological Vector Space is Open, $U$ is open.

So $U$ is an open neighborhood of $x$.

$\blacksquare$