Translation of Open Set in Topological Vector Space is Open

Theorem

Let $\struct {X, \tau}$ be a topological vector space.

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

Let $x \in X$.

Then:

$U + x$ is open.

Proof

Let $T_{-x}$ be the translation by $x$ mapping.

From Translation Mapping on Topological Vector Space is Homeomorphism, we have:

$T_{-x}$ is a homeomorphism.

So, since $U$ is open, we have:

$T_{-x} \sqbrk U$ is open.

That is:

$U - \paren {-x} = U + x$ is open.

$\blacksquare$

This article, or a section of it, needs explaining. In particular: Review notation -- there is a caveat that $T_{-x}$ and $T_x$ get confused, and on the translation by $x$ mapping this is specifically defined as $T_x$ not $T_{-x}$. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.6$: Topological vector spaces