Kolmogorov's normability criterion
# Kolmogorov's normability criterion

In mathematics, Kolmogorov's normability criterion is a theorem that provides a necessary and sufficient condition for a topological vector space to be normable; that is, for the existence of a norm on the space that generates the given topology. The normability criterion can be seen as a result in same vein as the Nagata–Smirnov metrization theorem and Bing metrization theorem, which gives a necessary and sufficient condition for a topological space to be metrizable. The result was proved by the Russian mathematician Andrey Nikolayevich Kolmogorov in 1934.

## Statement of the theorem

Kolmogorov's normability criterion — A topological vector space is normable if and only if it is a $T_1$ space and admits a bounded convex neighbourhood of the origin.

Because translation (that is, vector addition) by a constant preserves the convexity, boundedness, and openness of sets, the words "of the origin" can be replaced with "of some point" or even with "of every point".

### Definitions

It may be helpful to first recall the following terms:

* A topological vector space (TVS) is a vector space $X$ equipped with a topology $\tau$ such that the vector space operations of scalar multiplication and vector addition are continuous.
* A topological vector space $(X,\tau )$ is called normable if there is a norm $\|\cdot \|:X\to \mathbb {R}$ on $X$ such that the open balls of the norm $\|\cdot \|$ generate the given topology $\tau .$ (Note well that a given normable topological vector space might admit multiple such norms.)
* A topological space $X$ is called a $T_1$ space if, for every two distinct points $x,y\in X,$ there is an open neighbourhood $U_{x}$ of $x$ that does not contain $y.$ In a topological vector space, this is equivalent to requiring that, for every $x\neq 0,$ there is an open neighbourhood of the origin not containing $x.$ Note that being $T_1$ is weaker than being a Hausdorff space, in which every two distinct points $x,y\in X$ admit open neighbourhoods $U_{x}$ of $x$ and $U_{y}$ of $y$ with $U_{x}\cap U_{y}=\varnothing$; since normed and normable spaces are always Hausdorff, it is a "surprise" that the theorem only requires $T_1.$
* A subset $A$ of a vector space $X$ is a convex set if, for any two points $x,y\in A,$ the line segment joining them lies wholly within $A,$ that is, for all $0\leq t\leq 1,$ $(1-t)x+ty\in A.$
* A subset $A$ of a topological vector space $(X,\tau )$ is a bounded set if, for every open neighbourhood $U$ of the origin, there exists a scalar $\lambda$ so that $A\subseteq \lambda U.$ (One can think of $U$ as being "small" and $\lambda$ as being "big enough" to inflate $U$ to cover $A.$)
