It is always possible to define a topology in a vector space endow with a semi-norm?

Question

If $(X,\|\cdot\|)$ is a semi-normed vector space.

It is always possible to define a topology on $X$? If it is true What is the definition of a closed subspace of $X$ with respect to $\|\cdot\|$?

I guess that a subspace $M$ of $X$ is closed with respect to the semi-norm $\|\cdot\|$ if and only if every $(x_n)_n\subset M$ such that $\|x_n-x\|\to 0$ then $x\in M$.

Answer 1

A semi-norm $p$ directly induces a topology: the neighborhoods of a point $x$ are sets $U(x, r) = \{y:p(x-y)<r\}$. The topology gives you the concepts of open and closed sets. And yes, a closed subspace $M$ can be characterized by sequences: $M$ is closed iff $p(x_n, x)\to 0$ with $x_n\in M$ implies $x\in M$.

However, the above topology is not Hausdorff unless $p$ is a norm. Indeed, every neighborhood of $0$ contains the set $\{x:p(x)=0\}$. As Wikipedia notes,

A locally convex space is Hausdorff if and only if it has a separated family of seminorms. Many authors take the Hausdorff criterion in the definition.

The point being, there is not much to do with the space in its current form. If you are willing to take the quotient $X/\{x:p(x)=0\}$, that's one way to rectify the situation; the quotient is Hausdorff.

But I don't think there is any canonical way to get a Hausdorff topology on a seminormed space $X$ itself. See a related question Turning a semi-norm into a norm.