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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 $x\in M$.

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  • $\begingroup$ I would guess the same. $\endgroup$ – Berci Feb 23 '18 at 19:38
  • $\begingroup$ I've often seen closed subspaces defined in terms of their complement being open, as well. There are likely multiple equivalent definitions. $\endgroup$ – Adrian Keister Feb 23 '18 at 19:58
  • $\begingroup$ Is the definition of closedness remains that $X$ is a topological vector space or only a semi-normed vector space? $\endgroup$ – Student Feb 23 '18 at 20:08
  • $\begingroup$ The question you have to ask is: how can we define a topology given a semi-norm? $\endgroup$ – Rodrigo Dias Feb 23 '18 at 20:34
  • $\begingroup$ See this section about locally convex topological vector spaces en.m.wikipedia.org/wiki/Locally_convex_topological_vector_space $\endgroup$ – Rodrigo Dias Feb 23 '18 at 20:38
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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.

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  • $\begingroup$ Thank you. It is true that two equivalent semi-norms generate the same topology? $\endgroup$ – Student Feb 25 '18 at 7:09
  • $\begingroup$ Yes. If $p\le Cq$, then a $p$-neighborhood of $x$ contains a $q$-neighborhood of $x$, for another "radius". So the $p$-topology is contained in the $q$-topology (meaning every $p$-open set is $q$-open). Reversing the roles of $p$ and $q$, we get that the topologies are equal. $\endgroup$ – user357151 Feb 25 '18 at 8:09
  • $\begingroup$ I the definition of closed subspace we don't need that the topology is Hausdorff?? Thank you $\endgroup$ – Student Feb 26 '18 at 5:19
  • $\begingroup$ Once you have a topology - any kind of topology - you have the concept of closed sets - any kind of sets. In particular, you can tell, by looking at a linear subspace, whether it's closed or not. $\endgroup$ – user357151 Feb 26 '18 at 5:43

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