3
$\begingroup$

I am looking for a topological vector space $(X,\tau)$ enjoying the following conditions:

1- $(X,\tau)$ is not locally convex.

2- There exists a metric $d$ on $X$ and a sequence $\{X_n\}$ of subsets of $X$ such that $X=\cup X_n$ and the topology $\tau$ on $X_n$ is relatively second countable and $d$-metrizable for every $n$.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ For instance $X:=L^p([0,1])$, for $0<p<1$, is a TVS whose only convex open set is $X$ itself; it is a separable complete metric space with the distance $d(f,g):=\int_0^1|f-g|^pdt$, and you can take the $X_n$ to be subspaces of dimension $n$ $\endgroup$
    – Pietro Majer
    Commented Apr 26, 2018 at 20:33
  • 1
    $\begingroup$ @PietroMajer your cover is not countable. $\endgroup$
    – Tomasz Kania
    Commented Apr 26, 2018 at 21:10
  • 3
    $\begingroup$ Can't you take $X_1=X=L_p$ for $0<p<1$? $\endgroup$
    – Tomasz Kania
    Commented Apr 26, 2018 at 21:12
  • 2
    $\begingroup$ Sorry, I though I saw a closure- In fact, yes, in this example $X$ itself is second countable, so I don't quite understand the role of the $X_n$ in the question $\endgroup$
    – Pietro Majer
    Commented Apr 26, 2018 at 21:20
  • 1
    $\begingroup$ I don't understand the votes to close... $\endgroup$
    – Yemon Choi
    Commented Apr 29, 2018 at 0:33

1 Answer 1

Reset to default
1
$\begingroup$

If $X=L^!([0,1])$ then $X=\cup_1^\infty nB$, where $B=\{x:\int_0^1 |x(t)| dt\le 1\}$. Let $\sigma$ be the vector topology on $X$ defined by the F-norm $\|x\|=\int_0^1 \frac{|x(t)|}{1+|x(t)|}dt$. If $\tau $ is the strongest vector topology on $X$ coinciding with $\sigma$ on $B$, then the condition 2 is fullfilled. A linear functional $f$ is $\tau$-continuous iff the restriction $f|B$ is sequentially $\sigma$-continuous. Then $f=0$ by A. Alexiewicz, On the two-norm convergence, Studia Math. 14 (1954), 49-56 (see p. 54).

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ This was observed in the comments. $\endgroup$
    – Tomasz Kania
    Commented Jun 23, 2018 at 10:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .