I'm trying to understand the part of the Wiki article on Frechet Spaces which says that you can construct a Frechet Space by starting with a vector space $X$ and a countable family of seminorms $\|\cdot\|_k$, and imposing a couple of properties, the first of which says that

if $x\in X$ and $\|x\|_{k}=0$ for all $k\geq 0$, then $x=0$.

The article goes on to say that this implies that the topological vector space is Hausdorff.

The topology is induced from the family of semi-norms as follows: for any subset $U\subset X$, $U$ is open if and only if for any $u\in U$, there exist $K\geq0$ and $\epsilon>0$ such that $\{v; \|v-u\|_k<\epsilon\ \forall k\leq K\}$ is a subset of $U$.

In particular, given $x,y\in X$, an open set $O_1$ containing $x$ is open if and only if there exist $K_1\geq 0,\epsilon_1>0$ such that $\{v; \|v-x\|_k<\epsilon_1\ \forall k\leq K_1\}\subset O_1$, and an open set $O_2$ containing $y$ is open if and only if there exist $K_2\geq 0,\epsilon_2>0$ such that $\{v; \|v-y\|_k<\epsilon_2\ \forall k\leq K_2\}\subset O_2$.

How can I use the property state above to prove that for any $x,y$, there exist such sets $O_1$ and $O_2$ which are disjoint?

  • $\begingroup$ It's easier to just use that all open balls w.r.t. all semi-norms form a subbase for the topology. $\endgroup$ – Henno Brandsma Oct 8 '17 at 14:00

If $x \neq y$ then $x-y \neq 0$ so that there is some $k$ with $r=\|x_k -y_k\|_k >0$ Now note that $B_k(y)=\{z:\|z-y\|_k < {r \over 2}\}$ and $B_k(x)$ are subbasic open and disjoint.

  • $\begingroup$ Why does it matter that they're subbasic? For the space to be Hausdorff, don't we only require the exitence of disjoint open sets? $\endgroup$ – man_in_green_shirt Oct 9 '17 at 8:41
  • 1
    $\begingroup$ @man_in_green_shirt “subbasic” just implies open (because they are in “my” subbase). It’s an explicit reason why the sets are open. $\endgroup$ – Henno Brandsma Oct 9 '17 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.