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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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