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Here's some context before my question.

Let $\mathbb{V}$ be a topological vector space, which is Hausdorff and such that its topology is generated by some arbitrary family of seminorms $\{\rho_{\alpha}\}_{\alpha \in I}$; this means that $\mathbb{V}$ is locally convex. Now, if $I$ turns out to be countable (or if we can reduce the family $\{\rho_{\alpha}\}_{\alpha \in I}$ to a countable one, while keeping the same topology in $\mathbb{V}$), we can define a metric in $\mathbb{V}$ by $$d(u, v) = \sum_{i = 1}^{\infty} \frac{1}{2^i} \frac{\rho_i(u - v)}{1 + \rho_i(u - v)},$$ where $\{\rho_i\}_{i \in \mathbb{N}}$ is some enumeration of $\{\rho_{\alpha}\}_{\alpha \in I}$, so that its topology is metrizable. I've been told that the converse is also true, which leads to my question.

QUESTION: Let $\mathbb{V}$ be a topological vector space having a metrizable topology, generated by some metric $d$. How can I prove that $\mathbb{V}$ admits a countable family of seminorms generating its topology? Also, do I need to impose the condition that $d$ is translation invariant (since this happens in the above construction)?

Thanks.

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    $\begingroup$ There is a classical result which says that if $X$ is a locally convex space with a countable family of neighborhoods at $0$ then we can find a metric $d$ compatible with the topology on $X$ and invariant by translation. So in particular if we assume the space metrizable then we can find an invariant translation metric which gives the same topology to $X$. If we assume $X$ locally convex then all the balls for $d$ will be convex, so in this case we can take $\rho_n(x):=\sup\{\alpha>0, \alpha x\in B(0,n^{-1}\}$. $\endgroup$ Commented Feb 12, 2012 at 17:51
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    $\begingroup$ A reference for the first result is Rudin's book, Functional Analysis. $\endgroup$ Commented Feb 12, 2012 at 17:56

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We first show the following result:

Let $X$ locally convex topological vector space with a countable family of neighborhood of $0$. We can find a metric $d$ compatible with the topology of $X$ and invariant by translation.

Sketch of proof: We can take a basis a neighborhoods $\{V_n\}$ such that each $V_n$ is convex and for all $n$: $V_{n+1}+V_{n+1}+V_{n+1}+V_{n+1}\subset V_n$. Let $D$ the set of rational numbers of the form $r=\sum_{n=1}^{+\infty}c_n(r)2^{-n}$ where $c_i(r)\in \{0,1\}$ and $c_i=0$ except for a finite numbers of index $i$. We put $A(r)=X$ if $r\geq 1$ and for $r\in D$ we put $A(r):=\sum_{n=1}^{+\infty}c_n(r)V_n$, $f(x):=\inf\{r\in D: x\in A(r)\}$ for $x\in X$and $d(x,y)=f(x-y)$.

Now we put $\rho_n(x):=\sup\{\alpha>0, \alpha x\in B(0,n^{-1})\}$. Since the balls $B(0,n^{-1})$ are convex, $\rho_n$ is a seminorm. Since the topology given by these seminomrs is the same as the topology given by $d$, we are done.

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    $\begingroup$ It is probably worth pointing out that the local convexity assumption is essential (it is stated in the title but not in the question itself). The space $L^0[0,1]$ with the topology of convergence in measure is completely metrizable while not locally convex -- the only convex neighborhood of zero is the space itself. In particular it doesn't admit any continuous semi-norm or functional. $\endgroup$ Commented Feb 14, 2012 at 13:19
  • $\begingroup$ @t.b. Indeed, I forgot the local convexity assumption in my question. That's an interesting example, btw. $\endgroup$ Commented Feb 15, 2012 at 1:19
  • $\begingroup$ except for the zero functional, of course. Nice comment. $\endgroup$ Commented Jun 27, 2018 at 14:39

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