While proving the Banach-Alaoglu theorem, one needs to prove the topology induced by a countable family of seminorms $\rho_n$ on a vector space $X$ is metrizable if $X$ is Hausdorff with that topology. This is done by explicitly constructing a metric that gives the topology, precisely:


I can't see how I would prove this is indeed a metric. I see it is 0 iff all terms are zero thus iff all $\rho_n(x-y)=0$ thus iff $x=y$ since that is equivalent to 0 $y$ being in every neighborhood of $x$. I can see it is symmetric by symmetry of the seminorms. But what about the triangle inequalities? I tried some inequalities, but all I managed was:


but that RHS is clearly less than $d(x,y)$. How do I prove this?

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    $\begingroup$ Consider the map $t \mapsto \frac{t}{1+t}$. Prove that it is strictly increasing (on $[0,+\infty)$). Use that fact. $\endgroup$ – Daniel Fischer Oct 3 '15 at 15:53
  • $\begingroup$ Btw here is what got me stuck on this. $\endgroup$ – MickG Oct 3 '15 at 15:54
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    $\begingroup$ First, more generally, prove that $d(x,y)/(1+d(x,y))$ is a metric if $d(x,y)$ is. $\endgroup$ – Thomas Andrews Oct 3 '15 at 15:58
  • $\begingroup$ $f(t)=\frac{t}{1+t}$, then $f'(t)=\frac{1}{1+t}-\frac{t}{(1+t)^2}=\frac{1}{(1+t)^2}\geq0$, strictly greater actually, so $f$ is strictly increasing on $(-1,+\infty)$. And if I step back once in my inequality chain, I find a place to easily apply this. $\endgroup$ – MickG Oct 3 '15 at 15:58
  • $\begingroup$ @DanielFischer Don't you need some convexity condition on that map? Just being strictly increasing isn't enough, or $t\to t^2$ would work. $\endgroup$ – Thomas Andrews Oct 3 '15 at 15:59

My inequality chain was:

\begin{align*} d(x,z)+d(z,y)={}&\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x-z)}{1+\rho_n(x-z)}+\sum_{n=1}^\infty2^{-n}\frac{\rho_n(z-y)}{1+\rho_n(z-y)}={} \\ {}={}&\sum_{n=1}^\infty2^{-n}\left(\frac{\rho_n(x-z)}{1+\rho_n(x-z)}+\frac{\rho_n(z-y)}{\rho_n(z-y)+1}\right)={} \\ {}={}&\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x-z)+2\rho_n(x-z)\rho_n(z-y)+\rho_n(z-y)}{1+\rho_n(x-z)+\rho_n(z-y)+\rho_n(x-z)\rho_n(z-y)}\geq{} \\ {}\geq{}&\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x-y)}{1+\rho_n(x-z)+\rho_n(z-y)+\rho_n(x-z)\rho_n(z-y)}, \end{align*}

where the first = is by definition, the second one is because the series converge (they are, in general, less than or equal to the geometric series of ratio $\frac12$ which converges) so I can permute the terms, the third one simply sums the fractions, and the inequality throws away some positive terms from the numerator. If I only remove the two, I have a sum of $2^{-n}\frac{\alpha_n}{1+\alpha_n}$, where:


by the triangle inequality on the first two terms and throwing away the third which is positive, so those fractions are, since as Daniel pointed out $\frac{t}{1+t}$ is strictly increases, strictly greater than $\frac{\rho_n(x-y)}{1+\rho_n(x-y)}$, which completes the proof.

Thomas's comment notes something that seems not too useful here, but is anyway interesting: if $d$ is a metric, $d'(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is too. That is because the triangle inequality is proved as below, and the other properties of a metric are obvious.


which as before is at least $\frac{d(x,y)}{1+d(x,y)}=d'(x,y)$, since $\frac{t}{1+t}$ is strictly increasing.

And $\frac{t}{1+t}$ is strictly increasing on all of $(-1,+\infty)$ because its derivative is:


for all $t>-1$.

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    $\begingroup$ I'd say it's easier in the other direction, $$\begin{aligned}\frac{\rho(x-z)}{1+\rho(x-z)} &\leqslant \frac{\rho(x-y) + \rho(y-z)}{1 + \bigl(\rho(x-y) + \rho(y-z)\bigr)}\\ &= \frac{\rho(x-y)}{1+\rho(x-y)+\rho(y- z)} + \frac{\rho(y-z)}{1+\rho(x-y)+\rho(y-z)}\\ &\leqslant \frac{\rho(x-y)}{1+\rho(x-y)} + \frac{\rho(y-z)}{1+\rho(y-z)}\end{aligned}$$ $\endgroup$ – Daniel Fischer Oct 3 '15 at 16:19
  • $\begingroup$ @DanielFischer Seeing it, I agree :). It's just that not seeing this monotonicity I had tried that direction first mentally and failed, so I tried the other direction to see if I got somewhere and had the attempt on my notebook in this direction, so I was mentally oriented towards that direction :). $\endgroup$ – MickG Oct 3 '15 at 16:27

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