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Let $X$ be a Hausdorff vector space whose topology is induced by a countable family of seminorms $\{\rho_n\}_{n\in\mathbb{N}}$. Prove that the metric $$d(x,y)= \sum_{n=1}^{\infty}2^{-n}\frac{\rho_n(x-y)}{1+\rho_n(x-y)}.$$ induces the same topology.

Call $\mathcal{T}_1$ the topology generated by the seminorms. Then $U\in \mathcal{T}_1$ iff $U$ can be written as a union of sets of the form $$\bigcap_{j\in J} B_{r_j}^{j}(x_j),$$ where the $x_j$'s are in $U$, $J$ is finite, and $B_{r_j}^{j}(x_j):=\{v\in X; \rho_j(v-x_j)<r_j\}.$

Call $\mathcal{T}_2$ the topology generated by the metric. So $U\in\mathcal{T}_2$ iff it can be written as a union of $\epsilon$-balls.

I also know that $X$ is Hausdorff, so if $x,y\in X$ with $x\neq y$, then there exist disjoint $U_1,U_2\in\mathcal{T}_1$ such that $x\in U_1$ and $y\in U_2$.

I would like to show that if $U\in\mathcal{T}_1$, then $U\in\mathcal{T}_2$ (and vice-versa). Any tips on how to get started?

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    $\begingroup$ Both topologies are translation-invariant, so you just need to check that both have the same neighbourhoods of $0$. $\endgroup$ Oct 11 '17 at 16:09
  • $\begingroup$ Would you mind to edit the title to replace "norm" by "metric"? $\endgroup$
    – Hanno
    Oct 11 '17 at 16:12
  • $\begingroup$ Done - thanks @Hanno $\endgroup$ Oct 11 '17 at 16:16
  • $\begingroup$ Note that the Hausdorff condition is equivalent to the following: for each $x\in X$ non-zero, there is some $n\in\mathbb N$ such that $\rho_n(x)\neq0$. $\endgroup$
    – Aweygan
    Oct 11 '17 at 16:36
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One way to do this is to check that each neighborhood $U_1$ of $0$ in $\mathcal T_1$ contains some open neighborhood $U_2$ of $0$ in $\mathcal T_2$, and vice versa. This is equivalent to checking that the identity map $i:(X,\mathcal T_1)\to(X,\mathcal T_2)$ and its inverse $i^{-1}:(X,\mathcal T_2)\to(X,\mathcal T_1)$ are continuous at $0$. Since the topologies $\mathcal T_1$ and $\mathcal T_2$ are translation invariant, and $i$ is linear, the result follows.

Now each topology has a nice basis, so you can restrict your attention to these basic open neighborhoods of $0$.

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  • $\begingroup$ The first part is clear, but how does continuity follow from linearity and translation invariance? $\endgroup$ Oct 11 '17 at 17:51
  • $\begingroup$ Fix $x\in X$. Take a $\mathcal T_2$-open neighborhood $U_2$ of $x$, subtract $x$ to obtain $\mathcal T_2$-open neighborhood of $0$. Then there is a $\mathcal T_1$-open neighborhood $U_1$ of $0$ with $U_1$ with $U_1\subset (U_2-x)$, and thus $U_1+x\subset U_2$. Hence $i$ is continuous at $x$, and proving $i^{-1}$ is continuous at $x$ is obtained by switching subscripts. $\endgroup$
    – Aweygan
    Oct 11 '17 at 18:02
  • $\begingroup$ Yes, that's clear - sorry, I thought you meant that continuity at $0$ follows from linearity and translation invariance $\endgroup$ Oct 11 '17 at 22:41
  • $\begingroup$ No problem. If you need more help, let me know and I will edit. $\endgroup$
    – Aweygan
    Oct 12 '17 at 0:25
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    $\begingroup$ It's simpler than that: Given $\varepsilon>0$, there exist $K\in N$, $\delta>0$ such that $\bigcap\limits_{1\leq k\leq K}\{v:p_k(v)<\delta\}\subset B_\varepsilon(0)$. $\endgroup$
    – Aweygan
    Oct 12 '17 at 11:42
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(following on from Aweygan's answer)

Proof that $i$ is continuous:

Pick $M$ such that $2^{-M}\leq \frac{\epsilon}{2}.$ Call $K$ the smallest non-negative integer such that $\rho_K(x)$ isn't equal to $0$ for all $x\in B_{\epsilon}(0)$. Such a $K$ certainly exists, from the Hausdorff condition.

\begin{align*}d(v,0)=&\sum_{n\geq 1}2^{-n}\frac{\rho_n(v)}{1+\rho_n(v)}\\ =&\sum_{n= 1}^M2^{-K}\frac{\rho_K(v)}{1+\rho_K(v)}+\sum_{n\geq M+1}2^{-n}\frac{\rho_n(v)}{1+\rho_n(v)}\\ \leq & \sum_{n= 1}^M2^{-K}\frac{\rho_K(v)}{1+\rho_K(v)}+\sum_{n\geq M+1}2^{-n}.\end{align*} If $K>M$, then the first summation equals $0$ and the second one equals $2^{-M}\leq\frac{\epsilon}{2}$, so $d(v,0)<\epsilon$ and we are done.

If $K\leq M$, then call $K^*$ the $K$ such that, for a given $v$, $\rho_K(v)$ is maximised (this maximum is attained as there is only a finite number of $K\leq M$). Since $\frac{1}{1+x}$ is a monotone increasing function for $x>0$, we have that $$\sum_{n= 1}^M2^{-K}\frac{\rho_K(v)}{1+\rho_K(v)}\leq \frac{\rho_{K^*}(v)}{1+\rho_{K^*}(v)}\sum_{n= 1}^M2^{-K}\leq \frac{\rho_{K^*}(v)}{1+\rho_{K^*}(v)}.$$ If we choose $\delta$ such that $\frac{\delta}{1+\delta}<\frac{\epsilon}{2}$, then substituting above we see that $$d(v,0)\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon.$$

Proof that $i^{-1}$ is continuous:

We need to show that, given $K$ and $\epsilon$, there exists a $\delta$ such that $$B_{\delta}(0)\subset \bigcap_{k\leq K} \{v:\rho_k(v)<\epsilon\}.$$

For a given $v$, call $k^*$ the $k\leq K$ such that $\rho_k(v)$ is maximised. Then $$2^{-k^*}\frac{\rho_{k^*}(v)}{1+\rho_{k^*}}\leq d(v,0).$$ Choosing $\delta$ such that $$\frac{2^{k^*}\delta}{1-2^{k^*}\delta}<\epsilon$$ guarantees that $\rho_k(v)<\epsilon$ for all $k\leq K$.

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    $\begingroup$ Everything looks alright. $\endgroup$
    – Aweygan
    Oct 14 '17 at 7:02
  • $\begingroup$ Thanks, really appreciate you looking at it! $\endgroup$ Oct 14 '17 at 7:50

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