I know that if a locally convex space Hausdorff $(X,S)$ is first numerable then for the $\hat{0}\in X$ exists a countable local base $\{V_n, n \in \mathbb{N}\}$ and to each $V_n$ corresponds a seminorm $p_n \in S$. Then: \begin{equation} d(x,y) = \sum_{n = 1}^{\infty}\frac{p_n(x-y)}{2^n(1+p_n(x-y))}<1 \end{equation} Is a metric in X, and the topology induced by $d$ is the same that the locally convex topology. I've been trying to prove that $\tau_E \subset \tau_d$ but I do not achieve it. Where $\tau_E$ is the locally convex topology. Actualization: The containment $\tau_E \subset \tau_d$ is not the more difficult, here the proof. Let be $A \subset X$ an $\tau_E-open$ then for each $x \in A$ exists $j \in \mathbb{N}, q \in S:B_j + x \subset V_j + x\subset B_q + x \subset A$ . Let be $\epsilon = 2^{-(j + 1)}$, then: \begin{equation} \frac{p_j(x-y)}{2^{j}(1 + p_j(x - y))} \leq d(x,y) < \epsilon \Rightarrow p_j(x -y) < 1 \end{equation} So $B_{\epsilon}^d(x) \subset B_j(x) \subset A$, i.e. $A$ is $\tau_d-open$. The problem $\mathbf{\tau_d \subset \tau_E}$ remains unsolved for me.

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    $\begingroup$ Sorry, but what is the question? $\endgroup$
    – s.harp
    Mar 22, 2019 at 23:00
  • $\begingroup$ How to prove that $\tau_E \subset \tau_d$ $\endgroup$ Mar 22, 2019 at 23:04
  • $\begingroup$ What is $\tau_E$? $\endgroup$
    – K.Power
    Mar 22, 2019 at 23:27
  • $\begingroup$ The locally convex topology $\endgroup$ Mar 22, 2019 at 23:59

1 Answer 1


To show $\tau_E\subset\tau_d$, it suffices to show that if $x_n\to x$ in the $\tau_d$ topology, then it converges in the $\tau_E$ topology, for then the identity map $(X,\tau_d)\to(X,\tau_E)$ is continuous).

To this end, fix $m\in\mathbb N$. It suffices to show that there is some $N\in\mathbb N$ such that $x_n-x\in V_m$ whenever $n\geq N$, i.e., that $p_m(x_n-x)<1$ for $n\geq N$. Fix $\varepsilon\in(0,2^{-m-1})$. Then there is some $N\in\mathbb N$ such that $$\frac{p_m(x_n-x)}{2^m(1+p_m(x_n-x))}\leq d(x_n,x)<\varepsilon$$ for $n\geq N$. Rearranging, we obtain $$p_m(x_n-x)<\frac{\varepsilon 2^m}{1-\varepsilon 2^m}<1,$$ and the result follows.

  • $\begingroup$ Thanks, you have any suggestion for the containment $\tau_d \subset \tau_E?$ $\endgroup$ Mar 23, 2019 at 5:22
  • $\begingroup$ Given $\varepsilon>0$, obtain $M$ such that $\sum_{k=M+1}^\infty 2^{-k}<\varepsilon$, and you can control $p_k$ for $1\leq k\leq M$. $\endgroup$
    – Aweygan
    Mar 23, 2019 at 17:12

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