# About the locally convex topology

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.

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

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