If $d_1\sim d_2$ and $I:(E,d_1)\to(E,d_2) $, prove that $I$ is continuous.

Question

Edit: If $d_1\sim d_2$ and $$I:(E,d_1)\to(E,d_2) $$ $$x\mapsto I(x)=x. $$ I want to prove that $I$ is continuous.

Here is my trial: Since $d_1\sim d_2$, then $\exists \alpha, \beta>0$ such that $\alpha d_2\leq d_1 \leq\beta d_2$.

Let $\{x_n\}\subset E$ be Cauchy $\implies x_n\to x.$ So, $\forall \epsilon>0,\; \exists\;n_0$ such that $\forall\;n\geq n_0,$ $$d_1(x_n,x)<\epsilon.$$

So, $$\alpha d_2(I(x_n),I(x))= \alpha d_2(x_n,x)\leq d_1(x_n,x)< \epsilon,$$ which implies $$ d_2(I(x_n),I(x))< \frac{\epsilon}{\alpha}=\epsilon'.$$

Also,

$$d_1(x_n,x)\leq \beta d_2(x_n,x)= \beta d_2(I(x_n),I(x))< \frac{\beta}{\alpha}\epsilon\leq\epsilon.$$

Please, is this right? If no, where have I gone wrong?

Answer 1

You are were only missing to say that $d_1(x_n,x)<\epsilon$, for $n>n_0$.

A small detail that may not need to be mentioned but that needs to at least be kept in mind is that we are proving continuity of a function using sequences, because the topology comes from a metric. Depending on the definition of "continuous" used this can either be the definition of continuous or a theorem.

For other topologies checking the limits along sequences is not enough. However, for topologies arising from a metric, it is.