Show that if a topological space is metrizable then it is so in an infinite number of ways.
Show that a topological space $X$ is metrizable $\iff$ there exists a homeomorphism of $X$ onto a subspace of some metric space $Y$.
My effort:
1.Let $(X,\tau)$ be metrizable. Then there is a metric say $d$ on $X$ such that open sets in $(X,d)$ are same as that in $(X,\tau)$.
We know that if $d$ is a metric on $X$ then so is $kd$ for any $k(\in \Bbb R)>1$.Also $(X,d)$ and $(X,kd) $ generate the same collection of open sets. whose proof runs as follows:
Claim: $U$ is open in $(X,d)\iff U$ is open in $(X,kd)$
Proof:Let $U$ be open in $(X,d)$.Let $x\in U$ then for some $r>0;B_d(x,r)\subset U$ .Now choose $r_0=rk$.Then I want to show that $B_{kd}(x,r_0)\subseteq U$.
let $z\in B_{kd}(x,r_0)\implies kd(z,x)<r_0\implies d(z,x)<\dfrac{r_0}{k}=r\implies z\in B(x,r)\subset U\implies B_{kd}(x,r_0)\subset U$
Conversely let $U$ is open in $(X,kd)$..Proceeding in the same way we find that $U$ is open in $(X,d)$
Thus $(X,kd)$ and $(X,d)$ are homeomorphic.QED.
Thus a topological space is metrizable in an infinite number of ways.
Is the solution correct?
2.For the second problem,I am unable to find the mapping which will give the homeomorphism.Please give some hints.