Originally Posted by

**JaneBennet** There is no need to use contradiction for Part (1). It’s totally straightforward.

Let $\displaystyle x\in M$. You want to show that f is continuous at *x*. Given $\displaystyle \epsilon>0$, simply set $\displaystyle \delta=\epsilon$ and you have that $\displaystyle \forall\,y\in M$, $\displaystyle \mathrm{d}_M(x,y)<\delta\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x),\mathrm{f}(y))=\mathrm{ d}_M(x,y)<\delta=\epsilon$.

Part (2):

f is a function from *M* **onto** *N*; in other words, it is surjective.

Let $\displaystyle x_1,x_2\in M$. Then

$\displaystyle \mathrm{f}(x_1)=\mathrm{f}(x_2)\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x_1),\mathrm{f}(x_2))=0$

$\displaystyle \color{white}.\hspace{22mm}.$ $\displaystyle \Rightarrow\ \mathrm{d}_M(x_1,x_2)=0$

$\displaystyle \color{white}.\hspace{22mm}.$ $\displaystyle \Rightarrow\ x_1=x_2$

So f is injective as well. Hence it is a homeomorphism.