Why is ($\mathbb R$,usual) not homeomorphic to ($\mathbb R$,discrete)? 
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*Why is ($\mathbb R$,usual) not homeomorphic to ($\mathbb R$,discrete)?
($\mathbb R$,discrete) means $d(x,y) =1$ for any $x\neq y$ and $d(x,y) =0$ for all $x=y$, both $x$ and $y$ are in $\mathbb R$.


*Is ($\mathbb R$,discrete) homeomorphic to ($\mathbb R$,usual)? Why?


*$M$ is homeomorphic to $N$ means there exists a mapping $f$, both one to one and continuous while onto not necessary, from $M$ to $N$, is that right?

Because everything is clopen in $\mathbb{R}$ with the discrete topology.
$(0,1)$ is not closed in the usual.
2. The answer is no. Homeomorphisms are open maps. 
3. You need a continuous bijection whose inverse is also continuous.
