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?
 A: Unless you are using a convention that I am not familiar with, your definition of homeomorphism is not correct. A homeomorphism is a bijective continuous function with continuous inverse.
You can show that homeomorphisms form an equivalence relation. Thus $(\mathbb{R},$ usual$)$ being homeomorphic to $(\mathbb{R},$ discrete$)$ us equivalent to $(\mathbb{R},$ discrete$)$ being homeomorphic to $(\mathbb{R},$ usual$)$. In particular, your (1) and (2) are therefore the same question.
One reason why $\mathbb{R}$ with the discrete topology is not homeomorphic to $\mathbb{R}$ with the usual topology is that everything is disconnected in the discrete topology, while everything is connected in the typical topology. Connectedness is preserved by continuous maps, so there is no continuous map from typical onto discrete.
A: For 3)No a homeomorphism is continuous,bijective and the inverse is continuous.So the inverse of a homeomorphism is homeomorphism itself,which implies that the relation "is homeomorphic to" is an equivalence relation.For this reason 2) and 1) are equivalent.
2)In discrete topology every set with one element is open therefore every subset is open as union of open sets and for the same reason its complement is open.So all subsets of $\mathbb{R}$ are open and closed in discrete topology.
Of course this is not true in the usual topology sets with one element are not open.But if there was a homeomorphism from ($\mathbb{R}$,discrete) $\to \mathbb{R}$ then it should map the open set $\{x\} $ to an open set.But $f(\{ x\})=\{ f(x)\}$ which is not open,a contradiction.Thus there is no such homeomorphism.
A: 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.
