X be a countably infinite discrete topological space homeomorphic to the subspace Y of R with usual topology. which of the following can be Y? 
Let $X$ be a countably infinite discrete topological space which is
homeomorphic to the subspace $Y$ of $\mathbb R$ with usual
topology.Then which of the following can be $Y$?
$(i)\mathbb Q$
$(ii)\mathbb Z$
$(iii)\{\frac{1}{n}:n\in \mathbb N\}\cup\{0\}$
$(iv)\mathbb N$

$Y$ being a subspace of $\mathbb R$ is a hausdorff space.
All$(i),(ii),(iv)$ are hausdorff with the usual topology but $(iii)$ is not hausdorff as it is not possible to separate $\{0\}$ from $\{\frac{1}{n}:n\in \mathbb N\}$.But ,
$X$ is a countably infinite discrete topological space ,so it will also be a hausdorff space.
Hence it will not be possible to establish the homoemorphism between $X$ and $\{\frac{1}{n}:n\in \mathbb N\}$.
So,the only possibilities $Y$ to be $\mathbb Q$,$\mathbb Z$,$\mathbb N$
Please verify my arguments
 A: Hints:

*

*$f:(X,\tau)\to (Y, \tau')$ be a homeomorphism and $(X, \tau) $ is a discrete space.Then $(Y, \tau') $ is also discrete.

( What you know about the open sets of two homeomorphic spaces? )


*Any map define on a discrete space is continuous.

(What happen if a we take a bijective map and domain, codomain both are discrete space?)

Any two discrete space where the underlying sets have the same cardinality are homeomorphic.(Can you find a homeomorphism?)



*Suppose $(Y, \tau')$ is a topological subspace of $ (X, \tau)$.Then $A\in\tau'$ iff $\exists U\in\tau$ such that $A=Y\cap U$ .


*Prove $\Bbb{N}, \Bbb{Z}$ are discrete space as euclidean subspace.
It's enough to show that every singleton sets are open in the subspace topology. (Why?)
For an example $\{5\}=\Bbb{N}\cap(4, 6)$
$\{\frac{1}{n}:n\in\Bbb{N}\}\cup\{0\}, \Bbb{Q}$ are not discrete.
Is $\{0\}$ open in the euclidean subspace $\{\frac{1}{n}:n\in\Bbb{N}\}\cup\{0\}$ ? Do you know Archimedean property of real?
For $\Bbb{Q}$ use the density of rationals among reals.
