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?

Question

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

Answer 1

Hints:

1. $f:(X,\tau)\to (Y, \tau')$ 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? )

2. 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?)

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

4. 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.