$\mathbb{Q}$ can not be embedded in $\mathbb{Z}$

Question

Show that $\mathbb{Q}$ can not be embedded in $\mathbb{Z}$ (where both has the subspace topology of $\mathbb{R}$)

My attempt at a solution

Since Z is discrete, {k} is open in $\mathbb{Z}$ with subspace topology. Since f is continious $f^{−1} ( \lbrace k \rbrace )$ is open in $\mathbb{Q}$. $\mathbb{Q}$ has no isolated points. So $f^{−1} ( \lbrace k \rbrace )$ not singleton and not finite . I must show that there is not a homeomorphism between $\mathbb{Q}$ with $\mathbb{f(Q)}$

Answer 1

Since Z is discrete (meaning that every subset is open), {k} is open in $\mathbb{Z}$. Since f is continious $f^{−1} ( \lbrace k \rbrace )$ is open in $\mathbb{Q}$.

$\mathbb{Q}$ has no isolated points (meaning that no singleton set is open; in fact, no finite set is open). So $f^{−1} ( \lbrace k \rbrace )$ not singleton and not finite .

Any homeomorphism from $\mathbb{Q}$ to $\mathbb{f(Q)}$ would have to be a bijection and so the preimage of the singleton $ \lbrace k \rbrace$ in $\mathbb{Z}$(which is open) would have to be a singleton in $\mathbb{Q}$. It is a contradiction.

Hence $\mathbb{Q}$ can not be embedded in $\mathbb{Z}$.

Answer 2

Hint: suppose $f\colon {\bf Q}\to {\bf Z}$ is continuous, and $k$ is in the range of $f$. What can you say about $f^{-1}[\{k\}]$?