Countable Set, Surjective, Bijective

Question

Let $A \subset \mathbb{R}.$

Assume :

$(\star )$ Let A be an infinite set.

Let $f$ be a surjective function $f: \mathbb{N} \rightarrow A.$

The problem:

Show: There exists a bijective function $g: \mathbb{N} \rightarrow A.$

Latest attempt:

Let $A=${$a_1,a_2,a_3, .......$} , $n \in \mathbb{N}.$

Elements of $A$ may appear more than once, I.e. for some $m \not= n$ we have $a_n=a_m.$

Hence there is a subset of $T$ of $N$ such that

$T \sim A,$ i.e.

there is a bijection $g:T \rightarrow A.$

$T$ being a subset of $N$ implies $T$ is countable, i.e there is a bijection $h$:

$h: N \rightarrow T.$

The composition

$F: N\rightarrow A$ defined by

$F= g\circ h$ is a bijection , and we are done.

Note: Used elements of Rudin, Principles , 3rd Edition,Theorem 2.12

Cf. On countable sets

Note: $(\star)$ was not assumed (forgot) in the original version of the problem, hence the answer by Almagast.

Answer 1

The result is false. Take $A=\{1\}$. There is a surjective function $f:\mathbb{N}\to A$ namely $f(n)=1$ for all $n$, but no bijection.