# Countable Set, Surjective, Bijective

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

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

• This can't be true. The assumption implies that $A$ is countable and what you want to prove would imply that it is not. – Arnaud Mortier Feb 20 '18 at 10:30
• Take $A=\{0\}$ if you are not convinced. – Arnaud Mortier Feb 20 '18 at 10:32
• Existence of $f$ says $|A| \leq \aleph_0$. Existence of $g$ says $|A| = \mathfrak{c}$. But $\mathfrak{c}>\aleph_0$, contradiction. – Ivo Terek Feb 20 '18 at 10:32
• Mortier.Thanks, The only thing I try is to weed out the multiples to construct a injective fct. – Peter Szilas Feb 20 '18 at 10:34
• Sorry. A typo, N!! – Peter Szilas Feb 20 '18 at 10:35

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.
• @PeterSzilas But it is still hard to rescue this question. If you specify that $A$ is a countably infinite set, then you are specifying that there is a bijection between $A$ and $\mathbb{N}$ and there is nothing to prove. If you specify that there is an injection $g$ from $N$ to $A$ and a surjection $f$ from $N$ to $A$, then you are effectively asking for a special case of the Schroder Bernstein theorem. – almagest Feb 20 '18 at 17:26