elementary set theory - Prove that if set $A$ is denumerable and $xnotin A$ then $Acup {x}$ is denumerable.

Prove that if set $A$ is denumerable and $x\notin A$ then $A\cup\{x\}$ is denumerable.

See if my solution works:

Since $A$ is denumerable, there exists $f: \mathbb{N} \to A$ such that $f$ is a bijection.

Define $g:\mathbb{N} \to A$, $$g(n)=\begin{cases}x,&\text{if $n=1$,}\\f(n),&\text{ if $n\neq 1$}\end{cases}$$

Claim: $g$ is also bijective.

Let $n_1, n_2\in \mathbb{N}$ such that $g(n_1)=g(n_2)$.

If $g(n_1)=x$, then $n=1$. If $g(n_1)\neq g(n_2)$, then $g(n_1)=f(n_1)=g(n_2)=f(n_2)$. We have $f(n_1)=f(n_2)$. Since $f$ is a bijection, then $n_1=n_2$. Therefore $g$ is injective.

Let $b\in A\cup\{x\}$. There exists $n=1$ such that $g(n)=x=b$. If $n\neq 1$, then there exists $m\in\mathbb{N}$ such that $g(m)=f(m)=b$. Since $f$ is bijective, there exists $m\in\mathbb{N}$ such that $f(m)=b=g(m)$. So $g$ is surjective.

Therefore $g$ is bijective.

Therefore $A\cup \{x\}$ is denumerable.

Answer

If $A$ is denumerable then it has the form

$A=\{a_1,a_2,...,a_n,... \}$,

so $A\cup\{x\}$ can be written as

$A=\{x,a_1,a_2,...,a_n,... \}$,

then, if we write $b_1=x$, $b_2=a_1$, ... , $b_n=a_{n-1}$, ... ,

we have

$A=\{b_1,b_2,...,b_n,...\}$, which is a denumerable set