Show that the cardinality of ($X$ ∪ {$x$}) is equal to the cardianlity of ($X$)+$1$

Question

Let $X$ be a finite set, and let $x$ be an object which is not an element of $X$. Then $X$ ∪ {$x$} is finite and #($X$ ∪ {$x$}) = #($X$)+$1$. Note that #-here means cardinality.

Suppose the cardinality of $X = n$ then there is a bijection $f:$ {$i: 1\le i\le n$} $\to X$ where $f(i)=x_i$ for $x_i \in X.$

There is also a bijection from $f:$ {$i: 1\le i\le 1$} $\to {x}$ where $f(1)=x$ for $x \in {x}.$

Then we can naturally construct a bijection from $f:$ {$i: 1\le i\le n+1$} $\to$ ($X$ ∪ {$x$}) where $f(i)=x_i$ for $2 \le i\le n+1$ and $f(1) = x$.

Answer 1

Just make $g(n)=\begin{cases} f(n) & ,n\le\#X \\ x& ,n>\#X \end{cases}$