How to show if A is denumerable and $x\in A$ then $A-\{x\}$ is denumerable My thoughts:
If $A$ is denumerable then it has a bijection with $\mathbb{N}$
So therefore $A\rightarrow \mathbb{N}$. Then x is a single object in A and A is infinite.
So if a single object is removed from then $A$ is still infinite.

There exists a bijection between $A$ and $\mathbb N$. Let's denote it by $\psi:A\to\mathbb N$. Let $\widehat{\psi}$ denote the restriction of $\psi$ on $A\setminus\{x\}$. Then, $\widehat{\psi}$ is a bijection between $A\setminus\{x\}$ and $\mathbb N\setminus\{\psi(x)\}$. Here, $\mathbb N\setminus\{\psi(x)\}$ is the set of natural numbers except one particular element $\psi(x)$. We can establish a bijection $F:\mathbb N\to\mathbb N\setminus\{\psi(x)\}$ as follows:
\begin{align*}
F(n)\equiv\begin{cases}n&\text{if $n\in\mathbb N$ and $n<\psi(x)$,}\\n+1&\text{if $n\in\mathbb N$ and $n\geq\psi(x)$.}
\end{cases}
\end{align*}
It follows that $\mathbb N$ and $\mathbb N\setminus\{\psi(x)\}$ have the same cardinality and, since $\mathbb N\setminus\{\psi(x)\}$ and $A\setminus\{x\}$ also have the same cardinality, the claim follows.
