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.
 A: Suppose that $f:\mathbb{N}\to A$ is a bijection, where $f(n)=x$ for some $n\in\mathbb{N}$.
Define
$$
g(k)=\left\{\begin{array}{}
f(k)&\text{if }k\lt n\\
f(k+1)&\text{if }k\ge n\\
\end{array}
\right.
$$
Then $g:\mathbb{N}\to A-\{x\}$ is a bijection.
A: Your reasoning is informally correct.  If you want to give a careful answer to this question you should demonstrate a bijection from $A \setminus \{x\}$ to $\mathbb{N}$.
You have a given bijection between $A$ and $\mathbb{N}$.  Can you use this as a stepping stone to get from $A \setminus \{x\}$ to $\mathbb{N}$?
A: Let
$$f:A\to\Bbb N$$
be a bijection which sends $x\mapsto n$. Then the map
$$\begin{cases}f':\Bbb N\to\Bbb N\setminus\{n\} \\ k\mapsto k & k<n \\ k\mapsto k+1 & k\ge n\end{cases}$$
is a bijection.
Composing with the obvious restriction, $f\bigg |_{A\setminus\{x\}}$ gives us that
$$f'^{-1}\circ f\bigg|_{A\setminus\{x\}}:A\setminus\{x\}\to \Bbb N$$
is a bijection.
A: 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.
A: You need to construct a bijection $G$ from $\mathbb N$ onto $A-\{x\}$.
We know there is a bijection $F:\mathbb N \rightarrow A$
Let $G:\mathbb N\rightarrow A-\{x\}$ be defined by
$G(n) = 
\cases{
F(n+1)  & \text{if }  n\ge F^{-1}(x)\cr
F(n) & \text{if } n\lt F^{-1}(x)
}$
We need to show $G$ is one to one and onto $ A-\{x\}$
Suppose that $G(a) = G(b)$ for $a,b \in \mathbb N$.  If a and b are both greater or both less than $F^{-1}(x)$ then $a=b$ because F is one to one.  If $a$ is greater than $F^{-1}(x)$ but $b$ is less than $F^{-1}(x)$, then we have $f(a+1) = f(b)$ thus $a+1=b$ so that b is greater than a which is a clear contradiction.
Now we show that $G$ is onto $ A-\{x\}$
Suppose that $a\in A-{x}$
Then, either $F^{-1}(x)$<$F^{-1}(a)$ or $F^{-1}(x)\ge F^{-1}(a)$ 
In the former case, $G(F^{-1}(a))=a$ 
In the latter case, $G(F^{-1}(a)-1)=a$ 
This concludes the proof that G is a bijection.  Therefore $ A-\{x\}$ is denumerable.
