Do both sets have the same cardinality? Well I was trying to find out whether the two sets $[n, n+1]$ and $[n, n+1]\cup \{n+2\}$ has the same cardinality. If you add another infinite set(not any random one) to $[n, n+1]$, for example $[n, n+1]\cup[n+1, n+2]$, the cardinality does not change, because we can create a bijection using the function $f(x)=2x-n$. Similarly you can create a bijection with one of $[n,n+1]$'s subset as well, for example,  using the function $\frac{x-n}{2}$ we can create a bijection with $[n, n+\frac{1}{2}]$. I gave these two particular examples to show that a bijection can be create with one of this set's superset as well as a subset. But what happens when we add exactly one element? Does $[n, n+1]$ and $[n, n+1]\cup \{n+2\}$ have the same cardinality? 
 A: It is enough to show that $A=[0,1]\cup\{2\}$ and $B=[0,1]$ have the same cardinality. 
Let $f(2)=1/2$, $f(1/2)=1/4$, $f(1/4)=1/8$, and so on. For any other $x\in A$, let $f(x)=x$. It is not hard to verify that $f$ is a bijection from $A$ to $B$.
Remark: You may want to look up Hilbert's Infinite Hotel, for example here.
A: Yes, there exists a bijection between $[n, n+1]$ and $[n, n+1] \cup \{n+2\}$, but we need to use a small trick to do this.
Define the sequence $a_k=n+\frac{1}{2^k}$ for $k \geq 1$. Now, for this bijection, we want to move $a_2$ to $a_1$, $a_3$ to $a_2$, $a_4$ to $a_3$, etc.. Why do we do this? Because now, we still have covered the whole $a_k$ sequence since it's infinite, so there will always be a term behind to jump to the next term, but now we have created a space for $a_1$ because we have not mapped $a_1$ to anything. Now, we map $a_1$ to $n+2$.
To recap, our bijection is:
$$f(x)=\begin{cases}n+2 \text{ if } x=a_1 \\ a_{k+1} \text{ if } x=a_k \text{ for } k > 1 \\ x \text{ otherwise}\end{cases}$$
