Bijection between continuum and continuum plus one point I know that there exists a bijection between $[0,1]$ and $[0,1]\cup\{2\}$, but late last night I was not able to come up with a trivial solution. Will be glad if one can provide such example.
 A: Define $f:[0, 1]\to[0,1]\cup\{2\}$ as
$$f(x) = \begin{cases}
2 & x = 1\\
\dfrac{1}{n-1} & x\text{ is of the form }\dfrac{1}{n};\;n\ge 2\\
x & \text{otherwise}
\end{cases}$$
The idea basically is to take the countable subset $\left\{\dfrac{1}{n} : n \ge 1\right\}$ of $[0, 1]$ and put that in a bijection with $\left\{\dfrac{1}{n} : n \ge 1\right\} \cup \{2\}$ and keep everything else fixed.
A: Well, I'm not sure there is a simple formula for this.
The idea is to find an infinite countable subset $C$ of $[0,1]$, add the point $2$ to it and define a bijection between $C$ and $C \cup \lbrace 2 \rbrace$ using "Hilbert's hotel first trick". You then define the rest of the bijection as the identity on $[0,1] \setminus C$.
For instance, you could take $C = \lbrace \frac{1}{n} \,| \, n \geq 1 \rbrace$. You can choose the bijection between $C$ and $C \cup \lbrace 2 \rbrace$ as the one that sends $2$ to $1$, $1$ to $\frac{1}{2}$, $\frac{1}{2}$ to $\frac{1}{3}$, etc.
If you want to be pedantic, you can say that this is due to the equality $1+\omega = \omega$, i.e. "infinity of the natural numbers plus one point at the beginning is infinity of the natural numbers, because of this shifting trick".
