Help proving that semi infinite intervals are equipotent to the set of all real numbers Let $a∈\mathbb{R}$ (the set of all real numbers).
Define $[a, \infty) = \{ x \in \mathbb{R} \mid a \leq x \}$.
I need help proving that $[a,\infty)$ is equipotent to $\mathbb{R}$.
I understand that in order to do so, I can use the Schröder-Bernstein Theorem by stating that $\lvert [a,\infty) \rvert \leq \lvert \mathbb{R} \rvert$ and that $\lvert [a,\infty )\lvert \geq \lvert \mathbb{R}\rvert$. But I can't think of how to prove that there are injections from $[a, \infty)$ to $\mathbb{R}$ and from $\mathbb{R}$ to $[a,\infty)$.
Thank you for any help.
 A: An injection from $[a,\infty)$ to $\mathbb{R}$ is presumably not an issue.
For an injection from $\mathbb{R}$ to $[a,\infty)$, let's look at the simpler problem of a bijection from $\mathbb{R}$ to $(a,\infty)$. The function $a+e^x$  does the job. 
Or else you can proceed directly to find a bijection between $[a,\infty)$ and $\mathbb{R}$. This is trickier than using Schroeder-Bernstein. 
A: An explicit bijection from $A=[a,\infty)$ to $\mathbb{R}$ may be based on the partition of $A$ into half-open intervals $$[a,a+1),\ [a+1,a+2),\ [a+2,a+3),\ [a+3,a+4),...$$
Send these respectively to the following half-open intervals in $\mathbb{R}$:
$$[0,1),\ [-1,0),\ [1,2),\ [-2,-1),...$$
Since the list of half open intervals of $\mathbb{R}$ is a partition of $\mathbb{R}$, the map from $[a,\infty)$ to $\mathbb{R}$ so defined is a bijection.
We can describe the map as a function $f:[a,\infty) \to \mathbb{R}$ by noting that each $x \in [a,\infty)$ may be uniquely written in one of the two forms $a+2k+t$ or $a+2k+1+t$ where $k$ is a nonnegative integer $k=0,1,2,...$ and $0 \le t < 1$. [This is just a formal version of the partition $[a,a+1),[a+1,a+2),...$] Then $f$ is defined by $$f(a+2k+t)=k+t$$ and $$f(a+2k+1+t)=-k-1+t.$$
A: A previous poster mentioned using $f(x) = a+e^x$ to get a bijection with $(a,\infty)$. Then compose $f$ with the bijection $g:(a,\infty) \to [a,\infty)$ where 
$$g(x) = \begin{cases} x, & x \notin \mathbb{N}, \\ x-1, & x \in \mathbb{N}, \end{cases}$$
to finish the job.
