Show that open segment $(a,b)$, close segment $[a,b]$ have the same cardinality as $\mathbb{R}$ a) Show that any open segment $(a,b)$ with $a<b$ has the same cardinality as $\mathbb{R}$.
b) Show that any closed segment $[a,b]$ with $a<b$ has the same cardinality as $\mathbb{R}$.

Thoughts:
Since $a<b$, $a,b$ are two distinct real number on $\mathbb{R}$, we need to show it is 1 to 1 bijection functions which map between $(a,b)$ and $\mathbb{R}$, $[a,b]$ and $\mathbb{R}$.
But we know $\mathbb{R}$ is uncountable, so we show the same for $(a,b)$ and $[a,b]$?
and how can I make use of the Cantor-Schroder-Bernstein Theorem? The one with $|A|\le|B|$ and $|B|\le|A|$, then $|A|=|B|$?
thanks!!

Let's do a fundamentally different approach. Consider projecting from the open bottom half of the circle $x^2 + (y-1)^2 = 1$, as I drafted below.

In this, by projecting from $(0,1)$, we get a 1-1 correspondence from the angle  $(0, -\pi) \to \mathbb{R}$. There is a clear 1-1 correspondence from $(0,1)$ and $(0, -\pi)$.
Thus we have our bijection from $(0,1)$ to $\mathbb{R}$. For $[0,1]$, we could use big theorems, or we could just modify this one. So let's "make space" by saying that whatever was associated to $0$ is now associated to $2$, what was associated to $1$ is now associated to $3$, $2$ now goes to $4$, and so on, effectively freeing up the real numbers $0,1$ by displacing the natural numbers. Send $0$ to $0$, and $1$ to $1$, and now we have a bijection $[0,1] \to \mathbb{R}$.
Nice, direct, and constructive.
