Show that open segment $(a,b)$, close segment $[a,b]$ have the same cardinality as $\mathbb{R}$

Solution 1:

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.

Solution 2:

Hint: $(a,b)\subseteq[a,b]\subseteq\Bbb R$ so you only need to show an injection from $\Bbb R$ into $(a,b)$. Then the Cantor-Bernstein theorem can work its magic.

Show that every two open intervals have the same cardinality (exhibit a bijection between them), and so it is enough to show that there is an injection into $(0,1)$. Consider something along the lines of: $$x\mapsto\frac1{2+e^x}$$

Solution 3:

This can be done by a straightforward application of the Cantor-Schroeder-Bernstein theorem. All you have to do is find an injective function $f:(a,b)\to \mathbb [a,b]$ and then an injective function $g:[a,b]\to (a,b)$. Now, there is one screamingly obvious candidate for a injective function $f:(a,b)\to [a,b]$. As for the other direction, can you think of a function that will shrink the interval $[a,b]$ just a little bit so it fits into $(a,b)$? Think of the case $[-1,1]$ and $(-1,1)$ for intuition if needed.