There does exist this question here below, but my question is a bit different:
Prove that $|\mathbb{R}| = |(0, 1)|$.
Here is my proof below from the given hint:
Consider $f: (0, 1) \to \mathbb R$ as $f(x) = \tan\left(\left(\pi(x - \frac 12\right)\right)$.
Note, $f$ is continuous on $(0, 1).$ Let $x \in (0, 1).$ Then $\displaystyle{\lim_{x \to 1^-}f(x) = \infty, \ \lim_{x \to 0^+}f(x) = -\infty}$ and so $f(x) \in (-\infty, \infty) = \mathbb R.$ Now let $[x_1, x_2] \subset (0, 1).$ Then $[f(x_1), f(x_2)] \in f(0, 1).$ By Intermediate Value Theorem, for any $d \in [f(x_1), f(x_2)]$, there's some $c \in [x_1, x_2]$ s.t. $f(c) = d.$ Thus by definition, $f$ is surjective.
Again, let $x_1, x_2 \in (0, 1)$ s.t. $x \ne x_2.$ WLOG, suppose $x_1 < x_2.$ Since $f$ is strictly increasing on $(0, 1)$, we have $f(x_1) < f(x_2)$ meaning $f(x_1) \ne f(x_2)$ and so $f$ is injective by definition $\square$
I realize I have to prove $f$ is both continuous and strictly increasing on its domain and the fact that the given limits diverge, but for the purposes of this proof Im not worried about that.
My question:
Assuming the proof of surjectivity of $f$ is correct (?), how can we be sure there is no $y \in \mathbb R$ s.t. $y \not \in \bigcup f([a, b])$ for arbitrary $a, b \in (0, 1)?$ Are the reals and the union of all the closed subsets of $f(0, 1)$ equivalent? Thanks.
Edit:
Nevermind, for some reason I thought $|f(0, 1)| < |\mathbb R|$ even though right at beginning I showed $\mathbb R$ is actually the image of $f.$ Brainfart moment, sorry.
Essentially you have to use the fact that $\lim_{x\rightarrow 1^-} f(x)=+\infty$ and $\lim_{x\rightarrow 0^+} f(x)=-\infty$. By definition, they are saying for any $y$, you can find $a,b$ such that $f(a)<y<f(b)$. So $y\in f([a,b])$, which is a subset of the union of all such sets. Hence, any $y$ must be in $\cup f([a,b])$.