Prove: the cardinality of every interval in $ \mathbb{R}$ is equals to $ \vert \mathbb{R} \vert = 2^{\aleph_{0}}$
I know how to prove that the cardinality of every closed interval in $\mathbb{R}$ equals to $ 2^{\aleph_{0}}$, but I'm not sure how exactly the proof differs for an open interval / half open interval.
the proof of a closed interval would be:
first, I'd like to prove that: $[a, b] = [-1, 1]$.
we'll define the function $f: [-1 , 1] \to [a, b]$ by $f(t) = a+ (t+1) \cdot \frac {b-a}{2} $
i'll prove that $f$ is one-to-one and onto.
one-to-one
let $t_1, t_2 \in [-1,1]$
$$ f(t_1) = f(t_2)$$
$$ \iff$$
$$ a+(t_1 + 1) \cdot \frac {b-a}{2} = a+(t_2 + 1) \cdot \frac {b-a}{2}\\\ \Longrightarrow t_1 = t_2 $$
onto – of the definition of $f$.
because $f$ is a bijection that'd enough to prove that $\vert [-1,1]\vert \geq \vert \mathbb{R} \vert $.
Let $g: (-1,1) \to \mathbb{R}$.
$$g(x)=\left\{ \begin{array}{cc}
\frac{1}{2x-1}\ \ \text{if}\ & \frac{1}{2}<x\leq1\\
\frac{1}{2x+1}\ \ \text{if}\ & -1\leq x<\frac{-1}{2}\\
2x\ \ \text{if}\ & \frac{-1}{2}\leq x\leq\frac{1}{2}
\end{array}\right.$$
I'll prove that $g$ is onto.
Let $x \in \mathbb{R}$.
if $x <1$ then $\frac{1}{2} < \frac {1}{2} \cdot \left( \frac {1}{x} +1 \right) < 1 \ \text{and} \ f \left( \frac {1}{2} \cdot \left(\frac {1}{x} +1 \right)\right) = \frac {1}{2\cdot \left(\frac {1}{2} \cdot \left( \frac {1}{x} +1\right)\right)-1}
= x$
in the same way, if $x < -1 $ so $ \frac{-1}{2} < \frac {1}{2} \cdot \left( \frac {1}{x} +1 \right) > -1$
hence $f \left( \frac {1}{2} \cdot \left(\frac {1}{x} +1 \right)\right) = x$, otherwise, $-1 \leq x \leq 1$, therefore $f \left( \frac {x}{2} \right) = x$.
Best Answer
I would only prove it for the open interval $(a,b),$ (using e.g. a modification of $f:\mathbb{R}\to (-\pi/2,\pi/2), \; x\mapsto \arctan x$) and then you have $$|\mathbb{R}| = \vert (a,b) \vert \leq \vert (a,b] \vert \leq \vert \mathbb{R} \vert \quad \Rightarrow \quad \vert (a,b] \vert = \vert \mathbb{R} \vert, $$ where the inequalities follow since $(a,b) \subset (a,b] \subset \mathbb{R},$ and similarily for other half-open and the closed interval.