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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 $f \left( \frac {x}{2} \right) = x$.

Noa Even
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    Please note that if you add or subtrack to an uncountable set countable many elements , the cardinality does not change... – dmtri Aug 31 '18 at 09:27

2 Answers2

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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.

MSDG
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Or, have a look at this answer. Thus, $(a,b)$ and $[a,b)$ are in one-one correspondence... and use $\arctan $ for equivalence with $\mathbb R$.