How to prove $|[a, b]| = |\mathbb R|$, i.e. there are equally many points on a finite line segment than on the whole number line? Could one use the Cantor-Schröder-Dedekind theorem, which tells as that if we have injections $A\to B\to A$, then $|A| = |B|$? Then it only remains to prove that there is an injection $\mathbb R\to [a, b]$.
How to prove $|[a, b]| = |\mathbb R|$?
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elementary-set-theory
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1See [this question](http://math.stackexchange.com/questions/564056/can-there-exist-an-injective-function-from-mathbb-r-to-0-1). You can compose this injective with a bijective map from $(0,1)$ to $(a,b)$. The composed map will be an injective map from $\mathbb{R}$ to $[a,b]$. – 2017-02-22
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1Possible duplicate of [Proving that cardinality of the reals = cardinality of $[0,1]$](http://math.stackexchange.com/questions/660997/proving-that-cardinality-of-the-reals-cardinality-of-0-1) – 2017-02-22
2 Answers
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It’s easy to get a one-to-one onto map between $\Bbb R$ and an open interval $\langle a,b\rangle$, you can use $\arctan$ for instance, suitably scaled, to do it. Then there are standard tricks to get the endpoints into the act. Nothing as deep as Schröder-Bernstein should be necessary.
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0I'm not sure I would call Schröder-Bernstein particularly deep, though. – 2017-02-22
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0For my level of smarts, @HenningMakholm, it’s deep enough. – 2017-02-23
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There is an injection (bijection, in fact) from $\mathbb R$ to $(-1,1)$: $$f(x)=\frac{x}{1+|x|}$$
Now, transform the interval $(-1,1)$ into $(a,b)$.