So is the set $[0,1) \cap \mathbb{Q}$ countable and has countably infinite elements as $\mathbb{Q}$ is/has? Or in other words, is there a bijection between those sets?
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2Yes, all countable sets have the same cardinality. If you take a countably infinite set and then take a subset that is still infinite, you stay countably infinite (As that's the smallest) – Alan Aug 14 '21 at 11:26
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2If you consider $\frac{a}{b}$ with $a$ and $b$ coprime integers with $a$ non-negative and $b\ge 1$ then $\frac{a}{b} \to \frac{a}{a+b}$ is a bijection $\mathbb Q^{\ge 0} \to [0,1)\cap \mathbb Q$ – Henry Aug 14 '21 at 11:42
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If you want an explicit bijection, count $\mathbb Q$ by the standard cantorian diagonalization, count the ones between $0$ and $1$ by just skipping over any that aren't in that set, then match them (1st to 1st, second to second etc.).
Countably infinite subsets of countably infinite sets have to have the same cardinality, because subsets have less than or equal to cardinality, and the smallest infinite cardinality is $\aleph_0$
Alan
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Another approach: Denoting with $|X|$ the cardinality of a set $X$ we have \begin{align*} |[0,1)|\stackrel{(1)}{=}|(0,1)|\stackrel{(2)}{=}|\mathbb{R}|\tag{*} \end{align*}
(1) holds since changing an infinite set by a finite subset does not change its cardinality. See e.g. this answer by @Did.
(2) holds since e.g. the tangent function appropriately shifted and scaled gives a bijection.
We conclude from (*) \begin{align*} \color{blue}{|[0,1) \cap \mathbb{Q}|}=|(0,1) \cap \mathbb{Q}|=|\mathbb{R} \cap \mathbb{Q}|\color{blue}{=|\mathbb{Q}|} \end{align*}
Markus Scheuer
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