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If the set of real numbers $\Bbb{R}$ is continuous, and the set of integer $\Bbb{Z}$ is a discrete set, then is the set of rational number $\Bbb{Q}$ continuous or discrete? My question is stated in the context of analysis. Sorry if I can’t state my problem clear, this question just passed my mind. If it is just a nonsense question, please tell me right away. Thank you very much.

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    $\begingroup$ Please use MathJax to format your posts. $\endgroup$ Oct 12, 2017 at 3:09
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    $\begingroup$ “It is always possible to find another rational number between any two members of the set of rationals. Therefore, rather counterintuitively, the rational numbers are a continuous set, but at the same time countable.” —Wolfram MathWorld $\endgroup$ Oct 12, 2017 at 3:15

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This depends on the topology that we equip $\mathbf{Q}$ with. If it has its usual topology, i.e. the topology inherited from the standard topology on $\mathbf{R}$, then it is not discrete. A topological space $X$ is said to be discrete if given any $x\in X$ there exists an open set $U$ containing $x$ such that $U\cap X=\{x\}$. Given any $\frac{p}{q}\in \mathbf{Q}$, and an open neighborhood of radius $\epsilon$, we can find another rational $\frac{m}{n}$ satisfying $\lvert \frac{p}{q}-\frac{m}{n}\rvert<\epsilon$, so that $\mathbf{Q}$ is not discrete.

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To say that the reals are continuous does not mean anything. There is no such notion, strictly speaking. Functions are continuous. One could say that the reals are complete. The rationals then, although dense, are not complete.

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    $\begingroup$ Maybe a reasonable way to pose the question and maintain its spirit is, “can there exist continuous functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ ?” $\endgroup$
    – kdbanman
    Jan 21, 2020 at 19:57
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    $\begingroup$ @kdbanman But the answer to that question is "yes" for every space - the identity function is always continuous. And for a discrete space $X$, every function $X\to X$ is continuous! $\endgroup$ Jul 13, 2021 at 18:29
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    $\begingroup$ "for a discrete space $X$, every function $X \to X$ is continuous!" Thank you, I didn't realize this. I think this answer points to the right property then (i.e. completeness vs non-completeness.) $\endgroup$
    – kdbanman
    Jul 13, 2021 at 18:59
  • $\begingroup$ Sounds like the property he's asking for is connectedness. $\mathbb Q$ is not connected -- in fact it's totally disconnected. $\endgroup$
    – rollover
    Jul 10, 2022 at 8:54

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