Someone asked for an example of an infinitely uncountable set with only isolated points.

Someone answered $\mathbb{R}$ with the discrete topology and many agreed.

Then I read that: Any discrete subset of Euclidean space is countable, since the isolation of each of its points (together with the fact that the rationals are dense in the reals) means that it may be mapped $1-1$ to a set of points with rational coordinates, of which there are only countably many.

Which seems to contradict the example?

This is a very basic question but i seem to misunderstand something?

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    $\begingroup$ The two examples have different topologies. $\endgroup$ – André Nicolas Mar 20 '16 at 4:10
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    $\begingroup$ The discrete topology on $R$ is not a subspace of the usual topology on any $R^n$. Check the def'n of subspace, as opposed to subset. $\endgroup$ – DanielWainfleet Mar 20 '16 at 4:26
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    $\begingroup$ You need to be more discrete. $\endgroup$ – copper.hat Mar 20 '16 at 4:27
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    $\begingroup$ "Uncountable" implies "infinite," so "infinitely uncountable" is superfluous. $\endgroup$ – Akiva Weinberger Mar 20 '16 at 4:50
  • $\begingroup$ .... the rationals are not dense in reals with the discrete topology. But basically, R with discrete topology is not at all a subspace of euclidean space at all. $\endgroup$ – fleablood Mar 20 '16 at 5:45

The sentence

Any discrete subset of Euclidean space is countable

assumes that we are using the usual topology - that is, "Euclidean space" means "$\mathbb{R}^n$ with the standard topology." The standard topology is, of course, not discrete.

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    $\begingroup$ Yes, of course! LOL...its to late for doing math... $\endgroup$ – JKnecht Mar 20 '16 at 4:13

Any uncountable set would be cardinal to $R^n$, so it would obey the nested interval property. If there is a nested interval $x=∅$, then it cannot be cardinal, and therefore countable.

  • $\begingroup$ But then you should assume continuum hypothesis.. $\endgroup$ – CO2 Aug 8 at 15:43

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