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What are some examples for the latter claim?

I am comfortable with the fact that a discrete subset of $R$ is countable. And I think we might have some complicated space in which the discrete subsets can be uncountable. Am I correct?

High GPA
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    Take any uncountable set and endow it with the discrete topology. Or, if you want an example which ''arises in nature'' of a discrete uncountable subspace of a nondiscrete space, look here: https://math.stackexchange.com/questions/97648/why-is-l-infty-not-separable – arnett Apr 02 '22 at 06:51

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Just take an uncountable set equipped with the discrete topology. For instance, you could take the metric space $({\Bbb R}, d)$, where $d$ is the discrete metric $$ d(x,y)= \begin{cases} 1 & \text{if $x \not= y$}\\ 0 & \text{if $x = y$} \end{cases} $$

J.-E. Pin
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  • Thanks a lot for the answer (+1)! Is it possible that a discrete set is uncountable with some classic, non-discrete topologies which are more meaningful? – High GPA Apr 02 '22 at 11:39
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With regard to the OP's comment, it's probably more meaningful to look for topological spaces, which contain uncountable discrete subsets, but fullfill other countability conditions, eg., which are separable or Lindelof. Such a space cannot be metrizable, though.

The Stone-Cech compactification of the integers would be a standard example: It is compact, hence Lindelof. Since the integers are dense in it, it is separable. By a somewhat more complicated argument, it contains an uncountable, discrete subset (even of cardinality c (see, for instance, Engelking's book, 3.6.18).

An easier example is the one-point-compactification of an uncountable, discrete space. But this is not separable.

If you are looking for further examples, you can search for "uncountable spread".

Ulli
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