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A discrete random variable is countable (such as integers and natural numbers), whereas a continuous r.v. is not countable (like the real numbers $\mathbb{R}$).

If I have a dataset whose observations can only be real numbers between 0 and 1, which are respectively the lower and upper bounds of the r.v., is the r.v. discrete or continuous? Does the same answer apply to an r.v. whose bounds are -1 and 1?

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    $\begingroup$ The reals in an interval are not countable. $\endgroup$ – Glen_b Jul 31 '20 at 3:33
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It could be continuous. Think about uniform distributions in the intervals you mention. That would be a continuous distribution, agreed?

Setting aside a technical issue of absolute continuity that I think is inappropriate to address at this level, a continuous random variable has a continuous CDF.

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    $\begingroup$ This answer is incorrect (or at least needs careful interpretation to be correct). The variable could be either discrete or continuous or neither: boundedness is completely unrelated. See stats.stackexchange.com/questions/103969 for an explicit, illustrated description of a discrete variable supported on $[0,1].$ $\endgroup$ – whuber♦ Jul 31 '20 at 16:06
  • $\begingroup$ @whuber I'm now seeing some issues with my answer beyond Cantor distribution-type of pathology, but I think you'd agree that a distribution that can take any real number on $[0,1]$ can't be discrete, right? The example in your linked post wouldn't have support on $\sqrt{2}/2$, for instance. $\endgroup$ – Dave Jul 31 '20 at 16:12
  • $\begingroup$ It depends on what you mean by "take any real number." If "take" means every real number has a nonzero probability, the situation is vacuous: there is no such distribution. Attempting to understand "take" as meaning "the density of each real number is positive" makes it self-fulfilling: by assuming a density even exists, you are assuming the distribution is continuous. Thus, I think we are compelled to understand "take" in the meaning of the support of the distribution, which is the smallest closed set having 100% probability. $\endgroup$ – whuber♦ Jul 31 '20 at 16:16
  • $\begingroup$ @whuber Yes, I take it to mean support on all of an interval like $[0,1]$. $\endgroup$ – Dave Jul 31 '20 at 16:25
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    $\begingroup$ Well, then, consider any countable subset of $[0,1]$ whose closure is $[0,1]$ (such as all rational numbers between $0$ and $1$), which by definition of "countable" may be indexed $\{x_1,x_2,\ldots,x_n,\ldots\}.$ Take any sequence $(a_n)$ of non-negative real values that has a positive sum $a.$ Define a distribution on $[0,1]$ by setting the probability of $x_i$ to $a_i/a.$ This is a discrete distribution whose support is $[0,1].$ That provides myriad counterexamples to your initial conclusion that the distribution must be continuous. $\endgroup$ – whuber♦ Jul 31 '20 at 17:28

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