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In my real analysis module we proved that both $\mathbb{Q}$ and $\mathbb{R \setminus Q}$ are dense in $\mathbb{R}$, meaning that between any two real numbers there exists a rational and an irrational number respectively.

In my own reading of set theory and the cardinality of infinite sets, I have learnt that $$\left| \mathbb{Q} \right| = \aleph_0 \text{ , } \left| \mathbb{R} \setminus \mathbb{Q} \right| = \aleph_1 \text{ , } \left| \mathbb{R} \right| = \aleph_1 \quad \colon \aleph_1 > \aleph_0$$
From this it appears that there is a contradiction, since given that the cardinality of the real numbers is strictly greater than the cardinality of the rational numbers, it doesn't follow that the rationals can be dense in the reals. That is to say that given that there are "more" real numbers than there are rational numbers, there simply aren't enough rationals to fill the gaps between the reals.
Given that the cardinality of the irrational numbers is the same as the cardinality of the reals their density does follow since there are enough to go around.

For the rational numbers though, clearly I've misunderstood something, however I can't seem to think of an explanation myself; any explanation I come up with doesn't manage to get past this discrepancy.

The best analogy that I've managed to come up with is to consider the number line as a fractal, in that however much you zoom in, it will always show a rational number in between two real numbers, however I don't think this is the best way of looking at it, and I wanted to find a more formal and rigorous explanation of where I'm going wrong in my understanding and why this isn't actually a contradiction.


Thanks in advance!



EDIT: Even if $\left| \mathbb{R} \right| = \aleph_1$ is not accepted, $\left| \mathbb{R} \right| > \left| \mathbb{Q} \right|$ being accepted (or that the rationals are countable whilst the reals are uncountable) still implies the same question.

preferred_anon
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Darya
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    Well, first of all, $\aleph_1=|\mathbb R|$ is not a true theorem. $\aleph_1$ is the smallest cardinality greater than $\aleph_0,$ but it it undecidable whether that cardinality is the same as the cardinality of real numbers. – Thomas Andrews Mar 18 '24 at 22:15
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    I agree that it is counter-intuitive that $\mathbb Q$ is both dense in $\mathbb R$ and yet is still countable, but that's all it is: counter-intuitive. There is no actual contradiction here. Also, the statement that $\mathbb R$ has cardinality $\aleph_1$ is independent of the usual axioms of set theory – i.e. it is known to be neither provable nor disprovable (assuming that the axioms of set theory are consistent). What we do know is that $\mathbb R$ has cardinality $2^{\aleph_0}$. – Joe Mar 18 '24 at 22:16
  • If you try to construct the intuitive bijection, you'll find it very hard. Literally impossible unless set theory is inconsistent. Infinite cardinals can be very confusing at first, but that is why we want proof, not intuition. – Thomas Andrews Mar 18 '24 at 22:17
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    Density here can be understood as a statement about limits of sequences, i.e. every element of $\mathbb{R}$ is the limit of a sequence in $\mathbb{Q}$. But now with a little careful treatment the contradiction disappears: There are more sequences with values in $\mathbb{Q}$ than there are elements of $\mathbb{Q}$, enough in fact to account for every element of $\mathbb{R}$ (if you consider the Cauchy sequence definition of $\mathbb{R}$ this is immediate). – Ben Steffan Mar 18 '24 at 22:18
  • It is totally wrong to denote the cardinality of $\mathbb{R}$ by $\aleph_1$. This is only the case if you assume the continuum hypothesis, which is not an axiom that is usually assumed in math. Also, I don't see why you find it not intuitive that the rationals can be dense, despite being of smaller cardinality. There are still many rational numbers on the real line, just even more irrational ones. – Mark Mar 18 '24 at 22:18
  • @Mark: Well, there is at least something to the intuition that cardinality is related to density: no finite set can be dense in $\mathbb R$. So I think it's reasonable to find it counter-intuitive that $\mathbb Q$ is dense in $\mathbb R$ and yet is still countable. – Joe Mar 18 '24 at 22:20
  • The analogy you've written sounds good to me. It is in fact the way that I have always understood the density of the rationals in the reals. – While I Am Mar 18 '24 at 22:26
  • It also may be worthwhile to read into the notion of separability of a topological space, which generalizes this feature of the rational numbers. I always find it helpful to see examples where something counterintuitive is not true, so that I can contextualize why it is true in the case which I cannot believe. – While I Am Mar 18 '24 at 22:28
  • Note that even the set $B = {x = k/2^n \mid k \in \mathbb{Z}, n \in \mathbb{N}}$ is dense in $\mathbb{R}$. There are no two reals between which there is no multiple of a pure power of $1/2$. – Brian Tung Mar 18 '24 at 22:38
  • I think the way you're defining density is causing the confusion. "Between any two reals, there's a rational" suggests some sort of relationship between elements. But really, density says that in any neighborhood around a real, there's at least one rational. That same neighborhood contains uncountably many other real numbers. I think visualization is also likely to be misleading. – Alex K Mar 18 '24 at 22:56
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    I see nothing at all wrong with the O.P.'s definition of a dense subset X of the reals as one such that between any two reals there is some x ∈ X. (Though this definition does not generalize per se to the definition of a dense subset of other spaces.) – Dan Asimov Mar 18 '24 at 23:00
  • By the way, you don't even need all the rationals to get a dense subset of the reals. You can consider, for instance, just the rationals of form k/2^n for all integers k, n. – Dan Asimov Mar 18 '24 at 23:00
  • @DanAsimov: GMTA. :-) – Brian Tung Mar 18 '24 at 23:10
  • This has been asked before. One, two, three, and especially four, among many others. – Arturo Magidin Mar 18 '24 at 23:30
  • Welcome to Math.SE. I think this is a good question, but it might have been asked before. I added some tags to explain that you are looking for an explanation of why your intuition fails, rather than a proof of the stated property. – preferred_anon Mar 18 '24 at 23:32
  • For example there is this question which asks basically the same thing but with less detail: https://math.stackexchange.com/questions/4670427/density-of-rationals-in-the-reals-compared-with-respective-cardinalities – preferred_anon Mar 18 '24 at 23:34
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    @AlexK The notion of a dense set has two definitions. One in ordered sets and one in topological spaces. In $\Bbb R$ both definitions coincide and it is common to give only one of them and work with it, as it is common give the two and prove that coincide. – jjagmath Mar 18 '24 at 23:57

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The way I visualize it (vaguely) is that the irrationals are by definition the gaps between the rationals, not the other way around. As you move along the number line, you pass an irrational not every time you pass a rational but every time you finish passing an entire infinite set of rationals like $\{q\in \Bbb Q:q^2<2\}$. Since the rationals are densely ordered, this is happening all the time, and it turns out there are more of these sets than there are rationals.

The fact that there is a rational between any two irrationals doesn't imply that the sets have the same cardinality, because any such interval contains infinitely many elements of both sets; you can never "zoom in far enough" to see the elements "alternating" in a way that would provide a bijection.

Karl
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"Given that the cardinality of the irrational numbers is the same as the cardinality of the reals their density does follow since there are enough to go around."—This is an invalid argument. The interval $[0,1]$ also has the same cardinality as the real numbers but is certainly not dense in the reals.

I include this in an answer rather than just a comment because I think it's actually quite relevant to the crux of the misunderstanding.

  • Cardinality is a property of "bare" sets—sets with no other structure other than how many elements they have.
  • Density is a property of ordered sets—sets with the structure of an inequality relation.

So we shouldn't, for example, expect to be able to tell whether a subset of $\Bbb R$ is dense based only on its cardinality. (Sometimes we can—finite sets aren't dense—but that's an extreme case and an anomaly.)

Greg Martin
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I think a possible clue to your misunderstanding lies in the following sentence (quoting the OP, with emphasis added):

given that the cardinality of the real numbers is strictly greater than the cardinality of the rational numbers, it doesn't follow that the rationals can be dense in the reals

The structure "Given P, it doesn't follow that Q can be true" doesn't really parse. (In this case, P is the statement "$|\mathbb Q| < |\mathbb R|$", and Q is the statement "$\mathbb Q$ is dense in $\mathbb R$".) It seems to me that you are mixing together two different kinds of logical question:

  • Given P is true, does it follow that Q must be true? In other words, is Q a necessary consequence of P?

  • Given P is true, can Q be true? In other words, is Q consistent with P?

The answer to the first question is: No, just because $|\mathbb Q| < |\mathbb R|$, it doesn't follow that $\mathbb Q$ is dense in $\mathbb R$. There are in fact many sets that have the same cardinality as $\mathbb Q$ but are not dense in $\mathbb R$. (For example, the integers are such a set.). So we can say (accurately):

given that the cardinality of the real numbers is strictly greater than the cardinality of the rational numbers, it doesn't follow (from that fact alone) that the rationals must be dense in the reals.

But the (correct) observation that P does not, by itself, directly imply Q, should not be taken as evidence that P and Q cannot both be true. The answer to the second question is: Yes, it is entirely consistent to assert both that $|\mathbb Q| < |\mathbb R|$ and that $\mathbb Q$ is dense in $\mathbb R$. You seem to feel it is inconsistent, but you have not actually derived a contradiction from the simultaneous assertion that $|\mathbb Q| < |\mathbb R|$ and that $\mathbb Q$ is dense in $\mathbb R$.

I see this sort of argument among my students all the time: "I can't prove that P implies Q, so P and Q must be inconsistent." Not necessarily! It might be that Q is a consequence of P and some other important facts that have to be used as part of the proof. That's precisely the case here: the density of the rationals in the reals is actually a consequence of the fact that the reals are an Archimedean field. Without that property, we wouldn't be able to conclude that $\mathbb Q$ is dense in $\mathbb R$; cardinality arguments alone aren't enough to prove it.

mweiss
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Density is not about how many numbers are you considering, is about how those numbers are distributed. $\Bbb Q$ and $\Bbb Z$ have the same cardinality, but one is dense and the other is not. The cantor set $C$ and the set of irrationals have the same cardinality, and one is nowhere dense (its closure have empty interior) and the other is dense (its closure is the whole $\Bbb R$)

jjagmath
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