Density of $\mathbb{Q}$ in $\mathbb{R}$ seemingly contradictory to infinite set cardinality

Question

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.

Answer 1

"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.)

Answer 2

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.

Answer 3

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.

Answer 4

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 $\Bbb R$)