# Why aren't all dense subsets of $\mathbb{R}$ uncountable?

1) we say that $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable.

That implies $\mathbb{R}-\mathbb{Q}$, that is irrational numbers are uncountable.

2) Archimedian property of $\mathbb{R}$ suggests that there exists a rational between any two numbers. i.e. $\mathbb{Q}$ is dense in $\mathbb{R}$. Then how come $\mathbb{Q}$ is countable while irrational is uncountable?

• That's the way it is – Norbert Feb 27 '12 at 18:55
• If you regard $\mathbb{R}$ as a closure of $\mathbb{Q}$, this gives some insight why $\mathbb{R}$ might be uncountable. Informally, $\mathbb{R} \subset 2^{\mathbb{Q}}$ – Tom Artiom Fiodorov Feb 27 '12 at 19:04
• @Mathemagician1234: Yes, the irrationals are also dense in $\mathbb{R}$; but then, so is $\mathbb{R}$ itself. Why do you think the question is "moot"? – Arturo Magidin Feb 27 '12 at 19:41
• @Mathemagician1234: I think it's clear that the intention of the question is "Why aren't dense sets of $\mathbb{R}$ necessarily uncountable?" – Nate Eldredge Feb 27 '12 at 20:39
• @Mathemagician1234: I have edited the title, but I have to disagree (strongly) it was that ambiguous. – Aryabhata Feb 28 '12 at 20:18

I think the confusion is that your brain tries to use finite reasoning in a place where it doesn't apply. It's true that if you have a finite sequence of balls, say, colored red and blue, such that between every two red balls is exactly one blue ball and between every two balls is exactly one red ball, then by symmetry there must be (essentially) the same number of red and blue balls.

Your brain wants to make this work with infinite collections also: If between every two rationals lies an irrational, and between every two irrationals lies a rational, then surely there are the same number, right? But try replacing the observation with the fact that between every two rationals lies infinitely many irrationals, and between every two irrationals lies infinitely many rationals. Now must there be the same number of each? Certainly the water is more murky, and once you know of the existence of different sizes of infinity you realize that the answer is of course not! There may be a different size of infinity's worth of rationals between every two irrationals than the size of infinity's worth of irrationals between every two rationals, breaking the symmetry.

Of course, this answer is filled with anachronisms (if you already know about sizes of infinity, you've probably had to have already come across the fact that there are more reals than rationals), but I think it still helps illustrate why one's brain gets confused. (And in the brain's defense, it's pretty cool and unintuitive stuff.)

• Lol, defending our brain is pretty cool. +1 – Patrick Da Silva Feb 27 '12 at 19:07
• To be fair to intuition, it's not that the stuff is inherently unintuitive, it's just that it's new (and has many qualitative dissimilarities from the familiar notion it's put into analogy with). Once you study the topic for a while, you build intuition for it, just like any other topic. – Hurkyl Feb 29 '12 at 3:36

One reason -- at least for the plausibility of $\mathbb{R}$ having countable dense subsets -- is that density has a fairly weak connection to the cardinality of a topological space.

If a (Hausdorff) topological space $X$ has a dense subset $D$ of infinite cardinality $\kappa$, consider the mapping $X \to \mathcal{P} ( \mathcal{P} ( D ) )$ defined by $$x \mapsto \mathcal{A}_x = \{ A \subseteq D : x \in \overline{A} \}.$$ Since $X$ is Hausdorff, then given distinct $x, y \in X$ there are disjoint open sets $U , V$ such that $x \in U$ and $y \in V$. Then $x \in \overline{U} = \overline{D \cap U}$, but $y \notin \overline{U}$. It thus follows that $D \cap U \in \mathcal{A}_x \setminus \mathcal{B}_x$; i.e., the mapping above is one-to-one. From this we have that $|X| \leq 2^{2^\kappa}$.

Even more, this inequality is the best possible. Given any infinite cardinal $\kappa$, let $X$ be any set of cardinality $\kappa$ with the discrete topology. Then there is a space, called the Stone-Čech compactification of $X$ and denoted $\beta X$, which has the following properties:

• $\beta X$ is Hausdorff (and compact);
• $| \beta X | = 2^{2^\kappa}$; and
• there is a homeomorphic copy of $X$ embedded in $\beta X$ which is dense in $\beta X$.

While this is clearly not an explanation of why $\mathbb{R}$ has countable dense subsets, since $\aleph_0 < 2^{\aleph_0} = | \mathbb{R} | < 2^{2^{\aleph_0}}$, appeals to cardinality and density alone cannot discount their (possible) existence.

To answer the title question, it's because you can approximate any real number to within arbitrarily small error using a rational number (just truncate the decimal expansion far enough out)

If you want an intuitive reason why $\mathbb{Q}$ is countable while the set of irrationals is uncountable, it's because irrational numbers necessarily have infinite, non-repeating decimal expansions. So the intuitive difference is the same as the difference between the cardinality of the set of finite strings of natural numbers (a countable union of finite sets, one for each length), and the cardinality of the set of infinite strings ($10^{\mathbb{N}}$).

In general, if a topological space has a countable dense subset we call it "separable." Separable metric spaces have a countable basis: one such basis for the metric topology on $\mathbb{R}^n$ is given by the balls of rational radius, centered at points with rational coordinates. More explicitly, every open subset of $\mathbb{R}^n$ is actually a COUNTABLE union of open balls. This fact (or just separability itself) can be used to show that any uncountable subset of $\mathbb{R}^n$ necessarily has a cluster point.

• "Because you can approximate any real number to within arbitrarily small error using a rational number (just truncate the decimal expansion far enough out)" doesn't really answer the difference between the rationals and the irrationals, which is what the asker sought. After all, you can also approximate any real number to within arbitrarily small error using an irrational number (just truncate the decimal expansion far enough out and append $\pi$)! – msh210 Feb 28 '12 at 20:06
• I think my second paragraph provides the reason for why they should be different: it's the difference between the set of finite strings and the set of infinite strings over a finite language. My first paragraph is answering the title question. – William Feb 28 '12 at 20:09
• Ah, yes. I wasn't considering the title. – msh210 Feb 28 '12 at 20:10
• I made an edit to make it more obvious :P Thanks for pointing out the ambiguity. – William Feb 28 '12 at 20:11

Between any two rationals, there is an irrational. Between any two irrationals, there is a rational. It would be nice to use this to compare, but it's tricky, because between any two rationals, there are more rationals as well.

So, to really compare things fairly, we need to come up with some new device to talk about the numbers "between" rationals in a way that there are no extra rationals getting in the way.

To start, suppose we have two rationals $a < b$. Well, there is a rational $a_1$ between them in the way, so we can refine our view from $(a,b)$ to $(a_1,b)$. Or, we could have refined to $(a_0,a_1)$. Let's pick the first. Then there's a rational $a_2$ in $(a_1, b)$, so let's restrict our attention again, to $(a_1, a_2)$. Then....

Well, this is a bit of a mess, how can we organize it? Well, how about the following? Let's split all of the rational numbers into two (non-empty) sets: the "left" set and the "right" set, so that every number in the left set is smaller than every number in the right set.

Now, we can ask what's between the left set and the right set, and there are no rational numbers left to get in the way! This is called a "Dedekind cut".

It turns out there are exactly three sorts of Dedekind cuts:

• There is a rational number $q$ so that $L = (-\infty, q] \cap \mathbb{Q}$ and $R = (q, \infty) \cap \mathbb{Q}$
• There is a rational number $q$ so that $L = (-\infty, q) \cap \mathbb{Q}$ and $R = [q, \infty) \cap \mathbb{Q}$
• There is an irrational number $r$ so that $L = (-\infty, r) \cap \mathbb{Q}$ and $R = (r, \infty) \cap \mathbb{Q}$

So we've finally gotten to our goal of finding a device to talk about numbers that are between rationals, without other rational numbers getting in the middle. And, as we had hoped, there cannot be more than one irrational in-between.

But, surprise! There are uncountably many Dedekind cuts, so there really are more ways to talk about things "between" rationals than there are rationals themselves.

Fortunately, if we repeat the above for irrational numbers, we find that the number of Dedekind cuts of irrationals is the same as the number of Dedekind cuts for rationals, so at least we get those numbers the same.

Countable does't mean that set is small(that is it is not dense in uncountable)...

Actually countable means we have method of enumerating... That is we have bijective map $A(\subset \mathbb N) \to \mathbb Q$.

Dense is topological property, It doesn't related with counting( Don't confuse with Baire Category theorem)....

From your observation, you can feel that countable set may be "so much big"(in dense sense)

Whatelse I can write.... Just I will repeat the sentence of "Norbert" that That is the way it is..

If I understand your question correctly, I think you are wondering that since whenever I take two irrationals I can find a rational "in between", and since there are uncountably many pairs or reals for which I can ask for such a rational "in between" whether this would not mean there should be uncountably many such rationals to make sure we can always find one right ??

The reason this argument fails is that you do not need to use distinct rationals in each case when you are looking for one that lies between two given real numbers; i.e. you can use a given rational multiple times, in fact uncountably often!

Here's another example that may shed some light on your reasoning. Let $A\subseteq \mathbb{N}$ be the set $\{2, ... , 100, 102, ... , 1000, ... , 10^n + 2, ... , 10^{n+1}, ... \}$. Both $A$ and $A^c$ are infinite (large). However $$\lim_{n\to\infty}\left|\frac{A\cap \{1, ... ,n\}}{n}\right| = 1.$$ but $$\lim_{n\to\infty}\left|\frac{A^c\cap \{1, ... ,n\}}{n}\right| = 0.$$ In other words, $A$ is much more prevalent than $A^c$ even though both are infinte. This is supposed to illustrate the way different notions of largeness and smallness can be completely unrelated. The same relationship holds holds for (topological) density and cardinality..