# Uncountable closed set of irrational numbers

Could you construct an actual example of a uncountable set of irrational numbers that is closed (in the topological sense)?

I can find countable examples that are closed, like $\{ \sqrt{2} + \sqrt{2}/n \}_{n=1}^\infty \cup \{ \sqrt2 \}$ , but how does one construct an uncountable example?

At least one uncountable example must exist, since otherwise the rational numbers form a Berstein set and are non-measurable.

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– Asaf Karagila Dec 25 '12 at 20:29
I'd consider this a duplicate of this question, although it asks for the complementary set. – Marc van Leeuwen Dec 25 '12 at 20:31
Also related to math.stackexchange.com/questions/195313/… (and the accepted answer here to an answer there ) – Henry Dec 26 '12 at 10:00

If enumerate the rationals $\mathbb{Q}=\{q_1,q_2,\ldots\}$ and define the intervals $I_n=(q_n-2^{-n},q_n+2^{-n})$, we have that the set $S=\bigcup_{n=1}^\infty I_n$ contains every rational number and $$m(S)\leq\sum_{n=1}^\infty m(I_n)=\sum_{n=1}^\infty 2^{n-1}=2$$ Therefore, the complement of $S$ in $\mathbb{R}$ is of infinite measure (hence uncountable) and contains only irrationals.

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List the rationals as $r_0,r_1,\dots,r_n,\dots$.

Around the $n$-th rational, make an open interval of width $\epsilon/2^n$. Take the union of these. The complement does the job.

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Nicolas What math do you need to know to understand this?? – Don Larynx Oct 19 '13 at 2:55
It helps to have some intuitive feeling about measure theory. – André Nicolas Oct 19 '13 at 3:23
I know nothing of it. – Don Larynx Oct 19 '13 at 3:25
Don't need much, just that a countable set has measure $0$, and that the measure of a countable union is $\le$ the sum of the measures. – André Nicolas Oct 19 '13 at 3:29

The irrational numbers are homeomorphic to $\Bbb N^{\Bbb N}$. Let $C=\{0,1\}^{\Bbb N}$, viewed as a subset of $\Bbb N^{\Bbb N}$. By the Tikhonov product theorem $C$ is compact, and $\Bbb N^{\Bbb N}$ is Hausdorff, so $C$ is closed in $\Bbb N^{\Bbb N}$. (In fact $C$ is homeomorphic to the middle-thirds Cantor set.) Finally, $C$ is clearly uncountable. Now if $h:\Bbb R\setminus\Bbb Q\to\Bbb N^{\Bbb N}$ is any homeomorphism, $h^{-1}[C]$ is an uncountable closed subset of $\Bbb R\setminus\Bbb Q$, the irrationals. Being compact, it’s also closed in $\Bbb R$.

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Closed in $\mathbb{R}\setminus \mathbb{Q}$, yes, but why closed in $\mathbb{R}$? – WimC Dec 25 '12 at 20:22
@WimC: Closed in any Hausdorff space in which it’s embedded: it’s compact. – Brian M. Scott Dec 25 '12 at 20:23
A variation on this: if $C$ is a Cantor set (space homeomorphic to $\{0,1\}^\mathbb{N}$, or equivalently the Cantor middle third set) then $C$ is homeomorphic to $C \times C$ and the latter representation shows that a Cantor set can be written as a union of continuum many disjoint copies of a Cantor set. Only countably many of those (when we see this as subsets of $\mathbb{R}$) can contain rationals, so we have continuum many such compact sets inside the Cantor middle third set. – Henno Brandsma Dec 26 '12 at 18:21

Here's one example that requires neither measure theory nor an enumeration of the rational numbers. The set of positive irrationals is in bijection with the set of infinite sequences of positive integers through their continued fraction representation (which is unique for irrationals). Take the subset for which all of the terms of the continued fraction representation are $1$ or $2$. The complement in the set of positive reals is open, since for every irrational that has some continued fraction coefficient${}>2$ one easily finds a neighborhood where the coefficients up to and including this one are unchanged, and in a sufficiently small neighborhood of every rational number $\alpha$, some continued fraction coefficient (beyond the finite representation of $\alpha$ itself) is necessarily very big.

I may add that if the set ${\Bbb N_{>0}}^\Bbb N$ of infinite sequences of positive integers is equipped with the product topology for the discrete topology of $\Bbb N_{>0}$ (so that subsets of sequences in which a finite initial subsequence is fixed form a basis of open sets), then the continued fraction representation is a homeomorphism of the positive irrationals (with their topology restricted from $\Bbb R$) to ${\Bbb N_{>0}}^\Bbb N$. Therefore this is just a concrete version of the answer by Brian M. Scott (but with positive irrationals only, and with a more elementary argument to show the set is closed).

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I think with the Axiom of Choice one can construct such a set. For each irrational number $x$ in the interval $[0,1]$, associate a sequence of irrational numbers $\{x_n\}$ that converges to $x$. Do you think this is sufficiently constructive?

EDIT: @Tom Oldfield pointed out that terms from different sequences could converge to rational numbers, and this sounds rather difficult to prevent, so the set could well be not closed.

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No sequence could have a subsequence converging to a rational number since any subsequence of a convergent sequence converges to the same limit. I think that you would have to make sure that no sequence of numbers $y_n$, where each one may be taken from sequences for different $x$s converges to a rational number (and this is not true in general). – Tom Oldfield Dec 25 '12 at 20:08
@TomOldfield Yes you're absolutely right I got a little mixed up. I guess it would be too difficult to ensure that terms of different sequences don't converge to a rational. I will have another think. – user54147 Dec 25 '12 at 20:11
No worries, this was exactly my first thought! I think however you think about constructing one explicitly, you run into problems because there is no guarantee that an infinite union of closed sets is closed (For example, the countable union of $[\frac{1}{n}, 1-\frac{1}{n}]$ for natural $n$ gives $(0,1)$. I think the best way to do it is as in other answers where take unions to make an open set, who's complement has the desired properties. – Tom Oldfield Dec 25 '12 at 20:15
@TomOldfield It's actually quite nice how in this case that asymmetry between the properties of infinite unions of closed and open sets can help one solve the problem. Approaching it by taking unions of closed sets seems like it would be impossible to me now. – user54147 Dec 25 '12 at 20:22
exactly, this is exactly the reason I browse stackexchange! – Tom Oldfield Dec 25 '12 at 20:29

Here is an alternative construction that is a bit more general. It works for any comeager subset $X \subset \mathbb{R}$. Let $(U_i : i \in \mathbb{N})$ be a sequence of dense open sets such that $X \subset \bigcap_{i\in \mathbb{N}} U_i$. (If $X$ is the set of irrational numbers then we can take $U_i = \mathbb{R} \setminus \{q_i\}$ where $q_i$ is the $i^\text{th}$ rational number in some fixed enumeration.) We can define a family of non-degenerate closed intervals $I_s$, one for each finite binary sequence $s$, with the following properties:

• If the sequence $s$ extends the sequence $t$ then $I_s \subseteq I_t$

• If neither sequence $s$ or $t$ extends the other then $I_s \cap I_t = \emptyset$

• If the sequence $s$ has length $n$ then the interval $I_s$ has length $\le 2^{-n}$ and is contained in the set $U_n$

(It is easy to define $I_s$ by induction on the length of $s$. The exact choice of definition is not important.) We can use this "Cantor scheme" to define an embedding $f$ of the Cantor space $2^\mathbb{N}$ into $X$ by mapping an infinite binary sequence $x$ to the unique point $f(x)$ in the intersection of the decreasing sequence of closed intervals $(I_{x \restriction n} : n \in \mathbb{N})$.

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