# How many irrationals $x$ appear infinitely often in this set of open intervals?

The setup: Let $q_1,q_2,q_3,\dots$ be an enumeration of the rationals in the unit interval $I=[0,1]$. Then let $K_i=\{q_j\mid j \le i\}$, and $K=\bigcup_i{K_i} = \{q_i\} = \mathbb{Q}\cap I$.

Define a set of open intervals $S=\bigcup_i{S_i}$ where $S_i = \{(q-2^{-i},q+2^{-i})\mid q \in K_i\}$.

For any $x \in I$ count the number of elements of $S$ for which $x$ is a member:

$$\#(x) = |\{s \mid x \in s \in S\}|$$

Clearly $\#(x)=\infty$ for any rational $x$, but what about irrational $x$?

My question: for 'how many' irrational $x$ does $\#(x)=\infty$? The categories 'none', 'at least one', 'most', 'all' are good enough here.

Bonus question: can we construct or otherwise characterize such points $x$?

Motivation: We know that there is no sequence of continuous functions that pointwise converges to $\chi_K$, the characteristic function of the rationals in $I$. This has been asked and answered on this site at Exists continuous $f_n: [0,1] \to \mathbb{R}$ that converges pointwise, as $n \to \infty$, to $\chi_\mathbb{Q}$? and Existence of a sequence of continuous functions convergent pointwise to the indicator function of irrational numbers. There is also an elementary proof at https://math.stackexchange.com/a/1444407/1257.

Nevertheless, in the spirit of squaring the circle, let's try to build such a sequence and then troubleshoot. The construction is inspired by this article, but seems to be fairly common.

The members of the sequence are $$f_i(x) = \max(0,1-2^i d(x,K_i)),$$ where $d(x,A)$ is the distance between the point $x$ and the nearest point in $A$.

Note that the support of $f_i$ is $S_i$, a set of open intervals defined earlier.

The sequence ${f_i}$ obviously converges to 1 on the rationals, but because of the impossibility of converging to $\chi_K$ we can conclude that it is not converging to $0$ on least one irrational $x$. So at $x$ it either converges to a value other than $0$ (doubtful), or never converges at all. Either way, $x$ is in an infinite number of the intervals in $S=\bigcup_i{S_i}$. Which would mean that an answer to the main question stated above would be 'at least one'.

To me this is vaguely plausible, but also surprising because the measure of the intervals in $S_i$ is decreasing exponentially. In fact we could rejig the definitions to make the measures decrease at arbitrarily large rates. But my knowledge of real analysis is very limited, so it is quite possible I've made an error in reasoning, or that my intuition is insufficiently developed.

• – brainjam Apr 11 '18 at 17:25

Fix an enumeration of $S$ as $S=\{s_n:n\in\mathbb{N}\}$. Let $A_m=\bigcup_{n\geq m}s_n$ and $A=\bigcap_n A_n$. Note that $\#(x)=\infty$ iff $x\in A$. I will show there are ways in which $A$ is "small" and also ways in which $A$ is "large".
First, $A$ is small in the sense of measure. Indeed, if $\mu$ is Lebesgue measure, note that $\sum_n \mu(s_n)=\sum_i \frac{i}{2^{i-1}}<\infty$. This implies $\mu(A_m)\to 0$ as $m\to\infty$, since $\mu(A_m)\leq \sum_{n\geq m}\mu(s_n)$ and these tail sums must go to $0$ for the sum $\sum_n \mu(s_n)$ to converge. Thus $\mu(A)=0$.
On the other hand, $A$ is large in the sense of category. In particular, notice that for each $m$, $A_m$ is a dense open subset of $I$. So, $A$ is a set of second category in $I$. In particular, for instance, this means that for any countable set $C$, $A\setminus C$ is dense in $I$, so $A$ must contain uncountably many irrational numbers and the irrational numbers in $A$ must be dense in $I$.
• Nice answer. The infinite intersection of shrinking unions must be a 'pattern' because I see it used in an example at en.wikipedia.org/wiki/Baire_space#Examples. And the observation that $\#(x)=\infty$ iff $x\in A$ connects beautifully with my question. – brainjam Mar 31 '18 at 1:52
• Yes, you essentially rediscovered the standard way to build a second category set of measure $0$. – Eric Wofsey Mar 31 '18 at 1:56
• if our domain is the rationals in I, and we are trying for convergence to the characteristic function of, say, rationals with odd denominator, would we be successful? Because there wouldn't be second category sets and there may not be any x with even denominator for which $\#(x)=\infty$ (I could pose this as a separate question if you wish) – brainjam Apr 1 '18 at 21:40
• I think it could work, though I'm not sure your specific construction works (you might need to replace the numbers $2^i$ with something that grows faster). But more simply, you can just diagonalize to get whatever limit on the rationals you want (let $f_i$ be some continuous function that takes the value you desire on the first $i$ of the rationals). – Eric Wofsey Apr 1 '18 at 21:50
• So is it correct to say that for functions $\Bbb Q \to \Bbb R$ there are only two Baire classes: the continuous functions and everything else? – brainjam Apr 2 '18 at 4:12