Wikipedia:Reference desk/Archives/Mathematics/2019 August 7

= August 7 =

Lebesgue measure
Let A = the set of rational points in the unit interval (0,1). A has measure 0 because if r1,r2... is an enumeration of the rationals in A, fix &epsilon; and let $$E=\cup_{k=1}^\infty \{B(r_k; \varepsilon\cdot 2^{-k})\}$$. Now &lambda;(E) is at most 2&epsilon; and since &epsilon; can be arbitrary small, &lambda;(A) must be 0.

That means for a given small &epsilon; there must be points in (0,1) that are not in E.

Is there a way to show examples of such points? Since the rationals are dense, it's counterintuitive that a given real r might not be in one of the discs making up E. Thanks. 173.228.123.207 (talk) 04:08, 7 August 2019 (UTC)
 * Which numbers would be excluded would depend on the enumeration used, and in any case the details would get messy pretty quickly. One way to approach this is to define En to be the union of the first n intervals. The complement Cn of En is a closed subset of [0, 1]. the Cn's are non-empty otherwise the length of En≥1, and they are nested. The intersection is then non-empty; to see this pick an element from each set to form a sequence, apply Bolzano–Weierstrass to get a convergent subsequence, and the limit must be in the intersection because it is closed. Since the intersection of the Cn's is non-empty, the union of the En's, E, does not contain all of [0, 1]. This is, admittedly, a bit theoretical but it's difficult to be more specific without knowing the enumeration explicitly. There is no number that would work for every enumeration, but given an enumeration (and a value for ϵ) this process could be used to construct a sequence whose limit is not included in E.


 * If you'd prefer to have a specific value, I'll offer a similar set F which also includes an open interval around all rationals in [0, 1], and exhibit a number which is not in F. Specifically, for each q=a/b∈[0, 1]∩Q, a, b rel. prime integers 0≤a≤b≤1, b>0, define Fq = B(q, 1/(φb2)) where φ is the Golden ratio. Now define F to be the union of the Fq's. It turns out that φ is an example of a badly approximable number meaning it is not approximated well by rationals. (This is true of all quadratic surds but φ is particularly bad.) So I'll take r=1/φ=φ-1 to be my example as an element not in F, since it can be shown that for any q=a/b, |q-r|≥1/(φb2). This isn't that difficult to prove, but I'll skip the details and post a proof if requested; presumably it can be found in the literature.


 * Note that 2-i is used to ensure the length of E is less than 2ϵ, the lengths of the Fq don't approach 0 as quickly. Also note that the construction of E works for any countable set, not just the rationals, and one application is an alternate (i.e. non-Cator diagonal) proof that the reals are uncountable. --RDBury (talk) 08:19, 7 August 2019 (UTC)


 * Thanks! Some parts of that make me hesitate a bit, but I will study it some more. 173.228.123.207 (talk) 22:14, 7 August 2019 (UTC)


 * You might be interested in the Smith–Volterra–Cantor set, a nowhere dense closed set with positive measure. The set [0, 1]\E is apparently another example only constructed so it's disjoint with Q. So yeah, even closed subsets of [0, 1] can be pretty weird looking, but it's nothing compared to some of the things you encounter as you get deeper into topology. --RDBury (talk) 05:28, 8 August 2019 (UTC)
 * You can arrange that any given irrational is excluded, if you like. Suppose I want to exclude irrational $$\alpha$$.  You've got your numbering r1,r2...; I will define a new numbering q1,q2... as follows: qi = rj, where j is least such that rj has not already appeared in the q-ordering and $$|\alpha - r_j| > \varepsilon\cdot 2^{-i}$$.  Now build E using this new numbering, and it will exclude $$\alpha$$.--2404:2000:2000:5:0:0:0:C1 (talk) 01:41, 9 August 2019 (UTC)

Thanks again. 173.228.123.207 (talk) 05:57, 9 August 2019 (UTC)