1

I was reading up on some analysis and came across the incompleteness of the real space ]0;1[ ( or (0;1) in amaerican notations). It's easy to see that it's incomplete if you consider that any sequence that is of Cauchy must converge on an element of the set for it to be complete. But I also read that the nested interval thereom says that any set is complete iff the intersection of every nested closed interval with a strictly decreasing non 0 diameter reduces to a singleton. Since this is an if and only if statement, that implies that any incomplete set must contain a series of nested intervals that don't reduce to a singleton. How would I do to find one? I figured it's intersection must be around 0 (or 1 but it's a symmetrical problem), with 0 not being contained therefor proving the set is incomplete, but since the intervals must be closed, how do I construct a lower bound to my intervals so that they are contained in the set but also all contain 0? Or did I misunderstand something?

Sorry if this was already asked, new to the site but couldn't find anything similar.

Asaf Karagila
  • 393,674
Hongo
  • 11
  • 1
  • 1
    I am not sure where you've seen that this is equivalent to completeness. As far as i know, this can be used as an equivalent definition for completeness of the real numbers, but not for completeness of any metric space, at least not if you define the closed intervals to be of the form $[a,b]$. The catch is what is known as relative topology. If your universe is $(0,1)$ (i.e., you don't think of this as part of the whole real line), then intervals of the form $(0,x]$ are closed with respect to the relative topology on $(0,1)$ and then these nested intervals indeed have no intersection. – AnCar Nov 19 '23 at 08:59
  • 2
    @AnCar, in any metric spaces the same theorem is true, only nested closed balls are used whose radii tend to zero. And you correctly indicated such balls. – zkutch Nov 19 '23 at 09:16
  • @zkutch thanks, that makes sense, if one is allowed to define closed balls in the relative topology, this should still work. – AnCar Nov 19 '23 at 09:17
  • I'm not sûre if the theorem holds in any space, but I read this for the real space and it's higher dimensions equivalents. – Hongo Nov 19 '23 at 11:59
  • @Hongo. For any metrical space: Elements of the Theory of Functions and Functional Analysis by Kolmogorov, Fomin. 2004, page 76 – zkutch Nov 19 '23 at 23:19
  • Not sure where you picked up the idea that "$(0;1)$" was an "american" notation. It was the notation for open sets everywhere until sometime in the middle of the last century (except for the semi-colon instead of a comma - as far as I am aware, that is a fairly recent development). I'm not sure where the use of reversed square brackets orgininated, but it seems no more or less popular in the U.S.A, than elsewhere from what I've seen. – Paul Sinclair Nov 20 '23 at 03:17
  • Analysis teacher said in Europe we more oftenly use the reverse bracket notation, and that in English speaking countries they usually opt for parenthesis instead, but that's kind of irrelevant to the question... – Hongo Nov 20 '23 at 08:14

1 Answers1

0

The Nested Interval Theorem just says that if you have a sequence $\{[a_n, b_n]\}_n$ of bounded closed intervals on the real line with $b_n \ge a_n$ which satisfy the condition $[a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$ for all $n$, then their intersection will be non-empty. Further, if $(b_n-a_n) \to 0$ as $n \to \infty$, then the intersection will be a single point.

As your description of the result you are talking about is garbled, it is not clear what that result might be. The most likely possibility is

  • If $A \subseteq \Bbb R$, then $A$ is complete if and only if for every sequence $\{[a_n, b_n]\}_n$ of nested bounded closed intervals in $\Bbb R$, the intersection $A\cap\bigcap_n [a_n, b_n] \ne \emptyset$.

Note that $A\cap\bigcap_n [a_n, b_n] = \bigcap_n (A\cap[a_n, b_n])$

By this result, finding such a sequence of closed intervals to show $A = ]0,1[$ is not complete is simple: $\{\left[0, \frac 1n\right]\}_n$ works.

It turns out the theorem in question is

If $A$ is a metric space then $A$ is complete if and only if for every sequence $\{F_n\}_n$ of nested closed non-empty balls whose diameters converge to $0$, there is some $a\in A$ with $\bigcup_n F_n = \{a\}$.

For $]0,1[$, the sequence $F_n$ of closed balls of radius $\frac 1n$ centered on $\frac 1n$ consists of nested non-empty balls whose diameters $\frac2n$ converge to $0$, but their intersection is empty. The balls are the intervals $\left]0,\frac2n\right]$, which are closed in $]0,1[$.

Paul Sinclair
  • 43,643
  • But doesn't the interval have to be entirely contained in the original set? – Hongo Nov 20 '23 at 14:08
  • Not in the result I quoted. It is about intervals in $\Bbb R$, If you mean a different result, then please, please, for the love of God, STATE the result you are referring to clearly instead of that confused mash-up you gave in the question. – Paul Sinclair Nov 20 '23 at 14:11
  • Sorry if it's not clear, it's hard to explain in english, not having done any math in English so I'm translating the terms as closely as possible. But I'm not sure what else to state. – Hongo Nov 20 '23 at 14:49
  • The result I quoted in the bullet point is a corollary to the Nested Intervals Theorem, not the theorem itself. If that corollary is not the theorem you are referring to, then I cannot help without some better idea of what theorem you mean. It seems likely to me that the real problem is not translation, but rather that you have not figured out yourself what the theorem you are referring to is actually saying. – Paul Sinclair Nov 20 '23 at 14:57
  • In particular, if in your theorem, the intervals actually need to be contained in the set $A$, why does it talk about the intersection of the intervals with $A$? If they are contained in $A$, that intersection would be the entire interval. – Paul Sinclair Nov 20 '23 at 15:02
  • The theorem I am reffering to is :

    "If a metric space is complete, then for all decreasing, closed, non-empty sequences of intervals F_n who's diameter goes to 0, the intersection of F_n is a singleton"

    -Jacques Deixmer, Topologie générale

    And the corollary is that if the space is NOT complete, then there exists a sequence of intervals F_n who's intersection is of non 0 diameter but that contains no elements of the set. I.e. :

    \bigcap_n F_n = {\O} with F_k \subset F_{k+1}

    It is not specified whether or not F_k must be contained in the set A.

     – Hongo Nov 20 '23 at 15:26
  • I hope that's more specific, but to rephrase the question :

    If this theorem holds (and it's corollary), then how do I construct the sequence F_n to prove that the real set ]0;1[ is not complete?

    This question is trivial if F_n doesn't have to be contained in the set ]0;1[ , but it seems wrong to be able to use a sequence of intervals that aren't contained in the set.

    It also seems the lateX notations didn't follow through, is that not how to caligraphy the text?

     – Hongo Nov 20 '23 at 15:29
  • In that theorem, you mean "decreasing sequences of closed non-empty balls", not "intervals", as metric spaces in general do not have a concept of "intervals". In $\Bbb R$, balls and bounded intervals turn out to be the same thing, but they are defined differently. But in that theorem, $A$ is the entire metric space, not the subset of something larger, so yes, the balls are contained in $A$. However, for $]0,1[$ as a metric space, the sets $]0, 1/n]$ are closed balls (with center $\frac 1{2n}$ and diameter $\frac 1n$). – Paul Sinclair Nov 20 '23 at 15:39
  • Yes, sorry, closed balls, habit of working in 1 dimension. It's probably painfully obvious, but how is ]0;1/n] closed when the lower bound is open? – Hongo Nov 20 '23 at 15:59
  • It is closed in $]0,1[$. The complement of $]0,1/n]$ is the set $]1/n,1[$, which is open. Every point in the complement has a ball about it that completely misses $]0, 1/n]$. Thus $]0,1/n]$ is a closed set. This is the same as why $]-\infty, 0]$ is a closed set in $\Bbb R$. – Paul Sinclair Nov 20 '23 at 16:17
  • ok, thanks for your patience. – Hongo Nov 20 '23 at 16:42
  • Everyone struggles with the concept at first, but eventually you learn to see things in relation to their proper ambient space instead of $\Bbb R$. – Paul Sinclair Nov 20 '23 at 20:14