2
$\begingroup$

I'm new here and I'm hoping that maybe I could get some help with something my teacher told me. He said that it is possible to have a closed set nested within another closed set where the intersection between these two sets was empty. In fact, he said that it could be extended to a closed set within a closed set within a closed set etc...

This is all a part of a bigger proof, which I think I could solve on my own, but I was hoping for some examples where the above statement is true. I just cannot for the life of me come up with a working example. Is the same true for bounded sets as well?

  • 0
    How about a set of rationals in a closed interval and a set of irrationals formed by adding a small irrational number to each element in the set of rationals, and then shortening the interval of the irrational set so it fits inside the set of rationals?2011-04-04
  • 1
    Nested, meaning $C_1\subseteq C_2$? In that case, $C_1\cap C_2=C_1$, so the only way the intersection could be empty is if $C_1$ is empty. Similarly with any finite nested sequence of sets. For an infinite nested sequence, consider $[n,\infty)$, and let $n$ vary. For compact nested sets the intersection will be nonempty. (I'm commenting partly because I'm not sure if this answers your question.) @Matt: I don't understand what your comment is supposed to give an example of.2011-04-04
  • 3
    Can you clarify what you mean by "nested within". It's obviously not set containment for two sets, else the intersection is always nonempty. Maybe it means something like one being contained in a neighborhood of the other, or if we're restricting ourselves to $\mathbb{R}$ or some ordered space, maybe it means something specific like one always having an elements on either side of elements in the other?2011-04-04
  • 0
    Perhaps it means that $C_1\subseteq [\inf(C_2),\sup(C_2)]$?2011-04-04
  • 0
    @ Matt Gregory Perfect idea. I think this is exactly what I needed. I'll try to get more clarity from my teacher with regards to what everyone else has said. Thanks all!2011-04-04
  • 0
    @Jake: Could you please explain how Matt Gregory's idea answers your question? I still don't understand what your question means or what Matt's comment is supposed to give an example of.2011-04-04
  • 0
    @Jonas: Well, I was thinking that, first of all, he was working with sets of real numbers, since he's talking about boundedness. So, by "nested" he means nested intervals, rather than subsets. Furthermore, he didn't say the intervals have to be connected (I think that's the right word). So the elements of one set can exist within the gaps of the other set, making their intersection empty.2011-04-04
  • 0
    @Matt: Boundedness makes sense in a much more general context, e.g. in any Euclidean space, in any metric space (and more, but that's general enough here), but you're probably right about real numbers being intended. An interval in $\mathbb{R}$ is connected. What is your definition of interval? If you are just talking about arbitrary subsets of $\mathbb{R}$, wouldn't the example of $\{2,3\}$ being "nested" in $\{1,4\}$ be simpler? How is it that the sets you describe are closed?2011-04-04
  • 0
    @Jonas: Yeah, I know boundedness makes sense in more contexts, but I just figured he was talking in the context of calculus or advanced calculus. The sets I was thinking of would be something like $A = { x | x \in [0,1], x$ is rational $}$, and $B = { x + sqrt{1/1000} | x \in [sqrt{1/1000}, 0.9 + sqrt{1/1000}], x$ is rational $}$.2011-04-04
  • 0
    @Matt: But those sets aren't closed (and they are not intervals).2011-04-05
  • 0
    @Jonas: Well, they have the condition that their elements exist within a closed interval. That's all I was thinking. You're right, though. They're not closed, nor are they intervals.2011-04-05
  • 0
    Strange statement. What do you have is that if you have a sequence of properly nested intervals $(I_n)$ then their intersection consists of one point.2011-05-29

2 Answers 2