Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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? – Matt Gregory Apr 4 '11 at 20:27
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. – Jonas Meyer Apr 4 '11 at 20:43
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? – matt Apr 4 '11 at 20:44
Perhaps it means that $C_1\subseteq [\inf(C_2),\sup(C_2)]$? – Arturo Magidin Apr 4 '11 at 20:48
@ 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! – user9109 Apr 4 '11 at 20:57

The empty set works. The empty set is a closed subset of any closed set, and its intersection with any such set is empty.

share|cite|improve this answer

As I was going through unanswereds, I came across this question. I then posted a solution here, but it was terribly wrong (thanks Arturo). To redeem myself, I am heavily editing my previous answer to answer the question. I hope to answer Jonas Meyer's questions about what the answer means as well - that's how I justify this redemption.

First, I believe that the confusing part of the question was the word 'nested.' I interpret a closed set $C_1$ to be nested in a set $C_0$ if $C_1 \subseteq [\inf(C_0), \sup(C_0)], as suggested by Arturo. This is in response to the problem that if 'nested' implies any sort of containment, then there will be something in common between the nested set and the nesting set, unless the nested set is empty.

Now we can give a suggested answer. Suppose we let our '$C_1$' set to be the irrationals ($\mathbb{R} \backslash \mathbb Q$) in $[0,3]$. Then we might consider $C_2 = $ the set of rationals in $[1 - \frac{1}{2}, 2 + \frac{1}{2}]$. Then $C_2$ is nested within $C_1$, and clearly their intersection is empty.

We could then continue: $C_{2n}$ might be the set of rationals in $[1- \frac{1}{2n}, 2 + \frac{1}{2n}]$, and $C_{2n+1}$ might be the set of irrationals in $[1 - \frac{1}{2n+1}, 2 + \frac{1}{2n+1}]$. Then we have as many pairs of nested sets that we want. This is the key set of ideas within Matt Gregory's answer. But there is a problem with this answer (with respect to the demands of the OP) - none of these sets are closed. Of course, the OP seemed to accept this answer without any sort of fail, so it would seem that this might satisfy his needs.

But instead, we might consider the endpoints of the intervals I mentioned above. Thus $C_1$ would be the set $\{ 0, 3 \} $ and $C_2$ would be $\{\frac{1}{2}, \frac{5}{2}\}$, and so on. Then we still have that $C_2$ is nested within $C_1$. And for that matter we have that $C_k$ is nested within $C_j$ for all $k > j$. This is stronger than the above in a few ways.

Finally, in contradiction of Jonas T's final comment, there are infinite sets of nested intervals that do not end in a single point.

share|cite|improve this answer
Except that the "the intersection between [ $[0,1]$ ] and [$S$]" is not empty. – Arturo Magidin May 29 '11 at 0:34
@Arturo: Aha - I misunderstood the question. Somehow, I imagined this as two closed sets whose intersection was empty but formed a closed set. I will correct this shortly. – mixedmath May 29 '11 at 19:25
@Arturo - thank you again, I have corrected (I think) my post. – mixedmath May 29 '11 at 23:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.