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Given sequences such that $A_n = \{n,n+1,\cdots \}$, then it can be shown that $\bigcap\limits_{n=1}^\infty A_n = \emptyset$

Now, according to nested interval property if $\mathbf{I}_n = [a_b,b_n] = \{x \in \Re: a_n\leq x\leq b_n\}$, then $\bigcap\limits_{n=1}^\infty \mathbf{I}_n \neq \emptyset$

The above two statements looks very similar, but the results are just opposite. From what I can see is, $A_n$ is a countable infinite set whereas $\mathbf{I}_n$ is an uncountable infinite set. Is that the only difference, if it is true ?. Or is there anything more than that which relates two statements above ?

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First note that there are countably many $A_{n}$ and $I_{n}$ so the problem is not the countability or non-countability of the intersection. In other words the set containing all the $A_{n}$ and the set containing all the $I_{n}$ are both countable being indexed by the countable set $\mathbb{N}$.

Secondly it is true that the sets $I_{n}$ are uncountable whilst the sets $A_{n}$ are countable. This is a different statement from the above, for here we are talking about what the sets $I_{n}$ and $A_{n}$ contain, not what the sets that contain them contain.

However, the fact that the sets $I_{n}$ are uncountable whilst the sets $A_{n}$ are countable is not the reason for the discrepancy.

In fact it is possible to have a countable collection of countable sets $J_{n} = \{x \in \mathbb{Q} : 1 \leq x \leq 1 + \frac{1}{n}\}$ such that the intersection is non-empty, in this case equal to the set containing the number 1.

Conversely it is possible to have a collection of uncountable sets $B_{n} = \{x \in \mathbb{R} : x \geq n\}$ such that the intersection is empty.

Thus the crucial difference between your sets $A_{n}$ and $I_{n}$ is that the $I_{n}$ are ${bounded}$ and nested whilst the $A_{n}$ are ${unbounded}$ and nested.

It is this difference that is responsible for the different results for their respective intersections.

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  • $\begingroup$ Boundedness is not the key property here. The intervals $(0,1/n)$ are bounded and nested, but have empty intersection. $\endgroup$ – Mike Earnest Aug 1 '18 at 3:03
  • $\begingroup$ @Mike, in general yes you require intervals to be closed as well as bounded for nesting to have non-empty intersection, but in the specific case cited, both are already closed below and the difference "for their respective intersections" is down to the unboundedness (which implies openness) above of the $A_{n}$. The answer about compactness (of which closed and bounded is an example) gets my vote. $\endgroup$ – YeatsL Aug 1 '18 at 8:59
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There is a theorem which generalizes the nested interval property:

Theorem: (Nested Compact Sets Property) Let $K_1\supseteq K_2\supseteq K_3\dots$ be a nested sequence of nonepmty compact sets in a Hausdorff topological space. Then $\bigcap_n K_n$ is nonempty.

Proof: If $\bigcap_n K_n=\varnothing$ , then $\{K_1\setminus K_n\}_{n\ge 2}$ is an open cover of $K_1$ with no finite subcover. $\square$

However, the assumption of compactness is necessary. In a general topological space, a nested sequence of sets $A_n$ can have null intersection. Your sets $A_n$ are a typical counter-example. Note that the $A_n$ are not compact.

In summary, the commonality to your two examples is nested nonempty sets, and the difference is compactness.

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First of all note that both $\{A_n\}$ and $\{I_n\}$ are countable so this is not the problem.

The difference is in that $A_n$ is not a bounded closed interval, so we can not apply the nested interval theorem.

The nested interval theorem is about nested bounded closed intervals not arbitrary sets.

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  • $\begingroup$ how come $I_n$ is countable ?. it contains all real numbers in the interval. If i understand correctly it can not be marched 1 to 1 with natural numbers. $\endgroup$ – Shew Jul 31 '18 at 20:38
  • $\begingroup$ I mean the collection of $\{I_n\}$ not each set. $\endgroup$ – Mohammad Riazi-Kermani Jul 31 '18 at 20:42
  • $\begingroup$ Please check the edited version. $\endgroup$ – Mohammad Riazi-Kermani Jul 31 '18 at 20:44

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