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Cosider a set of a closed segments on a real line, each two of them have an intersection. Is it true that there exists a point, which is contained in all of that sets?

My suggestions: 1) First of all, we can build a construction, based on a principe of nested segments (Cauchy-Cantor). But, the thing is that that we can only cope with it in the case, where we work with a finite number of segments.

How to consider this in a common way?

Thank you in advance.

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  • $\begingroup$ Open or closed or arbitrary segments? $\endgroup$ – Hagen von Eitzen Oct 30 '14 at 17:04
  • $\begingroup$ Closed segments, actually ) $\endgroup$ – hyperkahler Oct 30 '14 at 17:04
  • $\begingroup$ Just to clarify for those who may be considering this question: Do you intend that all of the segments are finite closed segments with positive length? $\endgroup$ – MPW Oct 30 '14 at 17:04
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Let $[a_i,b_i]$ with $i\in I$, $-\infty< a_i\le b_i< +\infty$ be your segments. Wlog. $I\ne\emptyset$, and we can pick one segment $[a_0,b_0]$. Then by assumtion $a_i\le b_0$ and $b_i\ge a_0$ for all $i\in I$. Then $a=\sup\{\,a_i\mid i\in I\,\}$ and $b=\inf\{\,b_i\mid i\in I\,\}$ are both finite (namely $a,b\in [a_0,b_0]$). Assume $a>b$. Then for suitable $i\in I$ we have $a_i>\frac{a+b}2$ and for suitable $j\in I$ we have $b_j<\frac{a+b}2$, hence $[a_i,b_i]\cap [a_j,b_j]=\emptyset$, contrary to assumption. We conclude that $a\le b$. Then for all $i\in I$ we have $a_i\le a\le b\le b_i$, i.e. $$ \emptyset \ne [a,b]\subseteq \bigcap_{i\in I} [a_i,b_i]$$


If we allow the closed intervals to be unbounded, we are lost: Consider the closed unbounded intervals $[n,\infty)$, $n\in\mathbb N$. If we allow open intervals, we are also lost: Consider $(0,\frac1n)$, $n\in\mathbb N$.

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I am assuming that a segment is a compact interval.

Let $I_k$ be the segments. I claim that any finite intersection is non-empty.

This is clearly true for $I_1 \cap I_2$. Suppose this is true for $I_1 \cap \cdots \cap I_n$, but $(I_1 \cap \cdots \cap I_n) \cap \cdots I_{n+1} = \emptyset$. Then we must have $\inf I_{n+1} > \min_{k=1,...,n } \sup I_k$ or the corresponding condition on the other side. Suppose $k_0$ is an index for which the minimum is achieved, then this says that $\inf I_{n+1} > \sup I_{k_0}$ which means that $I_{n+1} \cap I_{k_0} = \emptyset$, a contradiction.

Since $I_{k_1} \cap \cdots \cap I_{k_n} \supset I_1 \cap \cdots \cap I_{\max(k_1,...,k_n)}$, it is clear that this is true for arbitrary finite intersections.

Clearly $I_1 \cap \cdots \cap I_n \subset I_1$, $I_1 \cap \cdots \cap I_n$ is closed. Since $I_1$ is compact, the collection $I_k$ satisfies the finite intersection property, we have $\cap_{k=1}^\infty I_k \neq \emptyset $.

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  • $\begingroup$ I wonder if you could re-select Hagen's answer please? He was first, you had previously selected his answer and I was adding my answer as an alternative, not to capture the vote. Much appreciated. $\endgroup$ – copper.hat Oct 30 '14 at 17:58
  • $\begingroup$ Unfortunately, i haven't got used to the interface of the service, so the click wasn't made intentionally. ) $\endgroup$ – hyperkahler Oct 30 '14 at 18:02
  • $\begingroup$ No problem, much appreciated! $\endgroup$ – copper.hat Oct 30 '14 at 18:02

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