$ \bigcap\limits_{n=1}^N I_n \neq \emptyset$ for all $ N \in \mathbb{N}$ implies that $ \bigcap\limits_{n=1}^\infty I_n \neq \emptyset $? How can I show that a sequence of closed bounded (Not necessarily nested) intervals $ I_1, I_2, I_3 ,\ldots$ with the property that $ \bigcap\limits_{n=1}^N I_n \neq \emptyset$  for all $ N \in \mathbb{N}$ implies that $ \bigcap\limits_{n=1}^\infty I_n \neq \emptyset $ ?
I'm being asked to determine if this is true.  I think that it is, because no matter how large $N$ is, we can always find an element in  $ \bigcap\limits_{n=1}^N I_n$.  So, we could simple let $N$ grow and there will always be an element in the intersection.  I was thinking about using induction, but this doesn't seem like an induction problem.  I am new to Real Analysis (self-study).  Someone tried to explain this to me using the bolzano weierstrass theorem, but I have not learned that. Any guidance will be appreciated.
 A: I will reduce the problem to the The Nested Interval Theorem (also known as Cantor's intersection theorem). Let's look at the following sets 
$$A_n=\bigcap\limits_{k=1}^{n}I_k$$
Each $A_n$ has the following properties


*

*$A_n \ne \varnothing $, this is given

*$A_n$ is closed

*$A_n$ is bounded
And 
$$A_{n+1}=\bigcap\limits_{k=1}^{n+1}I_k=\left(\bigcap\limits_{k=1}^{n}I_k\right)\bigcap I_{n+1}=A_n\bigcap I_{n+1}$$
or $A_{n+1} \subset A_{n}$ (because $\forall x \in A_{n+1} \Rightarrow x \in A_{n}$). According to The Nested Interval Theorem 
$$\bigcap\limits_{k=1}^{\infty}A_k \ne \varnothing$$
But $$\bigcap\limits_{k=1}^{\infty}A_k=\bigcap\limits_{k=1}^{\infty}I_k$$
A: The answer is: yes, it is true.
Sketch. Let $I_n = [\ell_n, \: r_n]$ for every $n$. Define $L = \{\ell_n: \: n \in \mathbb{N}\}$ and $R = \{r_n: \: n \in \mathbb{N}\}$, then let $\tilde{\ell} = \sup L$, $\tilde{r} = \inf R$. Show that $\tilde{\ell} \le \tilde{r}$. Hence deduce that any real number $x$ with $\tilde{\ell} \le x \le \tilde{r}$ lies in $\bigcap_{n = 1}^{\infty} I_n$.
A: Choose for each $k \in \mathbb{N}$ an $s_k \in \bigcap_{n=1}^k I_n$. Then the sequence $(s_k)$ is bounded, so it has a convergent subsequence $(s_{k_j})$. Call the limit of this sequence $s$. Now we want to show this limit is in the intersection of all $I_n$. Since all the intervals are closed, and the intersection of finitely many closed intervals is a closed interval,
$$s \in \bigcap_{n=1}^{k_j}I_n$$
for all $j$. Now take some $N \in \mathbb{N}$. Take some $j$ such that $k_j \geq N$ (this is possible, $j=N$ suffices for example, but this is also intuitively clear). Then 
$$s \in \bigcap_{n=1}^{k_j}I_n \subset \bigcap_{n=1}^{N}I_n \subset I_N.$$
So $s$ is in all $I_n$, so $s$ is in their intersection. So we found an element in their intersection, that is, the set $\bigcap_n I_n$ is not empty. 
A: Let $I_{n}=[a_{n},b_{n}]$, where $a_{n}\leq b_{n}$. We go to prove
that $\sup_n a_{n}\leq\inf_n b_{n}$ by contradiction. Denote $a=\sup_{n}a_{n}$
and $b=\inf_n b_{n}$. Suppose the contrary that $a>b$. Choose $l\in(b,a)$.
Then there exists $n_{1}$ and $n_{2}$ such that $a_{n_{1}}>l$ and
$b_{n_{2}}<l$. Take $N=\max(n_{1},n_{2})$. By assumption, $\cap_{k=1}^{N}[a_{k},b_{k}]\neq\emptyset$,
so there exists $x_{0}\in\cap_{k=1}^{N}[a_{k},b_{k}]$. Now $x_{0}\in[a_{n_{2}},b_{n_{2}}]$
implies that $x_{0}\leq b_{n_{2}}<l$. On the other hand, $x_{0}\in[a_{n_{1}},b_{n_{1}}]$
implies that $x_{0}\geq a_{n_{1}}>l$. Contradiction!
Hence $a\leq b$. We assert that $[a,b]\subseteq\cap_{k=1}^{\infty}[a_{k},b_{k}]$
and it will follow that $\cap_{k=1}^{\infty}[a_{k},b_{k}]$ is non-empty.
Let $x\in[a,b]$. Let $k\in\mathbb{N}$ be arbitrary. Then $a_{k}\leq a\leq x\leq b\leq b_{k}$
implies that $x\in[a_{k},b_{k}]$. Q.E.D.
A: Let $\displaystyle c_N = \inf \bigcap_{n=1}^N [a_n,b_n] $ and $\displaystyle d_N= \sup \bigcap_{n=1}^N [a_n,b_n].$
See if you can show that $c_1\le c_2\le c_3\le \cdots \le c_N \le \cdots \le d_M \le \cdots \le d_3\le d_2\le d_1$ for all $N,M.$
Then see if you can show that $\displaystyle \varnothing\ne \left[ 
\sup_n c_n, \inf_n d_n \right] \subseteq\bigcap_n [a_n,b_n].$
A: The intersection of bounded closed intervals is a closed interval. So 
$\tag 1 \bigcap_{k=1}^n I_k = [a_n,b_n] \text { for all } n \gt 0$
By assumption, this is nonempty with $a_n \le b_n$, defining two bounded monontonic sequences, $(a_n)$ and $(b_n)$.
The bounded increasing sequence $a_n$ has a limit $a$. It is easy to see that
$\tag 2 a \ge a_n \text{ for all } n \gt 0$
$\tag 3 a \le b_n \text{ for all } n \gt 0$
But then
$\tag 4 a \in  \bigcap_{k=1}^n I_k \subset I_n \text { for all } n \gt 0$
But (4) means that 
$\tag 5 a \in \bigcap\limits_{n=1}^\infty I_n $
We could also take the limit $b$ of the $b_n$ sequence, and it is easy to show that
$\tag 6  \bigcap\limits_{n=1}^\infty I_n = [a,b]$
A: If $\bigcap_{n=1}^\infty=\emptyset$, then $$I_1\subset \bigcup_{n=2}^\infty I_n^c,$$
where $I_n^c$, the complement of $I_n$, are open sets, so there is finite  natural number $2\leq n_1<n_2<\cdots<n_K$ such that 
 $$I_1\subset \bigcup_{i=1}^K I_{n_i}^c$$
since $I_1$ is bounded and closed which is equivalent to $I_1$ is compact.
Hence $$\bigcap_{n=1}^{n_K}I_{n}\subset I_1\bigcap(\bigcap_{i=1}^K I_{n_i})=\emptyset.$$
There is also a more general conclusion which can be proved with the same method:

If $\{K_\alpha\}$ is a collection of compact subsets of a Hausdorff space and if $\bigcap_\alpha K_\alpha=\emptyset$, then some finite subcollection of $\{K_\alpha\}$ also has empty intersection.

