Union of infinite number of sets is not empty

**Pythagorean12** · December 1st 2009, 01:52 PM

Let [

],

be closed intervals with [

]

[

]

for all

. Prove that

I know it has something to do with compactness, but I am not sure how to proceed.

**Drexel28** · December 1st 2009, 02:16 PM

Originally Posted by Pythagorean12

Let [

],

be closed intervals with [

]

[

]

for all

. Prove that

I know it has something to do with compactness, but I am not sure how to proceed.

What have you tried?!

**Plato** · December 1st 2009, 02:50 PM

The requirement that $\left( {\forall m,n} \right)\left[ {[a_n ,b_n ] \cap [a_m ,b_m ] \ne \emptyset } \right]$ implies that $\left( {\forall m,n} \right)\left[ {a_n \leqslant b_m } \right]$ .

So $A = \left\{ {a_n :n \in \mathbb{Z}^ + } \right\}$ has a least upper bound. Call it $\alpha$ .

Show that $\alpha \in \bigcap\limits_n {\left[ {a_n ,b_n } \right]}$ .

**Shanks** · December 2nd 2009, 12:51 AM

The statement implies that The bounded colsed interval collection has finite intersection property,thus the intersection is nonempty.

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