Prove 1 What is the intersection of all the open intervals c

Prove

1. What is the intersection of all the open intervals containing the closed intervals [0,1]?

2. What is the intersection fo all the closed intervals containing the open intervals (0,1)?

I dont have any idea for these problems.. help me..

Solution

1.)

A. x [0,1]. Therefore x any set that contains [0,1] Therefore, it\'s in the intersection because they ALL contain [0,1]

B.x < 0, x the intersection, because there exists some set containing [0,1] that doesn\'t include any values < 0.

C. x>1, x the intersection by the same reasoning.

2.)

This is a standard application of Baire\'s Theorem (BCT3):

Assume that indeed

(0,1)Q=nNUn,

where Un open dense in [0,1], and (0,1)Q={qn}nN.

Now define

Vn=Un{qn}.

Then clearly, Vn open and dense in [0,1], and

nNVn=,

which contradicts Baire\'s Theorem, as [0,1] is a complete metric spaces, and an intersection of countably many open and dense subsets of it is empty.

If you do not want to use Baire\'s Theorem, let Un\'s and Vn\'s defined as above. As V1 is open, then there is a non-empty interval (a1,b1), such that

(a1,b1)V1(0,1).

Choose now a closed interval [c1,d1](a1,b1), with d1>c1.

Next, as V2 is open and dense, then there is a non-empty interval (a2,b2), such that

(a2,b2)V2(c1,d1),

and choose a closed interval [c2,d2](a2,b2), with d2>c2.

Recursively we can thus obtain two sequences of intervals (an,bn), [cn,dn], nN, such that

[cn+1,bn+1](an+1,bn+1)Vn+1(cn,dn).

But nNVn= implies that nN[cn,dn]=, which is a contradiction.