A few proofs that any open interval (x,y) contains ...



A few proofs that:

Any open interval (x,y) contains infinitely many rationals.

Proof 1.

Lemma: For any n in N we can find n rationals in (x,y).

We’ve shown that (x,y) contains a rational say r1. Consider (r1, y) . This then contains a rational r2. Continuing in this way, the interval (ri, y) will contain a rational ri+1. We repeat this process until we have n rationals QED lemma. Note that this is really a sloppily written induction proof.

It’s tempting to stop right here but the main proof is not quite done.

Now if the number of rationals in (x,y) is finite, say K, we can find K+1 rationals using the lemma. C!

Proof 2.

We divide the open interval (x,y) as follows:

I0 = (L0,R0) where L0 = x and R0= L0 +(y-x)/2

I1 = (L1,R1) where L1 = R0 and R1= L1 +(y-x)/4

…

Ik = (Lk,Rk) where Ln = Rn-1 and Rn= Ln +(y-x)/2n+1

…

This infinite (but countable) collection of intervals consists of disjoint open subsets of (x,y) . Proof of this left to you. (Note that there are points in (x,y) NOT in any of the Ik.)

But each of these Ik contain a rational number. Therefore we have an infinite set of rationals in (x,y).

A subtle objection to this proof is that it uses the Axiom of Choice. (At this stage in your mathematical career, you should probably ignore this criticism.)

Proof 3.

Could there be only a finite number of rationals in (x,y)? Suppose there were, then let m=min{rationals in (x,y)}. (I’m using fact that every non-empty finite set contains a minimum. Can you prove that? Why does it apply?) Then (x,k1) doesn’t contain a rational – CONTRADICTION!

Proof 4.

Let r be a rational in (x,y) and let s be a rational in (r,y). The consider:

r1= r+(s-r)/2

r2= r+(s-r)/4

rn= r+(s-r)/2n

The ri are an infinite set of rationals in (x,y)

Proof 5.

Find a rational r in (x,y). Then consider: rn= r+1/n. For some N, n>N => r+1/n is in (x,y). Thus let S={r+1/n : n>N}. This is an infinite set of rationals in (x,y).

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