# Cantor’s Lemma¶

A (closed) interval, $$[a,b]$$ is a set of numbers: $$\{x: a\leq x\leq b\}$$. Intervals can contain other intervals: $$[a,b]\subset [c,d]$$ if and only if $$a\leq c$$ and $$d\leq b$$.

The next lemma allows defining limits without knowing what they are, or showing that a sequence of intervals has a point in common.

Assume $$a_n\leq b_n$$ are two sequences, $$a_n$$ monotonically increasing and $$b_n$$ monotonically decreasing. Also assume that the sequence $$b_n-a_n$$ converges to $$0$$.

These sequences can be thought of a sequence of intervals. The monotonicity requirements mean that each interval contains all following ones. The convergence requirement says that the limit of the lengths of the intervals is $$0$$.

Since $$b_n$$ is decreasing, $$b_n\leq b_1$$, and so $$a_n\leq b_1$$. Therefore, $$a_n$$ is bounded, and so has a limit, $$a$$. For similar reasons, $$b_n$$ is bounded and so has a limit, $$b$$.

By the results on the limits of addition and negation, $$b-a=0$$, so $$b=a$$. Call the limit $$c$$. Limits for monotnic sequences are the bounds, so $$a_n\leq c\leq b_n$$ for every $$n$$.

If $$d<c$$ the because $$c$$ is the least upper bound on $$a_n$$, for some $$n$$, $$d < a_n$$, and therefore $$d\notin [a_n,b_n]$$. Similarly, if $$d>c$$, some $$b_n$$ is smaller than $$d$$, and therefore $$d\notin [a_n,b_n]$$.

The only number that is common to all intervals is $$c$$.

This result is known as “Cantor’s Lemma”.

Here is an application of this lemma: if $$f:N\to [0,1]$$ is a function there is a number $$x$$, $$f(n)\neq x$$ for every $$n$$.

Define a sequence by induction.

$$a_0=0$$ and $$b_0=1$$.

look at $$f(n)$$. For $$n+1$$, define $$c_n=b_n-a_n$$. $$[a_n,b_n]$$ is the union of three intervals:

• $$[a_n,a_n+c_n/3]$$
• $$[a_n+c_n/3,a_n+2c_n/3]$$
• $$[a_n+2c_n/3,b_n]$$

Not all of them can contain $$f(n)$$, because the intersection of all three is empty. For $$[a_{n+1},b_{n+1}]$$, choose the smallest that does not contain $$f(n)$$

Note that by induction, $$0<=b_n-a_n = 3^{-n}<1/n$$ and so the conditions for Cantor’s Lemma are fullfilled.

Let $$c$$ be the common limit. By the $$n+1st stage, :math:c!=f(n)$$. This proves the result.

This result is often summarized as “there are more real numbers between $$0$$ and $$1$$ than