Theorem on uniqueness of limits

1. Introduction

This web page gives a very basic (almost obvious) theorem that seems at first to be uninteresting, but is in fact one of the keystones that the theory we will develop relies on. Also, together with later material on subsequences, it is one of the key methods for proving a sequence doesn't converge.

As with all the other key definitions and results you should at a minimum learn the statement of this theorem, and ideally learn the proof too.

2. Theorem on uniqueness of limits

The theorem on the uniqueness of limits says that a sequence ( a n ) can have at most one limit. In other words, if a n l and a n m then l = m .

Theorem on Uniqueness of Limits.

Suppose ( a n ) is a sequence and l , m are both limits of the sequence ( a n ) as n . Then l = m .

As a consequence, to show that a sequence ( a n ) does not converge to some number l (such as 1 2 say) it suffices to show that does converge to a different number such as 1 3 . This idea is very simple (almost obvious, in fact, but obvious is a dangerous word as occasionally obvious things turn out to be false). We will improve upon this idea considerably later on when we discuss subsequences so I don't provide any examples here at this stage.

Proof.

Remark.

As before, comments written in this font at a smaller size are not part of the proof but comments indicating some feature of the proof or how I was thinking about it. In this proof we'll assume that ( a n ) converges to both l and m and assuming that l m we'll get a contradiction. Note then that we can assume the following statements.

l m

ε > 0   N   n   ( n N | a n - l | < ε )

ε > 0   N   n   ( n N | a n - m | < ε )

Remark.

To make use of these last two statements we need to give them positive values of ε . Since l m and we are interested in when the sequence settles down about l or m away from the other, it makes sense to try the value ε = 1 2 | l - m | . (Other suitable guesses might be ε = 1 3 | l - m | , or ε = 1 4 | l - m | , etc. These all work too.)

Subproof.

Let ε = 1 2 | l - m | .

From the convergence of ( a n ) to l , let N 1 satisfy n   ( n N 1 | a n - l | < ε ) .

From the convergence of ( a n ) to m , let N 2 satisfy n   ( n N 2 | a n - m | < ε ) .

Remark.

Note that there is no reason to expect that N 1 = N 2 . So we should assume that these could be different, by giving them different names. Now, both our statements about N 1 and N 2 start n   and they are interesting when n N 1 and n N 2 , respectively. So we take n to be the larger of N 1 , N 2 .

Let n = max ( N 1 , N 2 ) .

Then | a n - l | < ε and | a n - m | < ε .

So | l - m | | a n - l | + | a n - m | < 2 ε , by the triangle inequality.

But this means that | l - m | < 2 ε = 2 1 2 | l - m | = | l - m | , which is impossible.

This completes the proof.