A Kick in the Discovery

We meet again in our real analysis series! In part 1 we learned about the supremum and infimum and then in part 2 we learned about sequence limits. Part 3 has been broken up into two parts: 3.1 is on the triangle inequality and 3.2 (which is this article) is building on what we did with limits last time. Here, we will practice writing proofs for limits by showing that when a sequence has a limit, it must be unique.

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Theorem (Sequence Limits Are Unique): Let be a convergent sequence. Then, is unique.

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Scratch Work: It might make intuitive sense that limits are unique, since we require that gets arbitrarily close to its limit and if had two limits it shouldn’t be possible for to simultaneously get close to two different numbers! This is the idea.

Let’s assume that has two different limits: and and try to get a contradiction.

Let’s pause for a moment and think about what our first step might be. We are saying that and simultaneously. This means, at some point, will be arbitrarily close to both and But this makes no sense if since at some distance scale they will be far apart. Maybe we can use the distance between and to get our contradiction.

Recall the definition of a sequence limit (it’s here or in the footnote)¹ states that for all there exists an blah blah blah… Let’s use this. Since if we subtract them while and adding and subtracting we have

It’s now prime time to use the triangle inequality!

But, since and there exists and such that

for all

and

for all

By letting we have both inequalities hold for all Thus,

for all

This implies that What???? Why’s that??? I hear you ask. Let’s prove it!

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Lemma (Analyst’s Proof that x = y): If for all then

Proof of Lemma: Let and Assume for the hope of a contradiction that, Then, for we have Our contradiction.

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Thus, it must be that

Before we write out our proof formally, question: did we really need to do a proof by contradiction and our assumption that

No, we didn’t.

This speaks to something that happens often when figuring out proofs. A proof by contradiction gives us a place to start; however, sometimes we don’t need to do a proof by contradiction. This is one of those cases. Challenge: Write out your own proof without the need for a contradiction using Lemma (Analyst’s Proof that x = y), then continue.

Proof that sequence limits are unique:

Let and let and By definition, there exists and such that,

for all

and

for all

Let then for all both inequities hold. Thus,

for all and by Lemma (Analyst’s Proof that x = y) we conclude Therefore limits are unique.

Closing Remarks

For some of you reading this (thank you first of all) you might have found it intuitive that limits are unique. But when building a theory, we cannot take anything for granted. Also, it gave us some practice proving stuff using the triangle inequality!

Also, for those who read part 2, recall we showed that the sequence doesn’t converge. A quicker proof than the one we showed before, will use that limits are unique and the idea of a subsequence. I leave this as something to think about in much the same way a tv writer leaves by writing TO BE CONTINUED at the end of a show for you to ponder.

In the meantime, keep practicing, keep learning, keep questioning, and as always keep discovering!

Footnote:

1. Definition (Sequence Convergence): Let be a sequence, then we say the limit of as approaches infinity, (or simply the limit of ) equals if for all there exists an such that for all
   Or, for those who love symbols:
   We say if
   
   *There are some other notations that are used instead of such as: *
   If the limit of exists, then we say that converges. ↩︎