Proving the Limit of a Sequence is Unique: Real Analysis Class

[B3NR4Y]

I'm in a real analysis class, and I love the material. But something that is coming up in homeworks worries me about my future math career, I find myself being tasked with proving things and using the internet to help me, finding the answer and completely understanding it, but that's not the problem. I have no clue where they got that from. It seems arbitrary when they pull some of this stuff out. For example:

I was told to prove the limit of a sequence is unique. I correctly started by assuming it isn't, which implies (by the definition of a sequence) x_n→L means ∀ε>0 ∃N∈ℕ such that n≥N ⇒ | x_n- L | < ε
x_n→M means ∀ε>0 ∃O∈ℕ such that n≥O ⇒ | x_n- M | < ε

The first thing they did was let n = max {N,O}. I understand why, the interesting stuff in the definitions happens when n ≥ N or O, so taking the max of these two ensures the interesting stuff is happening. But I don't think I could have come up with that myself. I'd like to believe i would, but I honestly don't know if I would. Since n is greater than or equal to N or O, both | x_n-L | < ε and | x_n-M | < ε are true, and I'm okay with this (it just comes from the definition), using the triangle inequality we can see that
| L - M | (which should be zero) ≤ | x_n - L | + | x_n - M | < 2ε
Now here's another part that messes me up, they chose [itex] \epsilon = \frac{| L - M |}{2} [/itex] at the very beginning of the proof! Which seemed arbitrary to me, so I ignored it until the end and this came up. I realized why they did that because it said | L - M | < | L - M |, which is a contradiction! Therefore the ε we chose has to equal zero, which is only possible if | L - M | = 0, therefore proving L = M. But how did they have such incredible foresight to see that [itex] \epsilon = \frac{| L - M |}{2} [/itex] at the very beginning? Just experience or is everyone in math a genius and I should just stick with physics (I'm dual majoring)?

 

[micromass]

Absolutely, absolutely stop looking up things from the internet. It will not help you at all. And the questions you ask here are typical from those who look up the proofs and don't find it themselves.

Nobody has had the forsight to see that ##\varepsilon = |L-M|/2##. In a proof, a choice of ##\varepsilon## happens in the beginning. But when finding a proof yourself, the right choice of ##\varepsilon## happens at the very end! This is why reading a proof without trying it yourself can be harmful, because it does not give you any idea where things came from.

 

[Infrared]

Since you don't quite get the intuition behind how the proof came about (yet!), here's the train of thought that might lead you to a similar solution:

We want to show that if [itex]x_n\to L[/itex] and [itex] x_n\to M[/itex], then [itex] L=M[/itex]. So, we look at the difference [itex] |L-M|[/itex]. To estimate this difference, we notice that [itex]L[/itex] and [itex]M[/itex] are "close" to [itex] x_n[/itex] for large [itex]n[/itex]. Since both [itex] |L-x_n|[/itex] and [itex] |M-x_n|[/itex] are small for large [itex] n[/itex], it makes sense to use the estimate [itex] |L-M|=|L-x_n+(x_n-M)|\leq |L-x_n|+|M-x_n|[/itex]. Suppose that we want to make the RHS smaller than some arbitrary positive number [itex]\varepsilon[/itex]. The easiest way to do this is to force both [itex]|L-x_n|[/itex] and [itex]|M-x_n|[/itex] to be smaller than [itex]\varepsilon/2[/itex]. Since we can find natural numbers [itex]N_1[/itex],[itex]N_2[/itex] (side note: don't name your bound "O" in this circumstance) such that [itex] |L-x_n|<\varepsilon/2[/itex] for [itex]n\geq N_1[/itex] and [itex] |M-x_n|<\varepsilon/2[/itex] for [itex]n\geq N_2[/itex]. With this in mind, [itex] |L-M|\leq |L-x_n|+|M-x_n|<\varepsilon/2+\varepsilon/2=\varepsilon[/itex]. Since [itex]|L-M|[/itex] is nonnegative and smaller than any positive real number it must be zero.

This condenses into the following proof:

Let [itex]\varepsilon[/itex] be an arbitrary positive real number. Since [itex]x_n\to L[/itex] and [itex]x_n\to M[/itex], there exist natural numbers [itex]N_1,N_2[/itex] such that [itex]|x_n-L|<\varepsilon/2[/itex] for [itex] n\geq N_1[/itex] and [itex]|x_n-M|<\varepsilon/2[/itex] for [itex] n\geq N_2[/itex]. Fix a natural number [itex]k\geq \max (N_1,N_2)[/itex]. Then, [itex] |L-M|\leq |L-x_k|+|M-x_k|<\varepsilon/2+\varepsilon/2=\varepsilon[/itex]. Since [itex]|L-M|<\varepsilon[/itex] for any [itex]\varepsilon>0[/itex], we conclude that [itex] |L-M|=0[/itex] and hence [itex] L=M[/itex].

Note that the nicety with choosing [itex]\varepsilon=\frac{|L-M|}{2}[/itex] is unnecessary and, indeed, makes the proof more complicated because it forces you into doing a proof by contradiction where such a choice of [itex]\varepsilon[/itex] is positive.

Hope this helps!

 

[B3NR4Y]

Since you don't quite get the intuition behind how the proof came about (yet!), here's the train of thought that might lead you to a similar solution:

We want to show that if [itex]x_n\to L[/itex] and [itex] x_n\to M[/itex], then [itex] L=M[/itex]. So, we look at the difference [itex] |L-M|[/itex]. To estimate this difference, we notice that [itex]L[/itex] and [itex]M[/itex] are "close" to [itex] x_n[/itex] for large [itex]n[/itex]. Since both [itex] |L-x_n|[/itex] and [itex] |M-x_n|[/itex] are small for large [itex] n[/itex], it makes sense to use the estimate [itex] |L-M|=|L-x_n+(x_n-M)|\leq |L-x_n|+|M-x_n|[/itex]. Suppose that we want to make the RHS smaller than some arbitrary positive number [itex]\varepsilon[/itex]. The easiest way to do this is to force both [itex]|L-x_n|[/itex] and [itex]|M-x_n|[/itex] to be smaller than [itex]\varepsilon/2[/itex]. Since we can find natural numbers [itex]N_1[/itex],[itex]N_2[/itex] (side note: don't name your bound "O" in this circumstance) such that [itex] |L-x_n|<\varepsilon/2[/itex] for [itex]n\geq N_1[/itex] and [itex] |M-x_n|<\varepsilon/2[/itex] for [itex]n\geq N_2[/itex]. With this in mind, [itex] |L-M|\leq |L-x_n|+|M-x_n|<\varepsilon/2+\varepsilon/2=\varepsilon[/itex]. Since [itex]|L-M|[/itex] is nonnegative and smaller than any positive real number it must be zero.

This condenses into the following proof:

Let [itex]\varepsilon[/itex] be an arbitrary positive real number. Since [itex]x_n\to L[/itex] and [itex]x_n\to M[/itex], there exist natural numbers [itex]N_1,N_2[/itex] such that [itex]|x_n-L|<\varepsilon/2[/itex] for [itex] n\geq N_1[/itex] and [itex]|x_n-M|<\varepsilon/2[/itex] for [itex] n\geq N_2[/itex]. Fix a natural number [itex]k\geq \max (N_1,N_2)[/itex]. Then, [itex] |L-M|\leq |L-x_k|+|M-x_k|<\varepsilon/2+\varepsilon/2=\varepsilon[/itex]. Since [itex]|L-M|<\varepsilon[/itex] for any [itex]\varepsilon>0[/itex], we conclude that [itex] |L-M|=0[/itex] and hence [itex] L=M[/itex].

Note that the nicety with choosing [itex]\varepsilon=\frac{|L-M|}{2}[/itex] is unnecessary and, indeed, makes the proof more complicated because it forces you into doing a proof by contradiction where such a choice of [itex]\varepsilon[/itex] is positive.

Hope this helps!

This does help! I suppose I could have come up with that myself, some things trip me up at first when I'm working the problem, in the heat of the moment, I forget that if the only condition you impose on ε is that it has to be positive, ε/2 is still a number greater than zero and is still greater than what we said it was greater than. I guess my brain has become lazy as a result of me looking up things when I get stuck. I am going to stop doing that now, as per micromass's suggestion.

Thank you!

 

[FactChecker]

There are some patterns of proof that will come up time after time. Most proofs are derived from those basic patterns. This is one of the basic patterns. Give it some time and you will catch on. It's like cooking. You start with the basic recipe and add spices as needed.