Real analysis question

[unborn meadow]

Hello, I need help to understand something in the context of proofs about sequences in real analysis. Maybe it's just a stupid question but I need to ask it xD.

We set for all epsilon > 0, exists N (natural number) such that n > N for all n (natural numbers). Then sometimes we get to inequalities that look like this:

| 1 / (n +1) | < epsilon

And we give explicit formulas to construct epsilon, like N = 1 / epsilon.

My question is, why is it ok to say N = 1 / epsilon?
For example, if epsilon = 32, then N = 1 / 32, but N is natural, so it can't be, because I assume I need to prove it for all epsilon and now I'm only proving it for epsilon <= 1, and ignoring epsilon > 1.

So I don't know if the author is being careless or abusing notation for something, but I don't know what something is.
Are we replacing N for a real number? Are we applying a function like ceil(N) = 1/epsilon? Am I misunderstanding something?

I'm not trying to be pedantic, I just want to make sure I understand the implications correctly since I study in a online uni, and my professor is a bit unavailable... xD

Thank you in advance!

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[neat tusk]

such that n > N for all n
This is saying that there exists a natural number less than all natural numbers. That's not true. In fact, it's trivially untrue. No element of any set can be less than every element of that set because it can't be less than itself.

[unborn meadow]

I messed up when writing it, I meant that for all e > 0 there exists a N such that for all n > N (not n > N for all n).

[neat tusk]

Yes, I understand. You were describing the definition of a limit at infinity.

I think specifically the limit of a sequence, which is why you're restricting to natural numbers, because those are the indices of the elements of a sequence.

I see your actual question now.

And it should technically be an inequality.

I mean, this whole thing is inequalities.

$\lim_{n \to \infty} a_n = L \Leftrightarrow \forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \in \mathbb{N} : n > N \Rightarrow |a_n - L| < \epsilon$

[tulip ermine]

Techie Literate

[neat tusk]

So for $a_n = \frac{1}{n + 1}$, we would have $\lim_{n \to \infty} a_n = 0$.

[tulip ermine]

Techie Literate

[neat tusk]

@unborn meadow Sorry, are you following?

[unborn meadow]

yes sorry, I was not looking at discord, I'm following for now, thanks

I have to learn latex xD

So, if we find that $N = \frac{1}{\epsilon}$, since $\forall \epsilon > 0 \exists N in \mathbb{N}$, we should be able to say, for every epsilon I can think I can get a N. But in this case, since N is natural, and epsilon is real, I cannot assign a real value to a natural variable.

[tulip ermine]

Sundered

[neat tusk]

I think the shortest version here is that N can be in R.

[unborn meadow]

I was trying the latex xD

[neat tusk]

Because all you require is n > N, and for every element of R there exists an element of N which is larger.

$\forall x \in \mathbb{R} \exists n \in \mathbb{N}: n > x$

[tulip ermine]

Techie Literate

[unborn meadow]

archimidean property?

[neat tusk]

Uh, maybe?

This actually seems simpler.

The point is just that you're looking for a lower bound on n.

And if there's a lower bound on n, and n is natural, then there's a natural lower bound on n.

Or, wait.

One modification.

We need n >= N.

I think.

Or, wait, no.

My mistake.

[unborn meadow]

In my book it uses n >= N but on internet I've seen n > N so I thought that maybe it doesn't matter, but it was a question I had too

[neat tusk]

No, it's my mistake. Either works.

Maybe you need an explanation of the intuition being captured here?

[unborn meadow]

hmm what I don't really understand is the justification for saying that N is natural and then saying N = 1/epsilon, if when I set epsilon = 32, then I have N = 1 /32 which is not a natural* number. But I think I understand what you mean.

N can be in R you said, so it makes sense, you can compare a natural to a real.

I wanted to gain intuition with this I guess, because nobody seems to explain this

[neat tusk]

Okay, so.

The idea of a limit of a sequence is that we can get as close as we like to the limit by going far enough in the sequence.

Epsilon and N exist to formally codify the notions of "as close as we like" and "far enough", respectively.

[unborn meadow]

I have to go now, I will read later, thank you btw

[neat tusk]

I mean, there's not much else to say unless you have questions.

[floral vigil]

@unborn meadow

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[unborn meadow]

+close

[floral vigil]

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