infinity in the extended reals

Question

I learned about the extended reals a month ago in my measure theory class, and while the syntax for handling $+\infty$ or $-\infty$ is well defined, the mathematical properties of $+\infty$ in the extended real numbers hasn't been distinguished from the $+\infty$ that is derived from the naturals and used for sequences among other things.

To make things precise I shall denote $\infty_{\mathbb{N}}$ to denote the infinity used for sequences and $\infty_{\mathbb{\bar{R}}}$ for the infinity used in the extended reals.

Now, I shall state what is common to both of them:

1. They are both non-computable.
2. The notion of a definite length or magnitude can't be made precise for either of them.
3. We can define a syntax of arithmetical rules that's similar for both of them. Ex:

$\forall x,y \in \mathbb{R}, \infty_{\mathbb{\bar{R}}} +x = y+ \infty_{\mathbb{\bar{R}}}$

$\forall x,y \in \mathbb{N}, \infty_{\mathbb{N}} +x = y+ \infty_{\mathbb{N}}$

However, $\infty_{\mathbb{N}}$ for sequences like $(x_n)_{n=1}^N$ where $x_N = \sum_{n=1}^{N} 2^{-n}$ can be defined within the context of a while loop that never ends. This can be made clear by the following pseudo-code:

while $N \in \mathbb{N}$:

$x_N = \sum_{n=1}^{N} 2^{-n}$

$N:=N+1$

This loop, which never terminates, is what computer scientists call an infinite loop. Now, what happens if $u_n = n \pi$ and $y_n = 2n$.

At every increment of the loop, $y_n$ is an integer whereas $u_n$ is irrational. Yet, an analyst would say that $\lim_{n\to\infty} u_n = \lim_{n\to\infty} y_n$. However, given that $\infty_{\mathbb{N}}$ originates from the positive integers does $\infty_{\mathbb{N}}$ simply refer to the set of naturals whose magnitudes we haven't bother to make precise?

Likewise, an analyst would say that:

$\lim_{n\to\infty} \frac{n+1}{n+2}=1$

but it would be more precise to say that $\lim_{n\to\infty} \frac{n+1}{n+2}=1 +\epsilon$ where $|\epsilon|$ is the smallest positive rational that's greater than zero. I think the same could be said about all other limits using $\infty_{\mathbb{N}}$. The epsilon almost vanishes but it doesn't. Unless $\infty_{\mathbb{N}}$ is a mathematical object that doesn't originate from the positive integers.

But if that's the case, why have mathematicians defined arithmetical operations for $\infty_{\mathbb{N}}$?

Now, as I have tried to reflect on $\infty_{\mathbb{\bar{R}}}$ it appears that there are problems with $\infty_{\mathbb{N}}$. These problems aren't unrelated but in my opinion must be resolved first. In fact, I don't think they are separate questions.

But, if it's possible to resolve the questions I've asked, I think it's natural to ask a further question. How is $\infty_{\mathbb{\bar{R}}}$ semantically different from $\infty_{\mathbb{N}}$? By this, I mean, what are its precise mathematical properties that distinguish it from $\infty_{\mathbb{N}}$?

Right now, if I go through the list of arithmetical operations that are defined for $\infty_{\mathbb{\bar{R}}}$, it appears that these aren't sufficient to show that $\infty_{\mathbb{N}} \neq \infty_{\mathbb{\bar{R}}}$.

Note 1: 5xum points out that my notion of a smallest positive rational that's greater than zero is clearer within the context of non-standard analysis, and not within standard analysis. As a result of this discussion I shall try to learn non-standard analysis.

Note 2: This is not clear in the two answers so far but Stefan Perko shows in the comments below that $\infty_{\mathbb{N}} = \infty_{\mathbb{\bar{R}}}$ for the following reason:

Given that $\infty :=\infty_{\mathbb{\bar{R}}}$ is the greatest element of $\mathbb{\bar{R}}$. Assume $\infty_{\mathbb{N}} \in \mathbb{\bar{R}}$ such that $\infty_{\mathbb{N}} \geq n$ for all naturals $n$. Then $\infty_{\mathbb{N}} \geq x$ for all reals $x$ by the Archimedian property. So it must be the same as $\infty$ (if it was a real there would be a greater real).

Answer 1

The thing is that $\infty_{\mathbb N}$ isn't really an element of any set. When talking about $\infty$ in sequences, saying $$\lim_{n\to\infty} a_n=L$$ simply means that for any $\epsilon>0$, there exists some $N\in\mathbb N$ such that $|a_n-L|<\epsilon$ if $n>N$.

So there is no $\infty_{\mathbb N}$ to speak of at all. The $\infty$ there is simply a symbol that describes a behavior, while in the extended reals, $\infty$ is an actual element of the set.

Answer 2

$\infty_{\mathbb{N}}$ and $\infty_{\overline{\mathbb{R}}}$ are not different notions - they're both just symbols standing in for "bigger than anything". Either can be used as an element of a set (for example, in the extended reals) but neither can be treated as "really" a number. For example, in the extended reals, no one ever tries to define $\infty - 1$, or (even worse) $\infty - \infty$. So think of the extended reals as the reals together with a couple of random objects - say, a horseshoe and a chipmunk - and a list of rules for what to do if you run into one of them in your calculations.

Another thing: "Origin" isn't really a thing to most mathematicians. The fact that your $\infty_{\mathbb{N}}$ is derived from the naturals and $\infty_{\overline{\mathbb{R}}}$ from the reals doesn't mean they're different. For example, $\pi$ can be derived from a circle or from the sum $\sum_{i=0}^{\infty}\frac{4(-1)^i}{2i+1}$; that doesn't mean we have two different versions of $\pi$, just two different ways of talking about the same thing.

And finally: $\infty_{\mathbb{N}}$ isn't just a stand-in for a natural we haven't specified yet - it's a very precise notion. $\lim_{n \to \infty}f(n) = L$ means "as $n$ grows without bound, $f(n)$ gets arbitrarily close to $L$". That "without bound" is $\infty_{\mathbb{N}}$ - it's not saying the magnitude we're interested in hasn't been specified, it's saying that there is no fixed magnitude.