There is no smallest infinity in calculus?

Question

Somewhat of a basic question, but I tried mixing set theory and calculus and the result is a giant mess.

From set theory (assume ZFC) we know there is a smallest infinite cardinal, $\aleph_0$, and that infinite numbers are well ordered, $\aleph_1 > \aleph_0$ etc

Now if we move to the world of calculus, even there, there is a difference between one infinity and the other.

$\lim_{x \to \infty} x = \infty$, and $\lim_{x \to \infty} e^x = \infty$, but they are not the same, you could say that the $e^x$ one is bigger, because $\lim_{x\to \infty} \frac{e^x}{x} = \infty$ as can be shown easily with L'Hôpital's rule.

This leads me to believe that unlike in set theory, in calculus there is no smallest infinity, since if $\lim_{x \to \infty} f(x)= \infty$, then $\lim_{x \to \infty} \log (f(x) = \infty$ but a smaller $\infty$.

So which version is "correct"? Is there really a smallest infinity like in set theory? or we can keep getting smaller and smaller to no end like in calculus? Or both are correct in different context? I'm a bit confused.

Which also leads to another question, when we say in calculus that some limit tends to $\infty$, which $\infty$ are we talking about? $\aleph_2$? $\aleph_0$?

Answer 1

There is no such thing as "infinity" in calculus - at least not as an object that is as "concrete" as, say, the number $42$. Things can get as complicated as you want later on, but in the beginning it might be best to think that "$\infty$" is just a symbol and expressions like "$\lim_{x\rightarrow\infty}x^2=\infty$" have a well-defined meaning in the sense that they are just abbreviations for longer sentences in which no more "$\infty$" will occur.

Infinites in set theory are completely different animals, though. Infinities like $\aleph_3$ are special sets which are "singled out" as yardsticks to measure the size (cardinality) of other sets. So, something like $\aleph_3$ really exists according to the axioms of set theory - as opposed to "$\infty$" which - see above - doesn't. (Or to put it more carefully - which has to be given a specific meaning in order to "exist" in a meaningful way. Others have explained ways to do this already.)

Answer 2

There is a reason set theory does not use the symbol $\infty$ - because it would be very easily confused by the way it has been used in limits and calculus in general for a long time. In particular, limits are talking about something that can best be called "linear infinity" or "topological infinity."

Essentially, we can add "values" $-\infty$ and $+\infty$, to the real line, and give them meaning in the sense of convergence, without making them numbers. There is a deep sense in which convergence to these new values is "the same as" convergence to other values on the real line. But the real key is that these two values are not numbers.

The sequences $\{\frac{1}{n}\}$ and $\{e^{-n}\}$ both converge to $0$ at very different rates, but we don't call those zeros different. How fast something converges can be wildly different, but that doesn't mean that the value to which they converge is different.

There is a sense in which convergence can be faster and slower, but that has nothing to do with infinity. You can do the same example above with:

$$\lim_{x\to 0} x = 0 =\lim_{x\to 0} e^{-1/x^2}$$

More generally, $f(x)\to+\infty$ is exactly the same as $\arctan f(x)\to\frac{\pi}{2}$, and similarly, $f(x)\to-\infty$ is exactly the same as $\arctan f(x)\to-\frac{\pi}2$. So the idea that there are different "values" of infinity in calculus is a bit of confusion. In this sense, convergence to "infinity" is the same as convergence to $1$ of a function with range the interval $(-1,1)$.

One typical way to define the real numbers is to define the notion of a Dedekind cut on the rationals, $\mathbb Q$. A Dedekind cut is a subset $C$ of $\mathbb Q$ with the following properties:

1. $\forall x\in C\,\forall y\in\mathbb Q(y<x\implies y\in C)$
2. $\forall x\in C\,\exists y\in C(x<y)$
3. $C\neq \emptyset$
4. $C\neq \mathbb Q$.

The set of real numbers is defined as the set of all Dedekind cuts.

But if we defined an Extended Cut with just (1) and (2), then $\emptyset$ corresponds to $-\infty$ and $\mathbb Q$ corresponds to $+\infty$. So in this sense, having the extended real line is "simpler" than having just the real line.

The main reason we start with Dedekind cuts is our obsession with the algebraic properties of $\mathbb R$. We can't define addition on the extended real line in a useful way, for example, because of the problem of $+\infty+(-\infty)$. But if we were only interested in the topological properties of the real line (continuity, limits, etc) then the extended real line actually makes more sense, and it has the advantage that it is simpler.

Answer 3

The first notion of infinity in calculus is geometric: the extended real numbers adds two points to the real line: $+\infty$ and $-\infty$ to make the real line a closed interval. This is entirely analogous to adding $0$ and $1$ to the interval $(0,1)$ to make the closed interval $[0,1]$.

There are no 'degrees' of infinity in this sense: there is only one positive infinite extended real number: $+\infty$. And similarly, there is only one negative infinite extended real number.

Your idea about $e^x$ being bigger than $x$ as $x \to +\infty$ is a topic called asymptotics. We can quantify in various senses the rate of growth of functions; a common method is by "big-oh" notation and related things.

But as they relate to $+\infty$, the right intuition is that both $\Theta(x)$ and $\Theta(e^x)$ are larger than any real number, but smaller than $+\infty$. You should think of them as both being approximately $+\infty$, with $\Theta(e^x)$ being a closer approximation.

Answer 4

In this context, $-\infty$ and $\infty$ can be thought of as the first and last elements of the extended real line respectively.

The extended real line has a perfectly well-defined topology: in fact it's a compactification of the real line. So in this sense, these infinities are filling in "holes" in the original topological space. When you discuss limits that diverge to infinity, they fit into limits in the extended real line just fine: those limits are actually converging to a point in the space.

But in elementary calculus, it's not ordinary to introduce the extended real numbers this way. Then the notation using $\pm \infty$ is just used formally to indicate a limit is increasing or decreasing without bound.

They do not arise from cardinality like the cardinal numbers you are talking about. They are labeled this way simply because that's where they're placed in the total ordering of the extended real line.

Answer 5

There is a context in which this question makes sense. That is, given any function $f$ with $\lim_{x \to +\infty} f(x) = +\infty$, must there be another one that goes to infinity more slowly?

Answer 6

Let's nail one thing straight away. $+\infty$ is not a number. $$\lim_{x \to a} f(x) = +\infty$$ is not an equality. It is a notational convention that describes the growth without bound of the function at the point $x=a$.

Answer 7

While as has been mentioned the two kinds of infinity are distinct, it is actually possible to unify them in the surreal numbers, which are an ordered field containing both the reals and all the transfinite ordinals. (Though since the ordinal arithmetic are very far from being field operations, e.g. $1+\omega=\omega\neq\omega+1$, the ordinals in the surreal field have differently defined operations.)

The surreals are defined in terms of certain pairs of subsets of surreals (which is noncircular because the construction is inductive,) so you get representations of the infinite ordinals in a natural way: for instance $\omega=\{1,2,3,...\}$. Then you get numbers like $e^\omega=\{e,e^2,e^3,...\}$. So we could think of $\omega$ as $O(x)$ and $e^\omega$ as $O(e^x)$, and so infinity-as-order-of-growth is a concept the surreals capture. But here $e^\omega$ is not one of the ordinals as embedded in the surreals, and so the ordinals themselves (let alone cardinals) don't capture this notion. So to some extent the idea of transfinite ordinals can be subsumed in the analytic notion of infinity, while the converse is entirely false.