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? $\aleph_2$? $\aleph_0$?

What do you mean, they aren't the same? $\lim_{x\to 0} 1-x=1=\lim_{x\to 0}1-x^2$, but it doesn't mean $1\neq 1$, just that these functions converge at different rates. — Thomas Andrews, Sep 17 '14 at 18:13
See Galileo's paradox. Notice also that inifinty in calculus is potential inifinty, not actual infinity. — Jean-Claude Arbaut, Sep 17 '14 at 18:23
Another related thread: http://math.stackexchange.com/questions/90758/the-aleph-numbers-and-infinity-in-calculus — Asaf Karagila, Sep 17 '14 at 18:54
I feel like this is best summed up by your first sentence: mixing calculus and set theory results in a big mess. — Teepeemm, Sep 17 '14 at 19:16
@Will: A field of really valued germs? Sounds like biological warfare to me. — Asaf Karagila, Sep 17 '14 at 19:22
@asaf, it was actually really valuable gems. I can see how confusion might arise. — Will Jagy, Sep 17 '14 at 19:30
Your argument could be also used to show that there are different zeros, some bigger than others: we know that, as $x\to 0$ both $f(x)=x^2 $ and $g(x)=x^3 $ tend to zero, but $f(x)/g(x)$ don't tend to $1$, hence they tend to "different" zeroes... — leonbloy, Sep 17 '14 at 19:59

score 22 · Accepted Answer · edited Apr 13 '17 at 12:19

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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.)

edited Apr 13 '17 at 12:19

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answered Sep 17 '14 at 18:17

Frunobulax

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Not sure I'd say that with absoluteness. $\infty$ in calculus can be defined as an entity. Adding $\pm\infty$ to the real line is not nonsense, and the sense in which $\lim_{x\to\infty} x^2=\infty$ can be formalized in terms of a metric on the extended real line. – Thomas Andrews Sep 17 '14 at 18:22
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(And non-standard analysis is quite different than adding $\pm\infty$ to the real line... – Thomas Andrews Sep 17 '14 at 18:23
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I know all this, but my answer was neither meant as a scientific paper nor did I want to demonstrate that I'm smarter than others. I just wanted to help someone who was obviously confused. I know from experience that it is sometimes helpful to not tell the whole truth at once. – Frunobulax Sep 17 '14 at 18:25
@Frunobulax I think the answer is generally helpful except for the first sentence. The symbol of infinity (and therefore its concept, in one form or another) certainly has concrete use in calculus, but interpreting what it means compared to other notions of infinity is certainly the trick to be accomplished. – rschwieb Sep 17 '14 at 18:33
@rschwieb: I've tried to phrase my answer more carefully without giving up its original pedagogical intent. – Frunobulax Sep 17 '14 at 18:43
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I still think $\pm\infty$ are easily seen as existing. For example, if you define the real numbers as Dedekind cuts, you could instead define the extended real line as subsets of $\mathbb Q$ and that satisfy the order properties of a cut without requiring either the upper bound nor that the set be non-empty. Then $\emptyset=-\infty$ and $\mathbb Q=+\infty$ and you define the real numbers as the "cuts" which are non-empty and have an upper bound. Certainly, no less "real" than $\aleph_3$. – Thomas Andrews Sep 18 '14 at 01:29
Perfect. Saying that "$= \infty$" is an abbreviation for a longer statement is exactly the correct way to think about infinity in this context (and what I was taught that it formally meant). Unless you know what you are doing, any other interpretation will lead you into trouble. – jwg Sep 18 '14 at 10:12
@jwg: I strongly disagree, in part. Saying that it is an abbreviation is the correct way to translate the mathematical ideas into precise statements (unless you know what you are doing) -- but it is very much the wrong way to think about it. – Sep 18 '14 at 17:56

Thomas Andrews · Answer 2 · 2014-09-18T01:52:54.487

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:

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

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.

Who says infinity isn't a value? It is not a real number, but it is a value, or you can make it one. Convergence to infinity is exactly the same as convergence to any other value, it turns out. Infinity in this sense, though, is entirely different from infinity in the sense of set theory. — Thomas Andrews, Sep 17 '14 at 18:26
@ThomasAndrews It's not exactly true I think. Unless you compactify the reals to get the extended number line, infinity is not really a number, it's a way to tell you can pick an arbitrarily large number in some limiting process. — Jean-Claude Arbaut, Sep 17 '14 at 18:29
@Jean-ClaudeArbaut Saying it is a value is not the same thing as saying it is a number. — Thomas Andrews, Sep 17 '14 at 18:38

score 9 · Answer 3 · 2014-09-18T00:04:16.403

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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.

edited Sep 18 '14 at 00:04

answered Sep 17 '14 at 18:27

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I think that last paragraph is nonsense. $O(x)$ is not a function, nor is $O(e^x)$. $O$ is something else entirely. Also, $0=O(x)$. – Thomas Andrews Sep 17 '14 at 19:47
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I second that. "Approximately $+\infty$" is gibberish. – TonyK Sep 17 '14 at 19:47
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Neither is $+\infty$ a number that one can approximate. – user164587 Sep 17 '14 at 19:48
@user164587 $+\infty$ can be made a value (though not a number) and we can define "approximate" in some sense fairly directly, but the above is still not very sensible. (Basically $\mathbb R\cup{\pm\infty}$ is the same, topologically, as $[-1,1]$ and we can talk about functions/series/whatever which lead to better approximations of $1$ in that domain.) – Thomas Andrews Sep 17 '14 at 19:51
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+0. I feel like describing something as "approximately $+\infty$" would be confusing to the target audience of this post. I give +1 for the content but -1 for the last paragraph. – user157227 Sep 17 '14 at 19:53
BTW, I +1d this answer, despite my comments, but there have been a spate of uncommented downvotes on this and my answer. – Thomas Andrews Sep 17 '14 at 20:03
@user164587 My old math teacher would always say infinity is as large as you like. Or any sufficiently large number. Seems like that is an approximation. – Michael Sep 17 '14 at 20:31
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@TonyK: "Approximately $+\infty$" is no less gibberish than "approximately zero". Aside from metrics, any notion of "nearness" one can apply to $0$ apply just as well to $+\infty$. e.g. topologically speaking, $+\infty$ has open neighborhoods, which are the intervals $(N, +\infty]$. From the standpoint of nonstandard analysis, the halo about $+\infty$ consists of $+\infty$ and all positive unlimited hyperreals, just like the halo about $0$ consists of all infinitesimals. – Sep 17 '14 at 20:35
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... and (also @user157227) IMO the opening poster is clearly developing a concept of the behavior of a function at $+\infty$ as being an actual mathematical object one can manipulate. In the grand scheme of things, $O(x)$ approximates $+\infty$ in conceptually the same way that $1 + o(x^{-1})$ approximates $1$ as $x \to +\infty$: mentally placing $O(x)$ as being "near $+\infty$" is where it belongs, and lets one directly express the OP's intuition of $O(e^x)$ being a larger thing than $O(x)$. – Sep 17 '14 at 20:39
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In fact, nonstandard analysis does decent job of capturing asymptotics as being actual numbers: the fact $x \in o(e^x)$ translates to $e^H / H$ being transfinite for all transfinite $H$. For well-behaved functions, we can reasonably replace the standard concept of asymptotic behavior with the notion of simply plugging in a transfinite number. – Sep 17 '14 at 20:45
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Yes, there is a sense in which "approximately infinite" makes sense. Still, it is really wrong to say "$O(x)$ is larger than any real number." First, because $0=O(x)$ and $-x=O(x)$, but also because $O(x)$ is not a value, so you can't say it is "larger than any value but smaller than $+\infty.$" That is meaningless. – Thomas Andrews Sep 17 '14 at 21:52
@Thomas: Ah I was thinking $\Theta$ in my head; I've fixed the notation. – Sep 17 '14 at 21:57
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But it's still very much an abuse of language. $\Theta(x)$ is a collection of functions, so what does it mean that it is "greater than" any number? It's not even clear what you'd mean if you said $f(x)=x$ is "greater than any real number." That's a severe abuse of language, unless you define what you mean by "greater than." At the very least, it requires comment that you are using some unusual notion of "greater than." – Thomas Andrews Sep 17 '14 at 23:10
@Thomas: It's an intuitive notion one makes precise as needed. Some particular ways to do so are that for all numbers $r$ and $f(x) \in \Theta(x)$, you have $r < f(x)$ for all sufficiently large $x$. Or $r = o(x)$. Or $O(r) \subseteq O(x)$ but not $O(x) \subseteq O(r)$. One can even define a general partial ordering based: $f \leq g$ iff $f(x) \leq g(x)$ for all sufficiently large $x$. Similarly, a partial ordering on $\Theta$ equivalence classes is given by $f \leq g$ iff $f \in O(g)$. But underlying all of this is simple ideas like "$e^x$ is asymptotically greater than $x$". – Sep 18 '14 at 00:10

rschwieb · Answer 4 · 2014-09-17T19:29:38.960

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.

Greatly interested in knowing what the issue with the post is that drew a downvote. — rschwieb, Sep 17 '14 at 19:29
There were a lot of downvotes to other posts, also without comment. — Thomas Andrews, Sep 18 '14 at 01:18

score 2 · Answer 5 · answered Sep 17 '14 at 19:01

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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?

answered Sep 17 '14 at 19:01

GEdgar

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The answer is yes @GEdgar. $g(x)=\log (f(x))$ is such an example. Though I would assume that if $f$ is not differentiable then this is somewhat problematic. – Oria Gruber Sep 17 '14 at 19:05
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And the point is: "Is there a smallest infinity in calculus?" is not a nonsense post, by someone without a clue. – GEdgar Sep 17 '14 at 19:18
@GEdgar. Nice point! And $\log(\log f(x) )$ is slower than $\log f(x)$. – mike Sep 17 '14 at 19:30
Note that log(f(x)) is not defined for all functions f(x). Therefore, while it may well be that there is no smallest infinity in calculus, citing log(f(x)) is insufficient to establish this unless you also establish that no f(x) for which log(f(x)) is undefined can be the smallest infinity in calculus. – Matthew Najmon Sep 19 '14 at 05:33
For functions going to $+\infty$, just use $\log(f(x) \vee 1)$. It still goes to $+\infty$, and is slower. – GEdgar Sep 19 '14 at 13:36

user164587 · Answer 6 · 2014-09-18T01:23:06.773

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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 $x=a$.

edited Sep 18 '14 at 01:23

answered Sep 17 '14 at 19:53

user164587

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That's a bit risky. Divergent is, in some places, defined as simply "not convergent," in which case converging to infinity is the just a subset of the divergent cases. For example, wikipedia has the series $1+(-1)+1+(-1)\dots$ as divergent. http://en.wikipedia.org/wiki/Divergent_series – Thomas Andrews Sep 18 '14 at 01:17
OK, "growth without bound", then. Although we are talking about limits of functions, not sequences. If the function is not continuous, the limit doesn't exist at all. – user164587 Sep 18 '14 at 01:22

score 0 · Answer 7 · answered Sep 17 '14 at 18:43

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.

There is no smallest infinity in calculus?

7 Answers7