Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For example, when we write $\lim_{x\rightarrow \infty} f(x)$ - which infinity is meant and why? Countable? If uncountable - which and why?

share|improve this question
add comment

5 Answers 5

up vote 49 down vote accepted

There is no need to ask about countable or uncountable infinity. The symbol $\infty$ here is used with the following precise meaning.

Let $f$ be defined on some ray $(a,\infty)$. We say that $$\lim_{x\to\infty}f(x)=A$$ if there exists $A\in\Bbb R$ such that for each $\epsilon >0$ there exists an $M>0$ for which $$x>M\implies |f(x)-A|<\epsilon$$

That is, for any $\epsilon >0$ we're given, we can take $x$ large enough to make $f$ as close as we wish to $A$.

The more "dramatic" (symbolically) $$\lim_{x\to\infty}f(x)=\infty$$ means precisely that

... for each $N>0$ there exists $M>0$ such that $x>M\implies f(x)>N$.

That is, we can make $f(x)$ as large as we want by taking $x$ large enough.

ADD Compare the above to the notation $$(a,\infty)$$ I used in it. We're not wondering about any "infinity" but just about the set of numbers $x>a$. The symbol $\infty$ is convenient and intuitive. Brian uses $(a,\to)$ instead!

share|improve this answer
add comment

None of the above. That's why it neither says $\omega$,nor $\aleph_0$, nor $\mathbf c$. Rather, $\infty$ should be considered as a symbol. There's no infinity really used in the definition: $$\lim_{x\to\infty}f(x)=c\iff\forall\epsilon>0\colon\exists M\in\mathbb R\colon\forall x>M\colon |f(x)-c|<\epsilon.$$

share|improve this answer
1  
I wonder at what time and how did Peter, Hagen and Brian match their versions on this problem...:) –  DonAntonio Apr 24 '13 at 21:34
    
@DonAntonio I guess all three took a decent analysis course –  artistoex Apr 25 '13 at 17:38
    
You think, @artistoex ? –  DonAntonio Apr 25 '13 at 18:27
    
Yes @DonAntonio. It took quite some time for mathematics to figure out just how to handle infinity in a strict manner. Strict meaning propositions involving the concept can be obtained through a deductive process. And if a mathematician is a person who obtains mathematical proposition through a deductive process, then he should of course know the technical sides of the various notions of infinity. So, Peter, Hagen and Brian have their versions matched independently from each other because they are mathematicians. –  artistoex Apr 29 '13 at 9:23
1  
Oh, I see @artistoex . Thanks for the info. –  DonAntonio Apr 29 '13 at 11:56
add comment

The $\infty$ in that limit does not refer to an infinite cardinality at all. The expression

$$\lim_{x\to\infty}f(x)=L$$

is simply an abbreviation for the following statement:

$\qquad\qquad$for each $\epsilon>0$ there is an $x_\epsilon\in\Bbb R$ such that $|f(x)-L|<\epsilon$ whenever $x\ge x_\epsilon$.

As you can see, there is no infinite anywhere in that statement. The ‘$x\to\infty$’ in the limit notation is a reminder that we’re talking about what happens when $x$ is very large; it does not refer to a specific entity $\infty$.

(There is in fact a way to interpret this $\infty$ as a specific entity: one can replace $\Bbb R$ with the so-called extended reals, which include two new points $\infty$ and $-\infty$. In effect this adds an endpoint at each end of the real line. But these objects, despite their standard names $\pm\infty$, are simply points in an extended space, not cardinal numbers that can be used for counting infinite collections.)

share|improve this answer
    
(+1) I seem to recall that not too long ago, a questioner was wondering why "$\infty$" wasn't a limit point of the natural numbers (or something like that). In the comments (under your answer to that question, I believe), you and I both discussed with the OP that, despite the unfortunate similarity in the notations $\lim_{x\to a}$ and $\lim_{n\to\infty},$ they don't mean quite the same thing. I'd love to link to that discussion/answer (for another question), but I can't find it (tried the search bar here, and several google searches, but no dice). Do you know if that still exists on the site? –  Cameron Buie Jun 15 '13 at 18:00
    
@Cameron: I remember the discussion, but not enough to be able to search for it with much chance of success. –  Brian M. Scott Jun 15 '13 at 18:30
    
That's what I suspected would be the case. Thank you anyway! –  Cameron Buie Jun 15 '13 at 18:45
1  
@skullpatrol: You can think of it as satisfying the inequality $x<\infty$ for each real number $x$, so you can think of it as being larger than any given real number in terms of the linear ordering of the real numbers, but you should not think of it as larger in terms of size: size isn’t really a relevant notion here. –  Brian M. Scott Jul 12 '13 at 9:41
1  
@skullpatrol: In a sense: in this context it’s more misleading than helpful to think in terms of distance. For the rest, I don’t really know what I can say. $\Bbb R$ has a natural linear order $\le$, and with respect to $\le$ it has no largest element. There is a way of adding a largest element that behaves nicely in a topological sense, but it doesn’t behave at all nicely in any algebraic sense. Fortunately, it’s the topological aspects that are relevant here. (But please don’t ask me to elaborate: that would require a fairly lengthy discussion and a fair bit of background.) –  Brian M. Scott Jul 12 '13 at 11:32
show 3 more comments

The infinity in the definition of limit has nothing to do with cardinality. Cardinalities are related to sizes of sets. Limits of functions have nothing to do with sizes, but rather with behaviour as $x$ approaches something. So, the relevant notion of infinity lurking under the surface is that of a directed set, and in the case of limits in metric spaces the simpler notion of a sequence. A directed set is a poset $(A,\le)$ such $A\ne \emptyset$ and for all $a,b\in A$ there exists $c\in A$ with $a\le c$ and $b\le c$. A particular type of ordered set is the set $\omega$ of the natural numbers, with their usual ordering.

As you probably know, in the context of metric spaces, a function $f:X\to Y$ satisfies $\lim _{x\to x_0}f(x)=L$ if, and only if, for all sequences $x:\omega \to X$ with $\lim _{n\in \omega}x=x_0$ holds that $\lim _{n\in \omega}f(x)=L$. It is the order type of $\omega$ that gives rise to the usual meaning of the expressions $\lim _{n\in \omega }x=x_0$ and $\lim _{n\in \omega }f(x)=L$.

In situations other than metric spaces to correctly capture limits of functions it is requires to consider 'sequences' over directed sets other than $\omega$. These are called nets. The reason that in metric spaces sequences suffice is that the metric takes values in $[0,\infty ]$ and limit behaviour is sensitive to 'getting close to $0$'. The set $\{\frac{1}{n}\mid n\in \mathbb N\}$ gives rise to a discrete directed set (after inverting the order) which is why sequences are all one needs to properly investigate 'getting close to $0$'.

So, the important underlying infinity when discussing limits is not at all related to cardinalities, but rather how a quantity is allowed to approach something - i.e., nets. Nets require directed sets, and the $\infty $ in the usual definitions of limits refers to the (infinite) directed set $\mathbb N$. In short, it is not cardinals but rather ordinals that are at play.

share|improve this answer
add comment

The infinity here is a point at infinity in the two-point compactification $\mathbb{R} \cup \{ -\infty, +\infty \}$ of the real line. This is just one of many meanings of "infinity" in mathematics; see, for example, this math.SE question.

share|improve this answer
    
For further expositions, readers may also find of interest the links in this answer. –  Bill Dubuque Jul 9 at 23:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.