What is infinity?

1. Sep 13, 2006

Skhandelwal

Does infinity have direction? Dimensions? As we all know it does, then doesn't it mean that infinity is actually limited? Just like values b/w 1 and 2 x values are infinity but have a total sum.

2. Sep 13, 2006

WhyIsItSo

There was a recent thread conceptually similar to this. Someone was having trouble with the concept of the set of integers being infinite.

Any integer is a finite value. Any integer plus any other integer is another finite value.

But that in no way means the range of possible integers is finite.

As I think I pointed out in that thread, anywhere you go in space has some specific (therefore finite) position, but the range of positions you might move to are still infinite.

3. Sep 13, 2006

Skhandelwal

Yes but since ranges or infinities are finite, why can't we add the ranges, thus adding the infinites?

4. Sep 13, 2006

Robokapp

I dont know if it helps the topic but...something to think about:

my brother (9th grade) was asked by the teacher:

Are there more numbers between 0 and 1 or between 0 and 2?

Now...there are infinitely many in both cases but I'd say since every number in the interval (0, 1) exists in the interval (0, 2) and not all numbers from interval (0, 2) exist in the interval (0, 1), the (0, 2) is greater than (0, 1). But you're comparing two infinities so...

One infinity greater than another infinity would make the samll infinity not really be an infinity...because an infinity is largest ammount possible...and you've just demoonstrated an ammount greater than it is existent...

5. Sep 13, 2006

CRGreathouse

Amusingly, you're wrong on both counts. There are the same number of numbers in (0, 1) as in (0, 2), since for every x in (0, 2) there is x/2 in (0, 1). But there can be infinite quantities which are smaller than other infinite quantities. There are an infinite number of rational numbers, but not as many as there are real numbers in (0, 1).

6. Sep 13, 2006

WhyIsItSo

What a fascinating topic this has turned into. But I think your logic is flawed.

In (0,1), that x/2 value you specify is already represented by some value y. By your logic, that value is counted twice!

7. Sep 13, 2006

Robokapp

the x/2 part...we're talking really small numbers. As small as they get...infinitely small, they still have halves. any real number has a real half...so the x/2 from (0, 2) interval corresponding to full x in (0, 1) seems to me like a bad concept. I'm interested to see how this topic develops also.

8. Sep 13, 2006

matt grime

Here is what I sugges happens when a mathematician looks at threads like this:

she throws hands in the air, sighs, shrugs, and walks away.

9. Sep 13, 2006

CRGreathouse

No, it's not counted twice. Let's use a concrete example: 4/3.

4/3 in (0, 2) maps to 2/3 in (0, 1). That's not counting 2/3 twice, though, because I'm not counting 2/3 in (0, 2) as 2/3 in (0, 1); I'm counting it as 1/3 in (0, 1).

10. Sep 13, 2006

Tchakra

CRgreathouse is right. on both counts.
the number of elements(called cardinality) in (0,1) is equal to the cardinality of (0,2) which is equal to the number of numbers in the real.

The problem many people have with ithe concept of infinity in mathematics is that they try to think of it as a "really big" finite number(ie bigger than any number you can imagine) whn in fact it is not.

infinities are of different size and some are bigger than others. PERIOD.

Think of infinity as a curve near an asymptote on a graph.
Lets visualize two increasing functions f annd g both starting at 0 and both are asymptotic at a. Say by construction that f(x)>g(x) for all values of x in [0,x]. Then it is evident that as x approaches a both approach infinity and f is ALWAYS bigger than g. Hence the infinity f is tending to is bigger than the infinity g is tending to.

Now explaining why there are the same number of elements in (0,1) and (0,2) is quite tricky if you have not learned about sets and cardinality. (and if you did you wouldnt be asking i guess.)

I just found out about this forum and i subscribed after reading the first thread.

Last edited: Sep 13, 2006
11. Sep 13, 2006

WhyIsItSo

First I should clarify I am not disputing the infinity issue itself, only your example here.

Your original argument that x/2 yields a new number is what I'm arguing against.

Let's take a simple, concrete example. I'm counting... 1, 2, 3, 4, 5, 6, 7, 8...

You say 8/2 gives another number. I'm saying no it doesn't, it is nothing more than a different way of representing a number we already have... 4.

The same applies to the OP subject. For any number "x" you pick in (0,2), then divide by two, I argue there already exists that number, and x/2 is nothing more than a different expression fo the same number.

It seems to me meaningless to talk about "bigger" or "smaller" infinities. Inifinity is infinity. $$2*\infty = \infty$$. What else can be said about it?

I would offer that you are mixing magnitude of the range with the number of values within that range. That doesn't work.

12. Sep 13, 2006

matt grime

Cantor. Cardinals. Look 'em up.

13. Sep 13, 2006

CRGreathouse

Careful there. x^2+1 > x for all x in $$[0, \infty)$$, but that doesn't mean it's tending toward a 'bigger' infinity (well, maybe as a hyperreal). In fact, it's a little hard to think of a way in which this is even well-defined.

There are just as many perfect squares {0, 1, 4, 9, 16, ...} as integers {..., -1, 0, 1, ...}, and just as many rational numbers {a/b: a an integer, b a nonzero integer} as integers. There are more reals, though, and yet more functions from the reals to the reals:

$$|\{0, 1, 4, 9, \ldots\}|=|\mathbb{Z}|=|\mathbb{Q}|=\aleph_0<|\mathbb{R}|=\mathfrak{C}<2^{\mathfrak{C}}$$

14. Sep 13, 2006

CRGreathouse

That dosn't apply directly to what we discussed (since 7, for example, is outside of (0, 1) and (0, 2)), but I can work with it. The cardinality of the even numbers $$\mathcal{E}$$ is equal to that of the integers $$\mathbb{Z}$$. For every even number $$e\in \mathcal{E}$$, there a corresponding integer e/2. Name an even number and I'll give you a corresponding integer; name an integer and I'll give you the even to which it corresponds. There's no overlap; 8/2 does give a new number.

No. Again, I've never mapped two numbers to the same number -- you're just pointing out that a number is in both sets, which follows trivially fro mthe fact that (0, 1) is a subset of (0, 2). Don't you realize that having the same cardinality as some proper subset is a definition of being infinite?

For a cardinal infinity $$\mathcal{I}$$, $$\mathcal{I}\cdot2=\mathcal{I}=2\cdot\mathcal{I}$$.

When you compare infinities as ordinals instead of cardinals (say, $$\omega$$ instead of $$\aleph$$ and $$\beth$$), even that doesn't hold any more. Since I'm less well informed about ordinal infinities, I'll let someone else discuss them or leave it at that.

This makes no sense to me.

15. Sep 13, 2006

Tchakra

i think you missed my constraint "asymptotic at a", which self implies that a is not $$\infty$$

16. Sep 13, 2006

WhyIsItSo

I stand corrected. What I was not able to discover was the point. As the wiki I read states, this is counterintuitive.

That the cardinality differs is obvious. That there is any meaning to "size" of infinity appears to be unproductive.

Can you give an example of a practical use?

17. Sep 13, 2006

Tchakra

I didnt know that anyone looked for practicality in mathematics at the level of asking about infinity, or that it conforms to intiuition.

18. Sep 13, 2006

CRGreathouse

First, I'm not sure what you mean by "asymptotic at a". If you mean "bounded above by a" that still doesn't help much -- f(x)=6 and g(x)=6-1/x are both bounded above by 6 with f(x) > g(x), but neither tends toward a higher infinity.

Wold you give an example of two such functions that are "asymptotic at a", one of which tends toward a 'higher' infinity? Maybe then I'll understand what you mean.

19. Sep 13, 2006

CRGreathouse

It's quite important in topology, which in turn has various applications.

20. Sep 13, 2006

WhyIsItSo

Thank you kindly.

21. Sep 13, 2006

Tchakra

I can't get to construct two good functions now, but
take, f(x)= -1/x - 2 and construct g(x)= atan(bx) + c where a, b and c such that g(-1/2)=0 and blows up at g(0).

Strictly in (-1/2,0), f(-1/2) = g(-1/2) and as x -> 0 f,g -> infinity

I dont know if in this case f>g, but as x -> 0 either f or g will overtake the other one hence at a given oint one will be ALWAYS bigger than the other one thus tending at a higher infinity.
i hope this clears my asymptotic way of explaining the different sizes of infinities.

22. Sep 13, 2006

HallsofIvy

Does "atan" mean arctangent? I was puzzled since you then mention "a, b, c". If "atan" does not mean a*tangent, then there is no a in your formula. In any case, I would not say that one function "tends to a higher infinity" than the other. I would say "one function tends to infinity faster than the other". But the "infinity" here has nothing to do with different cardinalities. The problem is that the word "infinity" has a number of different meanings in different form of mathematics.

23. Sep 13, 2006

CRGreathouse

Do you mean tan rather than atan?

I don't agree that one tends toward a higher infinity. If you want to think of numbers as a 'limiting' sequence like hyperreals (see a brief explanation of hyperreals and nonstandard analysis), then you'd be right, but I can't think of another way to understand what you wrote. When you write "f(x) --> infinity" it means that beyond some point f(x) is greater than any fixed value -- there's no 'number' it's headed toward in standard analysis. Before real things can be said about infinities, we need definitions -- intuition isn't worth much here.

24. Sep 13, 2006

Tchakra

I will admit defeat as much as i hate it , ok i was wrong.

I meant a*tan btw, but doesnt matter anyway.

25. Sep 13, 2006

Hurkyl

Staff Emeritus
I hate to risk confusing the issue, but that's not quite accurate. In standard analysis, we often use the Extended real numbers, which is formed by adding two "endpoints" (named $+\infty$ and $-\infty$) to the real line. This space is homeomorphic to a closed interval (just as the reals are homeomorphic to an open interval).

So, in the extended reals, when we talk about something like

$$\lim_{x \rightarrow +\infty} x^2 = +\infty$$

this is honest-to-goodness convergence. In fact, we usually extend elementary operations by continuity -- for example, $\arctan (+\infty) = \pi / 2$ is actually a rigorous statement.

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