Is Infinity Truly Limitless or Does It Have Boundaries?

[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.

 

[WhyIsItSo]

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.

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.

 

[Skhandelwal]

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

 

[Robokapp]

I don't 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 amount possible...and you've just demoonstrated an amount greater than it is existent...

 

[CRGreathouse]

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 amount possible...and you've just demoonstrated an amount greater than it is existent...

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).

 

[WhyIsItSo]

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).

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!

 

[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.

 

[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.

 

[CRGreathouse]

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!

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).

 

[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 wouldn't be asking i guess.)I just found out about this forum and i subscribed after reading the first thread.

 

[WhyIsItSo]

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).

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.

 

[matt grime]

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?

Cantor. Cardinals. Look 'em up.

 

[CRGreathouse]

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.

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}}

 

[CRGreathouse]

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.

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.

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.

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?

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?

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.

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

This makes no sense to me.

 

[Tchakra]

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.

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

 

[WhyIsItSo]

Cantor. Cardinals. Look 'em up.

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?

 

[Tchakra]

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?

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

 

[CRGreathouse]

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

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.

 

[CRGreathouse]

Can you give an example of a practical use?

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

 

[WhyIsItSo]

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

Thank you kindly.

 

[Tchakra]

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.

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 don't 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.

 

[HallsofIvy]

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 don't 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.

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.

 

[CRGreathouse]

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 don't 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.

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.

 

[Tchakra]

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

I meant a*tan btw, but doesn't matter anyway.

 

[Hurkyl]

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.

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.

 

[Robokapp]

I was confused about the arctan part until I looked at it otherwise, taking the Tan of pi/2 (or 90 degrees). If you take the limit of tan(pi/2) its a point of discontinuity though. If you come from letf you'd go up and if you come from right you'd go down. However something like...
\frac{1} {|x|} would fit your desired purpse better maybe? or even...

y = |tan(x)| at x=\frac {n\pi} 2

 

[shmoe]

Can you give an example of a practical use?

The reals are uncountable, the set of http://mathworld.wolfram.com/AlgebraicNumber.html" can be proven to be countable. This gives a relatively easy proof that transcendental numbers (=those that are not algebraic) not only exist (without ever having demonstrated a specific one!) but there are *lots* number of them. This leads to the frustrating situation where it's extremely difficult to prove that any given number is transcendental, yet in a sense we can make precise *almost all* real numbers are transcendental.

 

[CRGreathouse]

I hate to risk confusing the issue, but that's not quite accurate.

I agree with what you wrote, but I saw no reason to bring the extended real line into play here. For one thing it wasn't mentioned; for another, there are several good ways to extend it. The projective space (where there's just one 'point at infinity', or a line, etc.) would be an example.

To be sure, there are ways to make the statement sensible, but that doesn't mean we're disagreeing. By specifying a method of extending the reals we have the rigor I asked for. I wasn't saying it was impossible, just the opposite in fact: there are too many good ways to 'handle' infinities (cardinal, ordinal, 'calculus'/SA dual infinities, 'projective space' infinity, ...) so I had to have one specified.

 

[matt grime]

Can you give an example of a practical use?

Hmm, how practical is practical? They are very prevelant in set theory, topology, even discrete mathematics, analysis finds it important, as does number theory.

In many cases showing that two sets are not bijective is very useful. It would show that two groups are not isomorphic, and if you want to see how that it hard try to prove that the real numbers and the rational numbers (under addition) are non-isomorphic groups.

Philosophical issues also abound and it is not clear what (large) cardinal axioms one should assume, or one might wish to assume in your set theory.

In my area, there are constructions that require countable generating sets in whatever sense. These have been weakened to other cardinals of generating sets. These results are about cohomology theory, and physicists seem to care about cohomology a lot not just mathematicians. (Cohomology is the thing that tells you when an obstruction to sometihng occurs: the fact that there is a nowhere vanishing vector field on the sphere is supposedly a statement about cohomology groups, but not one that has anything to do with cardinality.)

I would imagine countable things are important in computer science. Actually ordinals as well. As a guess: it is possible to store information about countable number of objects if they are ordered and there is a relation between the order: it is possible to store the information that sepcifies n! for all n in two parts: 0!=1 and n!=n*(n-1)!. It wouldn't be possible to store information like that about the real number system since it is not ordered so well. Yes this is handwavy, quite probably wrong, but a desperate attempt to put some practical meaning into the practical answer.

 

[CRGreathouse]

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.

Thanks, that's what I was trying to say.