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Comparing Infinities

  1. Apr 28, 2010 #1
    I'm not sure if this is the right place to post this under, but here goes.

    I was curious as to know if anyone knew any method in which you could compare infinities. And if someone could explain (if there is such a method) how it works, it would be much appreciated. Also the counter to that, if not. Can someone explain why not.

    Thanks

    Si
     
  2. jcsd
  3. Apr 28, 2010 #2
    So you are asking in which ways you can compare infinities (so the countable infinity and the uncountable infinities).

    If this is so, then one way to describe the countable infinity is as the smallest infinity, which is the naturals. Any countable set can be put in a one to one correspondence with the set of natural numbers (the integer, rationals are also countable). The next biggest infinity is the Real numbers, which is the first uncountable set. The Complex number are also uncountable of same cardinality (means they are infinite of the same order). Also from set theory you have that the power set of any set (finite or infinite) is strictly larger than the original set, and in this manner you can build an infinite set whose cardinality is larger than the reals and so on.

    I would suggest reading on Cardinal Numbers to learn more on the subject. I am not sure what your Math background is, so the above may still be a bit unclear. But I will try to expand on the above if you need more clarification.
     
  4. Apr 28, 2010 #3
    I'm An A level student from the UK, and will be starting a theortical physics degree in october. So I've studied the basics of complex numbers and the alike. What you said makes sense, but i was wondering if there was any mathemtical way of for example saying how much larger on the scale of inifinties a set like the natural numbers is, in comparison to say the real numbers. Or any other set for that matter. Other then saying they are 'just' larger.
     
  5. Apr 28, 2010 #4
    Well in terms of cardinals, the set of countable infinity has cardinality Aleph-0. The set of reals has cardinality Aleph-1, power set of the reals is Aleph-2 and so on.

    Aleph-n is what is called a cardinal number. You could also look into reading about the Generalized Continuum Hypothesis, which states that 2Aleph-i=Aleph-(i+1). 2^N is how power sets are denoted, so basically the next largest infinity comes from taking the power set of the infinity you are looking at. I will remark that this G.C.H. cannot be proven as was shown by Gödel and Cohen within our set theory, however it is consistent as an axiom of our Set Theory.
     
  6. Apr 28, 2010 #5
    I don't think the quantitative comparison exists, only a qualitative one which arises when you consider the fundamental mathematics of the numbers. As far as I know you cannot relate them using formulas.

    There is a different perspective from asymptotic expansions where you can relate two different "scales" using limits to model boundary layer behaviour (ie. one scale is normal far from the boundary and one scale is normal close to the boundary), (looks something like lim(f(x)) as x-->inf = lim(f(X) as X-->0+, where X = a*x and a is your expansion parameter, some arbitrarily huge number) but this is a different story from set theory. This approach is probably much more useful to you from a physics perspective.
     
  7. Apr 28, 2010 #6

    Hurkyl

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    For the sake of completeness....

    This assumes the continuum hypothesis.

    This assumes generalized continuum hypothesis. (Or, at least, the first part of the GCH beyond the CH)
     
  8. Apr 28, 2010 #7

    Hurkyl

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    Comparing infinite cardinal numbers1 is often done by calculation. The ordering has several simple arithmetic properties2 such as:
    • [itex]2^\alpha > \alpha[/itex].
    • [itex]\alpha + \beta = \beta[/itex] whenever [itex]\alpha \leq \beta[/itex] and [itex]\beta[/itex] is infinite
    • [itex]\alpha \cdot \beta = \beta[/itex] whenever [itex]\alpha \leq \beta[/itex] and [itex]\beta[/itex] is infinite2

    And you can often get started via direct enumeration -- e.g. proving that [itex]X \leq \aleph_0[/itex] by finding a surjective partial map [itex]\mathbf{Z} \to X[/itex].

    Anyways, cardinal numbers have nothing to do with calculus -- limits, integrals, and stuff -- in particular, [itex]+\infty[/itex] and [itex]-\infty[/itex] are extended real numbers, not cardinal numbers. If you're interested in those, they have simple arithmetic too, but based on calculus and analysis, not set theory.


    1: the phrase, "comparing infinities" is grammatically incorrect, if not downright misleading
    2: if you assume the axiom of choice, anyways. Without choice, cardinal arithmetic becomes more complicated
     
    Last edited: Apr 28, 2010
  9. Apr 28, 2010 #8
    There is. Despite technical complications, the basic idea is an extension of the one used to compare finite sets and it's called the Von Neumann hierarchy.

    First, imagine the natural numbers, starting with 0, as a sort of measuring "rule"; now, if you have a finite set with, say, 5 elements, you may state that it "belongs" to the fifth level in the natural number scale. In fact, all sets with 5 elements will belong to this level; sets with less than 5 elements will be on the levels below (0,1,2,3,4) and with more, above (6,7,8,...).

    The Von Neumann hierarchy is as extension of this beyond the naturals, done with a particular class of sets called "ordinals" (the finite sets are the first ordinals, [itex]\mathbb N[/itex] is the smallest infinite ordinal, etc.), that may be seen as an extension of the natural numbers into the transfinite (cardinals are a subclass of the ordinals).

    It can be proved that, for every set X, there is a minimum level of the hierarchy to which it belongs (the hierarch is comulative: lower levels are included in the higher ones), so sets like
    [itex]\mathbb N[/itex], [itex]\mathbb Z[/itex] and [itex]\mathbb Q[/itex] all belong to the same level (the one indexed by [itex]\mathbb N[/itex], which is both an ordinal and cardinal), but sets like [itex]\mathbb R[/itex] and [itex]\mathbb C[/itex], despite having the same minimum level, are above the previous ones.
     
  10. Apr 28, 2010 #9
    I was just having a brief read up of the Generalised continuum hypothesis, and although it made sense in your explanation, it was slightly above my understanding when I read it from a different source. The other thing I picked up on, is whilst it is important to note that it cannot be proved, it also cannot be disproved. Which I thought was quite interesting. Having been shown this now, I'm going to make a note of this and revisit it when I get some more time, and a chance to look at this in more depth. So thanks ABarrios.

    And Yes Mikey that approach appeals more to me, as more often then not, in my physics experience, it's more exciting to look at the boundaries and to see how things change when we either pick a point on the boundary, or infinitesimally close to the boundary (depending on the scenario, finitely close).
     
  11. Apr 28, 2010 #10
    What makes you say that it is grammatically incorrect and downright misleading? Is it because there is almost two different meanings to the same question? In of which case I hadn't considered the other possibility within the question.
    Also could you explain what the following means?
     
    Last edited: Apr 28, 2010
  12. Apr 28, 2010 #11
    Yes.

    If [itex]\mathfrak{a},\mathfrak{b}[/itex] are cardinal numbers, [itex]\mathfrak{a}[/itex] is infinite, [itex]\mathfrak{a}>\mathfrak{b}[/itex] and [itex]\mathfrak{b}\neq 0[/itex], then [itex]\mathfrak{a}[/itex] is exactly [itex]\mathfrak{a}[/itex] times greater than [itex]\mathfrak{b}[/itex].

    That is [itex]\mathfrak{a}=\mathfrak{a}\times \mathfrak{b}[/itex].

    In the case you mention there are actually more real numbers than natural numbers, so the set of real numbers would be infinitely larger than the set of natural numbers by a factor [itex]2^{\aleph_0}[/itex]. Here [itex]\aleph_0[/itex] is the cardinal number of the set of natural numbers and [itex]2^{\aleph_0}[/itex] is then the cardinal number of the set of real numbers.
     
  13. Apr 28, 2010 #12

    disregardthat

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    For finite sets, two sets A and B of equal size is equivalent to that you can "rename" each element from a set A to a single unique element from B. This is the same as finding a bijective map from A to B. This concept is merely generalized to infinite sets. You can say that one set is "as large" as another if you can explicitly find a bijection between the sets.

    Without mentioning cardinals, you can also in analogy to finite sets define a ordering of sets in the following manner: If you can find a bijection between a set A to a subset of a set B, then you can say that the "size" of A is less than or equal to the "size" of B. However, if you also can prove that there cannot be a bijection between A and the whole of B, it makes sense to say that the size of A is strictly less than the size of B.

    Note however that even if there exists a bijection between a set A to a subset of a set B, that does not mean that there cannot exist a bijection between A and B.

    As previously mentioned here, uncountable sets are known to be "strictly larger" than countable sets in the sense of the ordering which is described above.

    It is important to note that infinite sets do not actually have a size in the same sense as finite sets do, we are merely drawing out the relevant notions from the concept of size from finite sets such that it may be useful when considering infinite sets(keyword: bijection, and it's important). Therefore I think it is at best not very pedagogical to present cardinal numbers as a first introduction to comparing infinite sets. In fact, I am not very fond of cardinal theory.
     
    Last edited: Apr 28, 2010
  14. Apr 28, 2010 #13
    I just noticed that Hurkyl already posted something very similar to my previous post, but I left it in anyway because it addresses the specific case you asked about. The examples Hurkyl gives assume infinite cardinals, whereas [itex]\mathfrak{b}[/itex] in my example could also be finite.

    Notice that [itex]0\times {\mathfrak b}=0[/itex] either way. The indeterminacy that occurs with limits in analysis disappears here, as also with [itex]{\mathfrak b}^0[/itex], which is always [itex]1[/itex], including the case when [itex]{\mathfrak b}=0[/itex].

    So if you're as good at arithmetic as I am, the arithmetic of infinite cardinals has a lot to be said for it.

    [itex]\flat\eighthnote\sixteenthnote[/itex] [itex]\aleph_0[/itex] [itex]\aleph_0[/itex]s are [itex]\aleph_0[/itex], [itex]\aleph_1[/itex] [itex]\aleph_0[/itex]s are [itex]\aleph_1[/itex], [itex]\aleph_2[/itex] [itex]\aleph_0[/itex]s are [itex]\aleph_2[/itex], [itex]\dots[/itex].
     
    Last edited: Apr 28, 2010
  15. Apr 28, 2010 #14

    Mentallic

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    May I ask, how are the rationals a countable infinite?
    There are a countable infinite for the naturals, but between each natural number there are also an infinite number of rationals. Oh and since the reals are uncountable, does this mean the irrationals are uncountable? But the irrationals seem to follow the same trend as the rationals do...
     
  16. Apr 28, 2010 #15

    CRGreathouse

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    The rationals are countable because you can list all rationals a/b with |a| + |b| = 1, then all rationals with |a| + |b| = 2, and so on.

    The irrationals are, as you say, uncountable.
     
  17. Apr 29, 2010 #16
    Yes. Between any distinct pair of irrationals there are infinitely many rationals, yet there are still infinitely many times as many irrationals as rationals. Goes to show you have to keep a tight reign on your intuition in this area.
     
  18. Apr 29, 2010 #17

    Hurkyl

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    An example is counting the rationals. We know that [itex]\mathbb{Z} \times \mathbb{Z}[/itex] is infinite countable. e.g. because [itex]|\mathbb{Z} \times \mathbb{Z}| = |\mathbb{Z}| \cdot |\mathbb{Z}| = \aleph_0 \cdot \aleph_0 = \aleph_0[/itex].

    However, we can partially map this set onto the rationals, by sending (m,n) to m/n, whenever the latter defined. Every rational number appears in this way (infinitely many times).

    Therefore, [itex]|\mathbb{Q}| \leq \aleph_0[/itex].

    However, because [itex]\mathbb{Q}[/itex] contains [itex]\mathbb{Z}[/itex], we know that [itex]|\mathbb{Q}| \geq \aleph_0[/itex].

    Therefore, [itex]|\mathbb{Q}| = \aleph_0[/itex].


    Because really you should be using the adjective "infinite" in this context; using the noun infinity awkward at best -- at worst, people parse it as a proper noun, and are misled think all uses of infinity/infinite are referring to the same thing.
     
    Last edited: Apr 29, 2010
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