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The set of all continuous functions

  1. Sep 21, 2011 #1
    I suppose my question is, "does the set of all continuous functions comprise a continuum?"

    How would one even start at trying to prove that? Any ideas or suggestions?
  2. jcsd
  3. Sep 21, 2011 #2
    A continuum as in "a compact, connected metric space"??

    And the continuous functions from where to where??
  4. Sep 21, 2011 #3
    Mapping from the reals to the reals... And by continuum I mean it's cardinality, aleph 1.
  5. Sep 21, 2011 #4
    You probably mean to say that the set of all continuous functions has cardinality the same as the reals?? Right? The reals do NOT in general have cardinality aleph 1!!
    So in general, you cannot prove that the continuous functions from the reals to the reals have cardinality aleph 1.

    Anyway, you can prove that the cardinality of continuous function is the same as the reals.
    I'll give you a hint. Consider the continuous functions [itex]\mathcal{C}(\mathbb{Z},\mathbb{R})[/itex] and [itex]\mathcal{C}(\mathbb{Q},\mathbb{R})[/itex].
  6. Sep 21, 2011 #5


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  7. Sep 21, 2011 #6
    Thanks guys... And yes I ahold have said 2^aleph naught (or just the cardinality of the reals. Tried to get too fancy).

    Thanks for the link marhman
  8. Sep 21, 2011 #7
    Thanks guys... And yes I ahold have said 2^aleph naught (or just the cardinality of the reals. Tried to get too fancy).

    Thanks for the link mathman but I'm gonna try working on it a bit before I view it.

    Micromass, My highest level of math is "bridge to advanced mathematics" course, learning some set theory and how to read and write proofs, but it ended just before Real Analysis. Do u think that is enough background to attempt this proof? If so, any more hints before I hit the whiteboard? (actually pulled over in my car now on my way home!)
  9. Sep 21, 2011 #8
    You just need one little fact: if [itex]f,g:\mathbb{R}\rightarrow \mathbb{R}[/itex] that coincide on [itex]\mathbb{Q}[/itex], then f=g. This is usually proven in a real analysis class, but you could try to prove this. This is the crucial fact to be able to tackle this thing.
  10. Sep 21, 2011 #9
  11. Sep 23, 2011 #10
    hey micromass,

    (first off I want to apologize for the lengthy response below. I'll understand if you feel I'm overstepping my forum bounds, just let me know.)

    So I mapped out a strategy for the proof and worked it out do a decent degree, but eventually checked out the link mathman posted. Seemed my ideas were similar. I went back to the drawing board about just what a function is (as they teach you in HS), with inputs and outputs. I realized the inputs would have c as the cardinality and so would the outputs, so I started to try to used some sort of Squeeze Theorem type of logic... in mathmans link it was the cantor-schroeder-bernstein theorem.

    So I outlined my proof pretty well I think, but I couldn't figure out where to use your theorem. So I eventually checked out mathman's link and to tell you the truth I still don't understand how that comes into play.

    Perhaps you can elaborate, I don't expect you to explain an entire proof to me though.

    But anyway, the answer is "c" as mathman and the link showed. So now I've been thinking and I have a different question for you... What would the cardinality be of the set of all integrable functions that map from the Reals to the Reals?

    Here's what I've come up with so far (I've just been outlining approaches, nothing concrete):

    "First, call the set of all integrable functions that map from the reals to the reals I.
    From previous discussion we know that the cardinality of the set of all continuous functions mapping from the Reals to the Reals is c. Now every continuous function is integrable [can't remember where I learned that, I suppose I'll have to find it and insert that theorem here] but not every integrable function is continuous. Therefore, the set of all continuous functions is a proper subset of the set of all integrable functions, where both sets map from R to R.

    By this we can determine that the cardinality of I is at least c."

    Am I on the right path here?

    Thank you kindly
  12. Sep 23, 2011 #11
    Well, it comes into play in the folowing sentence of mathman's link:

    What I said is equivalent with that sentence. Do you see why this sentence is true, or do I need to explain??

    Some remarks:
    1) You must be careful and state which kind of integrability your talking about. Is it Riemann integrablity? Lebesgue integrability?? The answers will (likely) depend on this.

    2) You say that every continuous function is integrable. This holds true, but only if the domain of the function is a closed interval (or some easy modification). So any continuous function [itex]f:[a,b]\rightarrow \mathbb{R}[/itex] is integrable. On the other hand, an easy function like [itex]f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x[/itex] is not integrable.

    3) I do like the approach you made that shows that the integrable functions have at least cardinality c. But it'll be somewhat harder to find an upper bound.
  13. Sep 23, 2011 #12
    What comes to mind is "42". Seems I am looking for an answer without first knowing the question! (if you haven't read the hitchhikers guide to the universe do so now!)

    okay, I'll look into it and really decide what type of integration I want to try to use. Thanks for the tip on the closed interval, it's been a while since calculus.

    with regard to mathmans post and the "rationals being dense so that determines the function" I really don't understand that very well. If you can elaborate on how the upper bound was determined I'd greatly appreciate. (I specifically don't understand the phrase "so that determines the function")
  14. Sep 23, 2011 #13
    The key to the upper bound is the function

    [tex]\mathcal{C}(\mathbb{R},\mathbb{R})\rightarrow \mathbb{R}^\mathbb{Q}:f\rightarrow f\vert_\mathbb{Q}[/tex]

    it suffices to show that this function is an injection (since [itex]\mathbb{R}^\mathbb{Q}[/itex] has cardinality c). For this, we must show that if [itex]f\vert_\mathbb{Q}=g\vert_\mathbb{Q}[/itex], then f=g. This holds since [itex]\mathbb{Q}[/itex] is countable.
  15. Sep 23, 2011 #14
    Perhaps you can just help me with some of the terminology... I'm sorry but I simply don't understand what a lot of the symbols you used are. I've never encountered functions written with a set as a superscript or a subscript. Does the "C" mean cardinality? Also, I'm assuming "dense" is equivalent to "countable"
  16. Sep 23, 2011 #15
    You mean [itex]f\vert_\mathbb{Q}[/itex]? That just means that we restrict the domain of f to [itex]\mathbb{Q}[/itex].

    Yes, c is the cardinaility of the continuum.

    No, saying that a set A is dense means that the closure of A is [itex]\mathbb{R}[/itex]. http://en.wikipedia.org/wiki/Dense_set
  17. Sep 23, 2011 #16
    What about when you wrote R^Q?

    (I never got into this stuff in my class. I suppose I will if I continue to Real Analysis?)
  18. Sep 23, 2011 #17
    That's just the set of all functions [itex]\mathbb{Q}\rightarrow \mathbb{R}[/itex].

    Yes, I'm sure you'll see it in real analysis. Or set theory of something of the like.
  19. Sep 23, 2011 #18
    ok, thanks for your help and patience micromass. I really appreciate it! take it easy.
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