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Math Help - constant and continuous

  1. #1
    Senior Member sfspitfire23's Avatar
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    constant and continuous

    Let f: \mathbb{R}\rightarrow \mathbb{R} be a continuous function such that f(x) is a rational number for every real input x. Show that  f must be a constant function.

    Heres what I have:

    Let x be any real number. There must exist a sequence of rational numbers \{r_n\}\rightarrow x. Since f is continuous, we can conclude that f(x)=\lim_{x\to c}(r_n)=c.

    Im iffy on my last sentence and am thinking there could be a better way to do this using the EVT?
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  2. #2
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    Suppose f(x)\neq f(y) where x<y. Chuse an irrational \xi between f(x) and f(y). Then by the IVT, there is point between x and y where f assumes the value \xi. But this contradicts the fact that f is always rational.

    Topologically, since R is connected and Q is totally disconnected, every continuous map from R to Q is constant. (This is a generalization of the IVT.)
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  3. #3
    Senior Member sfspitfire23's Avatar
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    So, if f is not constant on an interval [a,b] then it will hit at least one irrational contradicting that f(x) is rational, right? You're saying that, we can always choose an irrational for some c in [a,b] according to the IVT, but....blah I guess i'm having trouble putting the idea into words...ugh
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  4. #4
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    Ja. The IVT says that the continuous image of an interval is an interval. So since any interval that is not a single point contains an irrational, the image of f must be a single point.
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  5. #5
    Senior Member sfspitfire23's Avatar
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    Ah! got it
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  6. #6
    MHF Contributor Drexel28's Avatar
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    NOTE! OP! This is not meant to help with the question, this is just an interesting side note for maddas...though I'm not sure he'll care. I just noticed it.

    Quote Originally Posted by maddas View Post

    Topologically, since R is connected and Q is totally disconnected, every continuous map from R to Q is constant. (This is a generalization of the IVT.)
    There is a slightly weaker condition than totally disconnected that one can impose. Suppose that \varphi:G\to G' is continuous where G,G' are topological groups G is connected and and given any g\in G'-\{e\} there exists a separation of g and \{e\}. Then, \varphi must be constant.

    This follows since g,h)\mapsto \varphi(g)(\varphi(h))^{-1}" alt="\psi:G\times G\to G'g,h)\mapsto \varphi(g)(\varphi(h))^{-1}" /> is continuous \psi(\Delta)=\{e\} and so if \varphi weren't constant then \varphi(x)\ne\varphi(y)\implies \psi(x,y)\ne e but since we can separate \psi(x,y) and \{e\} I think you can see where this is going.


    EDIT: The above is a farce. Not because it's incorrect but because if you can separate \{e\} and any point then the space is actually totally disconnected. This follows since if x\ne y\implies xy^{-1}\ne e and so there exists a disconnection A\cup B=G such that xy^{-1}\in A,e\in B and so x\in Ay,y\in By there are open, non-empty and disjoint since the mapping L_y:G\to G:g\mapsto gy is a homeomorphism. Ahh! Topological groups always make me look stupid haha.
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