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Math Help - Topology questions

  1. #1
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    Topology questions

    I need help with these questions

    1. Does there exist a continuous function from (0,1) onto [0,1]
    2. Does there exist a continuous function from (-1,1) onto R
    3. Does there exist a continuous function from R onto (-1, 1)
    4. Does there exist a continuous function from R onto Q
    5. Does there exist a continuous function from Q onto R

    where R = real numbers
    and Q = rational numbers
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  2. #2
    Member mabruka's Avatar
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    a) \sin (\pi x)<br />
    b) \tan (\frac{\pi}{2}x)

    c) \frac{2}{\pi}\arctan (x)

    d) no, since R is connected and Q is disconnected

    because the continuous image of a connected set must be connected

    e)hmmm
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ogele View Post
    5. Does there exist a continuous function from Q onto R

    where R = real numbers
    and Q = rational numbers
    Of course not. A general theorem (requiring Zorn's lemma) implies that if f:X\mapsto Y is surjective then there exists an injection g:Y\mapsto X. Does there exist an injection g:\mathbb{R}\mapsto \mathbb{Q}.

    And since this thread is labeled topology I suppose that I can impose any topology on these sets.

    Give \mathbb{Q} the discrete topology and \mathbb{R} the indiscrete. Then any mapping f:\mathbb{Q}\mapsto\mathbb{R} is continuous.
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  4. #4
    Member mabruka's Avatar
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    I forgot about the label "topology"!!

    Ogele, since you did not specified the topologies it is understanded we have the usual topology in both the domain and range
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  5. #5
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    Smile Topology Questions

    Thanks for answering the question Mabruka and Drexel28
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    Of course not. A general theorem (requiring Zorn's lemma) implies that if f:X\mapsto Y is surjective then there exists an injection g:Y\mapsto X. Does there exist an injection g:\mathbb{R}\mapsto \mathbb{Q}.

    And since this thread is labeled topology I suppose that I can impose any topology on these sets.

    Give \mathbb{Q} the discrete topology and \mathbb{R} the indiscrete. Then any mapping f:\mathbb{Q}\mapsto\mathbb{R} is continuous.

    Or simpler, if we already got AC into the game: if there exists a surjective \mathbb{Q}\mapsto \mathbb{R} then \aleph_0=Card(\mathbb{Q})\geq Card(\mathbb{R})=2^{\aleph_0} , which of course is absurd.

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    Or simpler, if we already got AC into the game: if there exists a surjective \mathbb{Q}\mapsto \mathbb{R} then \aleph_0=Card(\mathbb{Q})\geq Card(\mathbb{R})=2^{\aleph_0} , which of course is absurd.

    Tonio
    Well, even simpler (without AC). We can use the well-ordering principle to prove that if f:\mathbb{N}\mapsto E is surjective then E is countable.
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    Quote Originally Posted by Drexel28 View Post
    Well, even simpler (without AC). We can use the well-ordering principle to prove that if f:\mathbb{N}\mapsto E is surjective then E is countable.
    Do you understand that
    Axiom of Choice \Rightarrow Hausdroff’s Maximal Principle \Rightarrow Zorn’a Lemma \Rightarrow well-ordering theorem \Rightarrow Axiom of Choice?
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    Do you understand that
    Axiom of Choice \Rightarrow Hausdroff’s Maximal Principle \Rightarrow Zorn’a Lemma \Rightarrow well-ordering theorem \Rightarrow Axiom of Choice?
    I'm kind of shocked you'd ask that. But, yes I do. I thought that the well-ordering principle of the natural numbers did not require the axiom of choice...but I could be wrong.
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  10. #10
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    Quote Originally Posted by Drexel28 View Post
    A general theorem (requiring Zorn's lemma) implies that if f:X\mapsto Y is surjective then there exists an injection g:Y\mapsto X. Does there exist an injection g:\mathbb{R}\mapsto \mathbb{Q}.
    Did you not post the above?
    So why are you surprised?
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  11. #11
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    Did you not post the above?
    So why are you surprised?
    Yes, but you can prove that if a set is the surjective image of a countable set it's countable using only the well-ordering principle of the natural numbers. Is that equivalent to the AOC?
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    Quote Originally Posted by Drexel28 View Post
    Yes, but you can prove that if a set is the surjective image of a countable set it's countable using only the well-ordering principle of the natural numbers. Is that equivalent to the AOC?


    I think this may be a confusion. The well ordering principle is usually taken to be: any non-empty subset of the naturals has a first element (in the usual, natural ordering), and this is logically equivalent to the (weak) principle of induction.

    Perhaps what Plato assumed, and this is what I thought at first, is Zermelo's well ordering theorem: every non-empty set can be well-ordered , and this in ZF is equivalent to Zorn's lemma and to AC.

    Tonio
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  13. #13
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    I think this may be a confusion. The well ordering principle is usually taken to be: any non-empty subset of the naturals has a first element (in the usual, natural ordering), and this is logically equivalent to the (weak) principle of induction.

    Perhaps what Plato assumed, and this is what I thought at first, is Zermelo's well ordering theorem: every non-empty set can be well-ordered , and this in ZF is equivalent to Zorn's lemma and to AC.

    Tonio
    Maybe. I meant the well-ordering of the naturals. That's all you need.
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