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Thread: Countability proof.

  1. #1
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    Countability proof.

    I need help on proving

    (1) prove that if B A and A is countable, then B is countable

    (2) prove that if B A, A is infinite, and B is finite, then A\B is infinite

    I don't even know where to start... THx
    Last edited by mr fantastic; Nov 29th 2010 at 11:06 AM. Reason: Re-titled.
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  2. #2
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    I have a theorem which says
    Suppose A is a set. The following statements are equivalent:
    1. A is countable
    2. Either A = empty or there is a function f:z+ -> A that is onto.
    3. There is a function f:A -> z+ that is one-to-one

    I am supposed to use those however I am stuck....
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    If $\displaystyle A$ is denumerable then $\displaystyle A=\{a_1,a_2,\ldots\}$ . If $\displaystyle B=\emptyset$ then, $\displaystyle B$ is finite . If $\displaystyle B\neq \emptyset$ let $\displaystyle n_1$ be the least positive integer such that $\displaystyle a_{n_1}\in B$, let $\displaystyle n_2$ be the least positive integer such that $\displaystyle n_2>n_1$ and $\displaystyle a_{n_2}\in B$ ...

    Could you continue?

    Regards.

    Fernando Revilla
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  4. #4
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    You can use the equivalent condition #3 to show (1). Namely, suppose $\displaystyle B\subseteq A$ and there exists a one-to-one function $\displaystyle f:A\to\mathbb{Z}^+$. Consider the restriction of f on B, i.e., the function whose domain is B and that acts just like f on any $\displaystyle x\in B$. Is this restriction still one-to-one?

    For (2), note that $\displaystyle A = B\cup(A\setminus B)$. What happens when $\displaystyle A\setminus B$ is finite?
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  5. #5
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    I get the (2) part , however could you explain more on (1) part please???
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  6. #6
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    could you explain more on (1) part please???
    Whom are you asking? Fernando suggests using condition #2 to prove (1), i.e., that B is countable. Going with #2, let's assume that there is an onto function $\displaystyle f:\mathbb{Z}^+\to A$ and that B is nonempty, i.e., there exists some fixed $\displaystyle x_0\in B$. I think it is easier to construct another function $\displaystyle g:\mathbb{Z}^+\to B$ as follows:

    $\displaystyle g(n)=
    \begin{cases}
    x_0 & f(n)\notin B\\
    f(n) & \text{otherwise}
    \end{cases}$

    You have to check that indeed $\displaystyle g:\mathbb{Z}^+\to B$ and that g is onto.

    If you are going ti use #3 to show (1), then assume that there exists an $\displaystyle f:A\to\mathbb{Z}^+$. Consider $\displaystyle g(x)$ such that $\displaystyle g(x)=f(x)$ for $\displaystyle x\in B$ and $\displaystyle g(x)$ is undefined otherwise. You need to show that $\displaystyle g:B\to\mathbb{Z}^+$ and that g is one-to-one.
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