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Thread: denumerability

  1. #1
    mathkid
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    denumerability

    Prove that if A is denumerable and B is a finite subset of A, then A \ B is denumerable.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathkid
    Prove that if A is denumerable and B is a finite subset of A, then A \ B is denumerable.
    Lets assume you are using the definition of denumerable as meaning
    countably infinite.

    Let $\displaystyle a_1,\ a_2,\ ...,\ a_n,\ ...$ be an enumeration of $\displaystyle A$.
    We can rearrange this enumeration so that <you may need to prove that this can be done>:

    $\displaystyle B\ =\ \{a_1,\ a_2,\ ...,\ a_{N_B}\}$

    ($\displaystyle N_B$ is the number of elements of $\displaystyle B$)

    Then:

    $\displaystyle A \backslash B\ =\ \{a_{N_B+1},\ a_{N_B+2},\ ...\}$.

    So $\displaystyle c_1,\ c_2,\ ...,\ c_N,\ ...$ where $\displaystyle c_n\ =\ a_{N_B+n}$ is an enumeration of $\displaystyle A \backslash B$

    RonL
    Last edited by CaptainBlack; Dec 6th 2005 at 11:26 PM.
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  3. #3
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    I like to use the definition of "countable" as "can be injected into the natural numbers". It is then immediate that a subset of a countable set is countable. "Denumerable" is countable and not finite, where "finite" means every injection of the set to itself is also a surjection. All that remains is to prove that the union of two finite sets is finite; use the lemma that a finite set injects into {1..n} for some natural number n.
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