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Thread: Proof - A is Infinite

  1. #1
    Newbie Pi R Squared's Avatar
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    Question Proof - A is Infinite

    I know that to prove A is finite, I say:

    $\displaystyle \exists$ a bijection $\displaystyle f : N_k \rightarrow A$

    So to prove A is infinite would I say:

    $\displaystyle \exists $ a bijection $\displaystyle f : \mathbb{N} \rightarrow A$

    Our professor just said "not finite." So I am a little confused as to what to negate.
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  2. #2
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    Quote Originally Posted by Pi R Squared View Post
    I know that to prove A is finite, I say:
    $\displaystyle \exists$ a bijection $\displaystyle f : N_k \rightarrow A$

    So to prove A is infinite would I say:
    Show that $\displaystyle (\forall~f)(\forall~k)[f : N_k \rightarrow A]$ then $\displaystyle f$ is not a bijection.

    That is $\displaystyle A$ is infininte if it is not finite.
    $\displaystyle A$ does not biject with any $\displaystyle N_k$.
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  3. #3
    Newbie Pi R Squared's Avatar
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    Question Proof - A is Infinite

    Quote Originally Posted by Plato View Post
    Show that $\displaystyle (\forall~f)(\forall~k)[f : N_k \rightarrow A]$ then $\displaystyle f$ is not a bijection.

    That is $\displaystyle A$ is infininte if it is not finite.
    $\displaystyle A$ does not biject with any $\displaystyle N_k$.


    I am understanding this a little, but still not enough to prove it. This is the problem I am working on:

    Proof: Suppose for the sake of obtaining a contradiction that if C is an infinite set then D is not an infinite set or $\displaystyle g \rightarrow C$ is not one-to-one.

    I clearly see the logic. But am confused as to how to relate $\displaystyle [f : N_k \rightarrow C]$ to $\displaystyle g \rightarrow C$.

    I guess I am confused as to how $\displaystyle [ N_k ]$ can be infinite...
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  4. #4
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    Quote Originally Posted by Pi R Squared View Post
    I guess I am confused as to how $\displaystyle [ N_k ]$ can be infinite...
    You really need to get this clear in your brain.
    There very basis of this approach to finite set is that $\displaystyle \left( {\forall j \in \mathbb{Z}^ + } \right)$ the set $\displaystyle N_j=\{1,2,\cdots,j\}$ is finite.
    That means that by definition $\displaystyle N_k$ cannot be infinite.
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  5. #5
    Newbie Pi R Squared's Avatar
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    Question I understand...

    Quote Originally Posted by Plato View Post
    You really need to get this clear in your brain.
    There very basis of this approach to finite set is that $\displaystyle \left( {\forall j \in \mathbb{Z}^ + } \right)$ the set $\displaystyle N_j=\{1,2,\cdots,j\}$ is finite.
    That means that by definition $\displaystyle N_k$ cannot be infinite.

    I understand...

    [tex]N_k[\math] is finite.

    So I am going to have to prove that if C is an infinite set then D is not, therefore proving by contradiction that the statement is true.

    I wonder if you could clarify this statement using onto and/or 1-1.

    Quote Originally Posted by Plato View Post
    does not biject with any .
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