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manjuthottam

  • 2 years ago

could someone explain what equinumerous means in mathematical Theoretical course?!

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  1. manjuthottam
    • 2 years ago
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    the definition states " two sets S and T are called equinumerous, and we write S ~ T, if there exists a bijective function from S onto T". Bijective definition is "a function is bijective if it is surjective and injective". I'm asked to "Prove that if (S \ T) is equinumerous to (T \ S), then S is equinumerous to T" how do i do that?

  2. KingGeorge
    • 2 years ago
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    What exactly do you mean by the S\T and T\S? Are those set minuses?

  3. manjuthottam
    • 2 years ago
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    yes

  4. manjuthottam
    • 2 years ago
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    set S without T

  5. KingGeorge
    • 2 years ago
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    So there are the same number of elements in S that aren't in T as there are elements in T that aren't in S. Let \(U=S\cap T\). So \(S\setminus T=S\setminus (S\cap T)\). Then \(S=S\setminus T +U\). Now let \(f:S\setminus T\to T\setminus S\) such that \(f\) is a bijection. Now define\[g:S\setminus T+U\longrightarrow T\setminus S+U\]by\[ g(s)=\begin{cases} f(s)\qquad s\in S\setminus T \\ s\qquad \;\;\;\;\,s\in U \end{cases}\]

  6. KingGeorge
    • 2 years ago
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    Note that if we restrict \(g\) to only the domain \(U\), we get a bijective function since \(g(g(s))=s\). I.e., \(g\) has an inverse. Similarly, if we restrict \(g\) to only the domain \(S\setminus T\), then we have a bijective function since \(f\) is bijective. So \(g\) is bijective over its whole domain, and is therefore a bijective function. Finally, since \(S\setminus T+U=S\) and \(T\setminus S+U=T\), \(S\) and \(T\) are equinumerous.

  7. KingGeorge
    • 2 years ago
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    Did that all make sense?

  8. manjuthottam
    • 2 years ago
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    oh yes thank you!!

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