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Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Discrete Mathematics Proof Help?

can anyone help me figure this out?

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2 Answers

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  • 1 month ago

    Inf.

    U Ai=A1 U A2 U A3 U....U Ai U...

    i=1

    Proof:

    Let P(i) be the proposition as stated above.

    .........1

    P(1)=U Ai=A1 is true.

    ........i=1

    Assume P(k) is true for some i=k, i.e.

    k

    U Ai=A1 U A2 U A3 U...U Ak is true

    i=1

    =>

    k

    U Ai U A(k+1)=A1 U A2 U A3 U...U Ak U A(k+1)

    i=1

    =>

    k+1

    U Ai=A1 U A2 U A3 U....U A(k+1)=>P(k+1) is true

    i=1

    Thus P(i) is true for all natural number i, where i is

    in (1, infinity) by math-induction.

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  • nbsale
    Lv 6
    1 month ago

     The Ai are nested (Aj is a subset of Ai if j >=i ), and the union is just the first one, (1,∞).

    The proof is short and trivial. Using induction is really overkill here.

    We simply need show A1 = U{i=1 to ∞} Ai.

    The usual way to show two sets are equal is to show each is a subset of the other.

    Clearly A1 is a subset of the union which includes A1.

    Now pick any x in UAi. Then x is in some set Aj = (j, ∞), making j < x

    Since j >=1, 1 <= j < x, so x is in A1. Therefore UAi is a subset of A1.

    Since we have shown each set is a subset of the other, the sets are equal.

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