Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Discrete Mathematics Proof Help?

can anyone help me figure this out?

Relevance
• 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.

• 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.