(New page: Category:Set Theory Category:Math == Theorem == Let <math>A</math> be a set in ''S''. Then <br/> A ∪ Ø = A ---- ==Proof== Let x ∈ ''S'', where ''S'' is the universal se...) |
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First we show that if A ∪ Ø ⊂ A. <br/> | First we show that if A ∪ Ø ⊂ A. <br/> | ||
| − | Let x ∈ A ∪ Ø. Then x ∈ A or x ∈ Ø. by definition of the empty set, x cannot be an element in Ø. So by assumption that x ∈ A ∪ Ø, x must be in A. So A ∪ Ø ⊂ A | + | Let x ∈ A ∪ Ø. Then x ∈ A or x ∈ Ø. by definition of the empty set, x cannot be an element in Ø. So by assumption that x ∈ A ∪ Ø, x must be in A. So A ∪ Ø ⊂ A. |
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Next, we want to show that A ⊂ A ∪ Ø.<br/> | Next, we want to show that A ⊂ A ∪ Ø.<br/> | ||
Latest revision as of 11:05, 5 October 2013
Theorem
Let $ A $ be a set in S. Then
A ∪ Ø = A
Proof
Let x ∈ S, where S is the universal set.
First we show that if A ∪ Ø ⊂ A.
Let x ∈ A ∪ Ø. Then x ∈ A or x ∈ Ø. by definition of the empty set, x cannot be an element in Ø. So by assumption that x ∈ A ∪ Ø, x must be in A. So A ∪ Ø ⊂ A.
Next, we want to show that A ⊂ A ∪ Ø.
We know this is true because the set resulting from the union of two sets contains both of the sets forming the union (proof).
Since A ∪ Ø ⊂ A and A ⊂ A ∪ Ø, we have that A ∪ Ø = A.
$ \blacksquare $
References
- B. Ikenaga, "Set Algebra and Proofs Involving Sets" March 1st, 2008, [October 1st, 2013]
