(New page: Category:Set Theory Category:Math == Theorem == Let <math>A</math> be a set in ''S''. Then <br/> A ∩ Ø = Ø ---- ==Proof== Let x ∈ ''S'', where ''S'' is the universal s...) |
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==Proof== | ==Proof== | ||
| − | + | First we show that A ∩ Ø ⊂ Ø. <br/> | |
| + | We know this is true because the set resulting from the intersection of two sets is a subset of both of the sets forming the intersection ([[Union_and_intersection_subsets_mh|proof]]). | ||
| − | + | Next, we want to show that Ø ⊂ A ∩ Ø.<br/> | |
| − | We know this is true because the set | + | We know this is true because the empty set is a subset of all sets ([[empty_set_contained_in_every_set|proof]]). |
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Since A ∩ Ø ⊂ Ø and Ø ⊂ A ∩ Ø, we have that A ∩ Ø = Ø. <br/> | Since A ∩ Ø ⊂ Ø and Ø ⊂ A ∩ Ø, we have that A ∩ Ø = Ø. <br/> | ||
<math>\blacksquare</math> | <math>\blacksquare</math> | ||
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[[Proofs_mhossain|Back to list of all proofs]] | [[Proofs_mhossain|Back to list of all proofs]] | ||
Latest revision as of 06:11, 6 October 2013
Theorem
Let $ A $ be a set in S. Then
A ∩ Ø = Ø
Proof
First we show that A ∩ Ø ⊂ Ø.
We know this is true because the set resulting from the intersection of two sets is a subset of both of the sets forming the intersection (proof).
Next, we want to show that Ø ⊂ A ∩ Ø.
We know this is true because the empty set is a subset of all sets (proof).
Since A ∩ Ø ⊂ Ø and Ø ⊂ A ∩ Ø, we have that A ∩ Ø = Ø.
$ \blacksquare $
