(New page: Category:ECE600 Category:Set Theory Category:Math == Theorem == Intersection is distributive over union ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) <br/> <math>A\cap (B\c...) |
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Intersection is distributive over union ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) <br/> | Intersection is distributive over union ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) <br/> | ||
<math>A\cap (B\cup C) = (A\cap B)\cup (A\cap C</math> <br/> | <math>A\cap (B\cup C) = (A\cap B)\cup (A\cap C</math> <br/> | ||
| − | where <math>A</math>, <math>B</math> and <math>C</math> are | + | where <math>A</math>, <math>B</math> and <math>C</math> are sets. |
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Combining the two results we have that: <br/> | Combining the two results we have that: <br/> | ||
| − | A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C) and (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C) | + | A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C) and (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C) |
⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)<br/> | ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)<br/> | ||
<math>\blacksquare</math> | <math>\blacksquare</math> | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | == References == | ||
| + | |||
| + | * W. Rudin, "Basic Topology" in "Principles of Mathematical Analysis", 3rd Edition, McGraw-Hill Inc. ch 2, pp 28. | ||
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[[Proofs_mhossain|Back to list of all proofs]] | [[Proofs_mhossain|Back to list of all proofs]] | ||
Latest revision as of 10:13, 1 October 2013
Theorem
Intersection is distributive over union ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
$ A\cap (B\cup C) = (A\cap B)\cup (A\cap C $
where $ A $, $ B $ and $ C $ are sets.
Proof
Let x ∈ A ∩ (B ∪ C). Then, x ∈ A and x ∈ (B ∪ C) ⇒ x ∈ A and at the same time, x ∈ B or x ∈ C, possibly both ⇒ either x ∈ A and x ∈ B or x ∈ A and x ∈ C (possibly both). Hence, x ∈ (A ∩ B) or x ∈ (A ∩ C), i.e. x ∈ (A ∩ B) ∪ (A ∩ C).
So we have that x ∈ A ∩ (B ∪ C) ⇒ x ∈ (A ∩ B) ∪ (A ∩ C), which is equivalent to saying that A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C).
Next we assume that x ∈ (A ∩ B) ∪ (A ∩ C). Then, x ∈ (A ∩ B) or x ∈ (A ∩ C)⇒ x ∈ A in addition to being in B, C or both ⇒ x ∈ A ∩ (B ∪ C).
This gives us that x ∈ (A ∩ B) ∪ (A ∩ C) ⇒ x ∈ (A ∩ B) ∪ (A ∩ C)A ∩ (B ∪ C), which is equivalent to saying that (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C).
Combining the two results we have that:
A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C) and (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C)
⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
$ \blacksquare $
References
- W. Rudin, "Basic Topology" in "Principles of Mathematical Analysis", 3rd Edition, McGraw-Hill Inc. ch 2, pp 28.
