Discrete MathShow that if A \not{\subset} B^{c}, then A \cap B=\phi. Hint

BolkowN 2021-08-22 Answered

Discrete Math
Show that if \(A \not{\subset} B^{c}\), then \(\displaystyle{A}\cap{B}=\phi\). Hint:Use the contrapositive.

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Expert Answer

delilnaT
Answered 2021-08-23 Author has 16248 answers

Step 1
Given: \(A \not{\subset} B^{c}\)
We need to prove that \(\displaystyle{A}\cap{B}=\phi\).
Proof by contradiction:
Suppose \(\displaystyle{A}\cap{B}\ne{}\phi\), then \(\displaystyle\exists{x}\in{A}\cap{B}\)
Therefore \(\displaystyle{x}\in{A}\) and \(\displaystyle{x}\in{B}\)
Step 2
Since \(\displaystyle{A}\subset{B}^{{{c}}}\) and \(\displaystyle{x}\in{A}\), therefore
\(\displaystyle{x}\in{A}\subset{B}^{{{c}}}\Rightarrow{x}\in{B}^{{{c}}}\). But \(\displaystyle{x}\in{B}\) as well. Which is not true because no element of a set B can be found in both B and \(\displaystyle{B}^{{{c}}}\), as \(\displaystyle{B}\cap{B}^{{{c}}}=\phi\). Therefore our assumption that \(\displaystyle{A}\cap{B}\ne{}\phi\) must be false.
Hence \(\displaystyle{A}\cap{B}=\phi\).

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