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What are the steps of proving this?

If A\cap B' = \varnothing then $A \subseteq B$

where B' is the complement of $B$.

3 Answers 3

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$x\in A \implies x\notin B^c \implies x\in B$

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    The first implication holds because $A$ and $B^\prime$ cannot have any elements in common (this is what $A \cap B^\prime = \emptyset$ means). The second implication holds because any element you pick must either belong to a given set or belong to its complement, but not both.2012-03-22
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Let me give it a try:

  • For every $A$, $A=b(A)\cup c(A)$ with $b(A)=A\cap B\subseteq B$ and c(A)=A\cap B'\subseteq B'.
    To prove this, note that B\cup B' is everything hence A=A\cap(B\cup B')=(A\cap B)\cup(A\cap B') by distributivity of $\cup$ with respect to $\cap$.
  • The hypothesis is that $c(A)=\varnothing$. Hence $A=b(A)$. Recall that $b(A)\subseteq B$, always. Hence $A\subseteq B$.

Note: The conditions that A\cap B'=\varnothing and that $A\subseteq B$ are in fact equivalent.

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Nana proves it directly. A proof by contradiction is supplied here.

If $A = \emptyset$, then $A \subseteq B$.

If $A \neq \emptyset$, let $x \in A$. We want to show that $x \in B$.

Suppose not, then x \in B'. We have x \in A \cap B' = \emptyset which is not possible.

Hence $A \subseteq B$.

Edit: I answered this question to test the proof approach and technique I learnt from school years ago. Would anybody point out anything that is wrong or inappropriate in the proof so that I can improve them? Thanks.