# How to prove that if $A\cap B' =\varnothing$ then $A \subseteq B$?

What are the steps of proving this?

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

where $B'$ is the complement of $B$.

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you really should accept some of the answers to your previous questions. – Nana Mar 22 '12 at 5:15

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

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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. – Austin Mohr Mar 22 '12 at 6:18

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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Since you are asking for the steps for proving the statement, I would like to say that write down the definitions of $$A\cap B^c=\emptyset ,$$ $$B^c,$$ and $$A\subset B$$ first. And then try to go on. (Here $B^c$ is your $B'$.)

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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.

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