If A∩B' = φ, then show that A = A∩B, and hence show that A⊆B.

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> [!TLDR]
> $A\cap B'=\varphi$ means no element of $A$ lies outside $B$; this forces $A=A\cap B$, and since $A\cap B\subseteq B$ always, $A\subseteq B$.
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> **Step 1: Show $A=A\cap B$.**
> We always have $A\cap B\subseteq A$ (an element in both sets is certainly in $A$).
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> For the reverse inclusion, let $x\in A$. Suppose, for contradiction, $x
otin B$; then $x\in B'$ (the complement of $B$). Combined with $x\in A$, this gives $x\in A\cap B'$. But we are given $A\cap B'=\varphi$, so no such $x$ exists — contradiction. Hence every $x\in A$ must satisfy $x\in B$, so $x\in A\cap B$. This shows $A\subseteq A\cap B$.
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Q.If A∩B' = φ, then show that A = A∩B, and hence show that A⊆B.

Concept understanding — Complement of a Set

Q.If A∩B' = φ, then show that A = A∩B, and hence show that A⊆B.

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