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Prove $(A\cap B=A) \land(A\cup B=B)\iff A\subseteq B$

To prove this both $(A\cap B=A) \land(A\cup B=B)\implies A\subseteq B$ and $A\subseteq B\implies(A\cap B=A) \land(A\cup B=B)$ must hold true. I will refer to them as $(1)$ and $(2)$ respectively.

  • To prove $(1)$:

Assume $(A\cap B=A) \land(A\cup B=B)$. Let $x\in A$. Then $x\in A\cup B=B$ so $A\subseteq B$.

Or in another way,

Assume $(A\cap B=A) \land(A\cup B=B)$. Let $x\in A$. Then $x\in A\cap B$ so $x\in A$ and $x\in B$. So $A\subseteq B$.

  • To prove $(2)$ we have to prove the following four things assuming $A\subseteq B$ and knowing $A\subseteq B$ means that $x\in A\implies x\in B$:

$$A\cap B\subseteq A\tag{3}$$ $$A\cap B\supseteq A\tag{4}$$ $$A\cup B\subseteq B\tag{5}$$ $$A\cup B\supseteq B\tag{6}$$

To prove $(3)$:

Let $x\in A\cap B$. Then $x\in A$, so $(3)$ is proven.

To prove $(4)$:

Let $x\in A$. Now we know from $A\subseteq B$, that $x\in B$, so $x\in A$ and $x\in B$ so $A\subseteq A\cap B$.

To prove $(5)$:

Let $x\in A\cup B$. Then $x\in A$ or $x\in B$, but $x\in A$ implies $x\in B$ so either ways $x\in B$. Hence $x\in A\cup B\implies x\in B$.

To prove $(6)$:

Let $x\in B$, then $x\in B\cup A$, so $B\subseteq A\cup B$.

Now that all the conditions have been proven we can conclude that $(A\cap B=A) \land(A\cup B=B)\iff A\subseteq B$ is true. $\blacksquare$

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    Do you want us to verify your solution? – 2017-02-26
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    @AkivaWeinberger Yes, I do want some validation as I'm not completely sure about the result. That's why I tagged it as proof-verification. – 2017-02-26
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    Your solution seems fine to me. – 2017-02-26
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    If x belongs to A then it maybe not belongs to A intersection B. – 2017-02-26
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    So in proving (1) your another way is not correct. – 2017-02-26
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    @Amar why should it be incorrect? He assumed $A=A\cap B$ thus $x\in A=A\cap B\implies x\in A\cap B$ – 2017-02-26

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