Let $A,B$ be sets. Show that the three statements $A\subseteq B$, $A\cup B = B$ and $A\cap B = A$ are logically equivalent.

elementary-set-theorysolution-verification

Let $A,B$ be sets. Show that the three statements $A\subseteq B$, $A\cup B = B$ and $A\cap B = A$ are logically equivalent.

MY ATTEMPT

  • $A\subseteq B$ $\Rightarrow$ $A\cup B = B$.

If $x\in A\cup B$, either $x\in A$ or $x\in B$. In the first case, due to the assumption, $x\in B$. In the second case, clearly $x\in B$. Thus, in both cases, one has that $x\in B$, which finishes the first part (the inclusion $B\subseteq A\cup B$ is obvious).

  • $A\cup B = B \Rightarrow A\cap B = A$.

If $x\in A\cup B$, either $x\in A$ or $x\in B$. In both cases, we conclude that $x\in B$, given that $A\cup B\subseteq B$. Thus, if $x\in A$, we conclude that $x\in B$, due to the fact that $A\cup B\subseteq B$. In other words, $A\subseteq A\cap B$. Since the inclusion $A\cap B\subseteq A$ is obvious, we are done.

  • $A\cap B = A \Rightarrow A\subseteq B$.

If $x\in A$, then $x\in A\cap B$, since we have that $A\subseteq A\cap B$. But then $x\in B$. In other words, if $x\in A$, we have proven that $x\in B$, and the inclusion $A\subseteq B$ holds.

I would like to know if I am reasoning correctly. Can someone check if my argument proceed?

Best Answer

Yes, that is convincing.

The second paragraph may need a little polishing.   The reasoning is valid, but a little confusing to follow. Something like:

Let us prove now that $A\cup B=B$ implies that $A\cap B=A$.   Firstly, $A\cap B\subseteq A$ is obvious.   Secondly, $A\subseteq A\cap B$ is proven thusly: anything in $A$ is in $A\cup B$, which equals $B$ by the premise, so therefore anything in $A$ is both in $A$ and in $B$.

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