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I'm proving that if $A , B\subset S$ then

$A\subset B \Leftrightarrow A\cup B=B$ $A\subset B \Leftrightarrow A\cap B=A$

I go as follows:

Suppose it is true that $A\subset B$. This implies that if $x\in A$ then it is also true that $x \in B$, this is to say

$\tag{1} x\in A\Rightarrow x \in B$

This means that, using the set builder notation

$B=\{ x:x\in B\}=\{x:x\in B \vee x\in A\}=A\cup B$

Now suppose $A\cup B=B$ is true. This means that

$A\cup B=B=\{ x:x\in B\}=\{x:x\in B \vee x\in A\}$

But then this means that if $x\in A$ then it is also true that $x\in B$, which means that $A\subset B$.

(The proof has been now corrected.)

  • 0
    I second Brian's objection to $A\wedge B\subset S$. There is no object $A\wedge B$ which is a subset of $S$.2012-04-11

1 Answers 1

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What you have now is basically correct, but I think that a somewhat different form of argument, which I’ll illustrate with the second equivalence, is a bit easier to follow.

Suppose first that $A\subseteq B$. If $x\in A\cap B$, then $x\in A$ and $x\in B$, so certainly $x\in A$. Thus, $A\cap B\subseteq A$. On the other hand, if $x\in A$, then by hypothesis $x\in B$ as well, so $x\in A$ and $x\in B$, and therfore $x\in A\cap B$. This shows that $A\subseteq A\cap B$. Putting the pieces together, we see that $A\cap B=A$.

Now suppose that $A\cap B=A$. If $x\in A$, then $x\in A\cap B$, so $x\in B$, and it follows immediately that $A\subseteq B$.

We conclude that $A\subseteq B$ iff $A\cap B=A$.