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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.)

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Your $(1)$ is equivalent to $B\subseteq A$, not $A\subseteq B$. –  Brian M. Scott Apr 11 '12 at 2:41
    
@Brian Yes. I meant the opposite, I'll correct it. –  Pedro Tamaroff Apr 11 '12 at 2:42
    
@Brain What about now? –  Pedro Tamaroff Apr 11 '12 at 2:43
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No, you should have from $(1)$ that $x\in A\implies x\in B\lor x\in A$ and hence that $$B=\{x:x\in B\}=\{x:x\in B\lor x\in A\}=A\cup B\;.$$ Your $\{x:x\in B\land x\in A\}$ is $A\cap B$. By the way, it’s very bad practice to write $A\land B\subseteq S$ when you mean $A,B\subseteq S$. –  Brian M. Scott Apr 11 '12 at 2:55
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$\land$ is \land; $\lor$ is \lor. If you right-click on any formula, select Show Math As, and then select TeX Commands, you can see how anything was done. –  Brian M. Scott Apr 11 '12 at 3:01

1 Answer 1

up vote 2 down vote accepted

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 $A\cap B=A$.

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