# Set theory homework

Can you help me with the following exercise? The main reason I can't do it is because I think it's impossible.

Given A and B sets, let X be a set with the following properties:

P1) $X\supset A$ and $X\supset B$

P2) If $Y\supset A$ and $Y\supset B$ then $Y\supset X$

Prove that $X=A\cup B$

From the way I see it, if X has the properties P1 and P2, $A\cup B \subset X$, but not necessarily $X \subset A \cup B$. That is, I think the properties mean X will contain $A\cup B$ but X can be much bigger than that. I don't see how $X\setminus A\cup B$ is necessarily empty. I don't understand the use of P2, either. How does P2 constrain X to exactly $A\cup B$?

Thus, I don't think I can prove what it asks because it's wrong. But I feel I'm missing something. Any help is appreciated.

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Does $X\supset A$ mean $A\subset X$ or $A \subseteq X$? –  Alraxite Aug 17 '13 at 20:16
Informally, the second property means any set containing $A$ as a subset and $B$ as a subset contains $X$. This (together with (1)) implies that $X$ is the SMALLEST subset containing both $A$ and $B$ as subsets, or $X = A \cup B$. –  Alex Wertheim Aug 17 '13 at 20:17
$X \supset A$ means X is just a superset of A, not a proper one. –  BeetleTheNeato Aug 17 '13 at 20:18

HINT: The first property alone is enough to ensure that $A\cup B\subseteq X$. Now let $Y=A\cup B$, note that $Y\supseteq A$ and $Y\supseteq B$, and see what the second property tells you.
To motivate this note that we want to prove $A\cup B \supset X$. The conditions we are given don't tell us about $A\cup B$, but they do tell us about a $Y\supset X$, so it is natural to give $Y=A\cup B$ a try. –  Mark Bennet Aug 17 '13 at 20:28
Then X must be equal to Y, is that right? Because since $Y = A\cup B$ and $Y\supset X$, there's no possibility of an $X\neq A\cup B$. I think this proof is too circular. Isn't it? Does it suffice? –  BeetleTheNeato Aug 17 '13 at 21:03
@BeetleTheNeato: You already knew that $A\cup B\subseteq X$, and now you know that $X\subseteq A\cup B$; this immediately implies that $X=A\cup B$. It's a general principle that for sets $S$ and $T$, $S=T$ if and only if $S\subseteq T$ and $T\subseteq S$. –  Brian M. Scott Aug 18 '13 at 0:59