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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
up vote 10 down vote accepted

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.

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Beautiful! I was typing up some clumsy response to this question but this is far superior. I've reached my vote limit for the day but I'll have to come back to this later. :) – Alex Wertheim Aug 17 '13 at 20:19
@AWertheim: Thanks! – Brian M. Scott Aug 17 '13 at 20:20
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

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