Elementary set theory - Confused about A $\cup$ B = B I am trying to solve: 
Proof that  $\exists!A\subset U(\forall B \subset U(A \cup B = B)$. 
But once I think about it, I can see not one but 2 solutions.
a) $ \varnothing \cup B = B $
b) $ B \cup B = B $
The solution is supposed to be a), what is wrong about b)?
Thanks!
 A: What's wrong with b) is that the question asks you to find some subset $A$ of $U$ such that, for each subset $B$ of $U$, $A\subset B$. Saying that the answer is $B$ makes no sense. What is $B$? On the other hand, $\emptyset$ makes sense (and it is correct since, indeed, $\emptyset\subset B$, for each $B\subset U$).
It's like someone asks you whether there is a natural number $n$ such that $n\leqslant m$ for each natural number $m$. Would you say that an answer is $m$? I guess not. I suppose that you would say that the answer is $1$ (or $0$, if, for you, $0\in\mathbb N$).
A: It needs to work for all possible sets $B$.  Not just one specific $B$.   
For example: if $U =\mathbb R$ the we need to find an $A$ so that $A \cup B = B$ for ALL $B$.  So for example $A \cup [5, 7.3) = [5, 7.3)$ and $A \cup \mathbb Q = \mathbb Q$ and $A \cup \{-27, 0, \sqrt{\pi}, e^{-\frac 23}\}= \{-27, 0, \sqrt{\pi}, e^{-\frac 23}\}$.
It should be clear from those examples that the only possible set that will do that is $A = \emptyset$.
We can formally prove that.  But notice because $A\cup [5,7.30)  = [5,7.30)$ then $2 \not \in A\cup [5,7.3)$ so $2 \not \in A$ and $A \cup \{-27, 0, \sqrt{\pi}, e^{-\frac 23}\}= \{-27, 0, \sqrt{\pi}, e^{-\frac 23}\}$ so $5\frac 12 \not \in A \cup \{-27, 0, \sqrt{\pi}, e^{-\frac 23}\}= \{-27, 0, \sqrt{\pi}, e^{-\frac 23}\}$ so $5\frac 12 \not \in A$, etc.  Can anything be in $A$?  Well, no because.... if $x \in U$ then $A\cup \{x\}^c = \{x\}^c$ and $x \not \in A\cup \{x\}^c$ so $x \not \in A$.
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By the way.  Using your incorrect  interpretation of the question: If $B\subset U$ then show there exists a $A$ so that $A \cup B = B$. (a different question than what was asked.) Then the answer to that can be any subset of $B$.  Not just $\emptyset$ or $B$ but for any $A \subset B$ then $A \cup B = B$.
But that's an entirely different question.
