Prove that $(\cap\cup\langle x,y\rangle) \bigcup \bigl((\cup\cup\langle x,y\rangle ) \setminus (\cup\cap\langle x,y\rangle)\bigr) = y$

$\langle x,y \rangle$ is a Kuratowski pair. Prove that $$\Bigl(\cap\cup\langle x,y \rangle\Bigr) \bigcup \Bigl((\cup\cup\langle x,y \rangle)\setminus(\cup\cap\langle x,y \rangle)\Bigr)=y$$

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I guess by Kuratowski pair you mean $\{ \{ x \}, \{ x, y \} \}$, right? – Martin Sleziak Jan 25 '12 at 15:28
Just write out the elements in the unions/intersections. For example, $\cap \cup \langle x,y \rangle = \cap \{x,y\} = x \cap y$. – Mikko Korhonen Jan 25 '12 at 15:33
Yes, that`s what I meant. – John Doe Jan 25 '12 at 15:35
Thanks for the help. – John Doe Jan 25 '12 at 16:04
What have you tried? It's not a very hard exercise. You just need to unfold the definitions. – Asaf Karagila Jan 25 '12 at 16:32

Note that \begin{align*} \cup\langle x,y\rangle &= \cup\{ \{x\},\{x,y\}\}= \{x\}\bigcup\{x,y\} = \{x,y\}\\ \cap\langle x,y\rangle &= \cap\{ \{x\},\{x,y\}\} = \{x\}\bigcap\{x,y\}=\{x\}. \end{align*}
So \begin{align*} \cap\cup\langle x,y\rangle &= \cap\{x,y\} = x\cap y,\\ \cup\cap\langle x,y\rangle &= \cup\{x\} = x,\\ \cup\cup\langle x,y\rangle &= \cup\{x,y\} = x\cup y. \end{align*}