Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
I guess by Kuratowski pair you mean $\{ \{ x \}, \{ x, y \} \}$, right? –  Martin Sleziak Jan 25 '12 at 15:28
2  
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

1 Answer 1

up vote 7 down vote accepted

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 $$\begin{align*} \cap\cup\langle x,y\rangle &= \cap\{x,y\} = x\cap y,\\ \cup\cap\langle x,y\rangle &= \cup\{x\} = x,\\ \cup\cup\langle x,y\rangle &= \cup\{x,y\} = x\cup y. \end{align*}$$

Things should be rather easy now.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.