So we know that the an ordered pair $(a,b) = (c,d)$ if and only if $a = c$ and $b = d$. And we know the Kuratowski definition of an ordered pair is: $(a,b) = \{\{a\},\{a,b\}\}$
http://en.wikipedia.org/wiki/Ordered_pair#Kuratowski_definition
The proof for the Kuratowski definition is in the wikipedia link.
Now, why is this alternate definition, $(a,b) = \{a,\{b\}\}$ incorrect?
I'm trying to follow the proof for $(a,b) = (c,d)$ iff $a = c$ and $b = d$ as given in the wikipedia link, only for this alternate definition for an ordered pair, in order to search for a contradiction. But I don't think I'm dong it right.
I started with...
- $(a,b) = (c,d)$
- Then $\{a,\{b\}\} = \{c,\{d\}\}$ based on the alternate definition
Now...
- Suppose $a \neq b$
- $\{a,\{b\}\} = \{c,\{d\}\}$
- But since it's an ordered pair, either of the following can be true?
- $a = c$ and $\{b\} = \{d\}$ ?
- OR $a = \{d\}$ and $\{b\} = c$ ?
Yeah I have no idea where to reallly go from here. Is that a contradiction in itself? I can't tell.
Thank you for the help.
Edit: Alright, I have developed a counter example based mostly off of Asaf Karagila answer (Thanks Asaf!). Essentially what I needed to do was prove that, by this definition, a != c or b != d, even when (a,b) = (c,d).
So using what Asaf told me, I set a = {x} and b = y. Which by the incorrect definition gives... (a,b) = {{x},{y}}
Then I set c = {y} and d = x, which gives (c,d) = {{y},{x}} which is equivalent to {{x},{y}}
So, (a,b) = (c,d) even though a != c and b != d, which is a contradiction. I cleared this method with my professor.
Thanks for the help everyone!