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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!

  • 1
    Think of the case where $a$ and $b$ are sets, maybe this will give you the idea of a counterexample...2012-10-05
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    What exactly does $\{a, \{b\}\}$ mean to you? It means the set that consists of the element $a$ and the element $\{b\}$. So, if we're dealing with real numbers, $a$ is a real number and $\{b\}$ is a set that contains the real number $b$. Is this really what you want to say? Do you mean $\{\{a\}, \{b\}\}$?2012-10-05
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    One point being, $a = \{d\}$ makes no sense. A real number equals a set that contains a real number? So, maybe you could distinguish the first element by having it be the one that is not inside set brackets. I mean, all you're doing is rewriting the ordered pair in a more complicated way. But, I guess you could do it.2012-10-05
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    @Graphth: Probably not -- $\{\{a\},\{b\}\}$ is _obviously_ not a viable ordered-pair construction, because it is symmetric in $a$ and $b$. Further, $a=\{d\}$ makes perfect sense -- it says that $a$ is the singleton set whose element is $d$. Who said $d$ was a real number?2012-10-05
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    @Graphth: But the point in set theory is not to build *upon* the real numbers, instead it builds *the* real numbers. *Everything* is a set. Besides, you want set theory for **more** than just real numbers, you want it for its own sake.2012-10-05
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    @everyone Alright, I stand corrected :)2012-10-05
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    If it was supposed to be {{a},{b}} I could have solved it much easier. The fact that it is {a,{b}} is what is confusing me. But that is what the problem states.2012-10-05
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    @Joey: You're asking different questions in the title of your question and in the question itself. The construction from the title, $\{a,\{a,b\}\}$ looks like it will work.2012-10-05
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    Oh dear, that is my bad. In the title I meant to say {a,{b}}. I accidentally put the CORRECT definition in the title. Sorry about that! Can I edit titles? There we go, I fixed the title. As if this problem wasn't confusing enough. Sorry about that.2012-10-05

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