releasing a second question concerning ordered stuff and set theory, which is very similar to my first question (Ordered Pairs and Set Theory) which I'm putting here for reference.

The next problem that I'm stuck at in my book is this: we have an ordered triple $(a,b,c)$ and we want to define it using set theory. Why would $\{\{ a\} ,\{ a,b\} ,\{ a,b,c\}\}$ not work? The book requires from me to prove why it is not valid.

Additional info: I don't know how this matters but I'm gonna put it up anyhow. It is written within the book that the correct way to show it is $\{\{ a\} ,\{\{ a\} ,\{\{ b\} ,\{ b,c\}\}\}\}$

EDIT: Now, I consider that the correct case should matter. Besides pointing out why the first case is wrong I'm curious as to why the second one is correct

  • $\begingroup$ Hrm - looks fine to me! You can infer the first element from the one element set, the second element from the first element and the two element set, and the third element from the first two elements and the three element set. $\endgroup$
    – user261214
    Dec 15 '15 at 15:46
  • $\begingroup$ The book literally says it's a no-go for the first set theory case and that the second system would be correct. I could assume that the book lies but I strongly believe that is not the case. Also I edited something in. $\endgroup$ Dec 15 '15 at 15:57
  • $\begingroup$ Aahh - I forgot about repeated elements. Thanks @Henry! $\endgroup$
    – user261214
    Dec 15 '15 at 15:59
  • $\begingroup$ I added in more information, and therefore expanded my question. $\endgroup$ Dec 15 '15 at 16:03

If you try to write $(a,a,b)$ as $\{\{a\},\{a,a\},\{a,a,b\}\}$, removing duplicate set elements simplifies it to $\{\{a\},\{a\},\{a,b\}\}$ and then to $\{\{a\},\{a,b\}\}$.

Similarly if you try to write $(a,b,b)$ as $\{\{a\},\{a,b\},\{a,b,b\}\}$, removing duplicate set elements simplifies it to $\{\{a\},\{a,b\},\{a,b\}\}$ and then to $\{\{a\},\{a,b\}\}$.

So this method risks different ordered triples appearing to be the same

  • $\begingroup$ Is that considered a proof? Or am I required to prove it some other way? $\endgroup$ Dec 15 '15 at 16:00
  • 2
    $\begingroup$ @DamjanBabić Here this counterexample is a valid proof. $\endgroup$
    – lisyarus
    Dec 15 '15 at 16:01

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