# n-tuples definition

In Kuratowski's definition of n-tuples:

$$(a,b,c,d) = \left\{ {\{ a\} ,\{ a,b\} ,\{ a,b,c\} ,\{ a,b,c,d\} } \right\}$$

This means: $$(a,b,c) = \left\{ {\{ a\} ,\{ a,b\} ,\{ a,b,c\} } \right\}$$ Also, $$(a,b,b,c) = \left\{ {\{ a\} ,\{ a,b\} ,\{ a,b,b\} ,\{ a,b,b,c\} } \right\} = \left\{ {\{ a\} ,\{ a,b\} ,\{ a,b\} ,\{ a,b,c\} } \right\} = \left\{ {\{ a\} ,\{ a,b\} ,\{ a,b,c\} } \right\}$$ Which implies: $$(a,b,c) = (a,b,b,c)$$ What am I missing here?

-
Where did you get this definition? –  Stefan Hamcke Sep 8 '13 at 23:09
Yeah, that definition isn't going to work, because even if you don't care that some $3$-tuples can be equal to some $4$-tuples. For example, you have the real problem that $(a,b,b)=(a,a,b)$. –  Thomas Andrews Sep 8 '13 at 23:13
I found this in a textbook called "Naive set theory". I don't know if I misunderstood the author's intention. –  Ahmed Ali Sep 8 '13 at 23:15

I think your definition is wrong. In my experience, Kuratowski's definition of $n$-tuples is defined inductively by $$(a,b) = \{ \{a\}, \{a,b\} \}$$ and, for $n>2$, $$(a_1, a_2, \dots, a_n) = (a_1, (a_2, \dots\, a_n))$$ This definition doesn't give rise to any nonsense.

For instance, $$(a,b,c) = \{ \{a\}, \{a, \underbrace{\{ \{b\}, \{b,c\} \}}_{(b,c)} \} \}$$ whereas $$(a,b,b,c) = \{ \{a\}, \{a, \underbrace{\{\{b\}, \{b, \underbrace{\{ \{b\}, \{b,c\} \}}_{(b,c)} \} \}}_{(b,b,c)} \} \}$$

-
The above aren't "falsehoods," they just have problems for real usage. :) –  Thomas Andrews Sep 8 '13 at 23:14
I'd say $(1,2,3)=(1,2,2,3)$ is false! Or did I misinterpret your comment? –  Clive Newstead Sep 8 '13 at 23:16
It's a definition. It isn't a falsehood, it just indicates that you can't use the definition for what you want it to be. A falsehood is a very particular thing in mathematics, and it is bad to confuse it with "intuitive" meanings. In particular, I'd call that a "problem" with the definition, not a falsehood. –  Thomas Andrews Sep 8 '13 at 23:17
@ThomasAndrews: Fair point. Perhaps I meant "should be false"! –  Clive Newstead Sep 8 '13 at 23:19
Thanks, it seems that I misunderstood the author's explanation for a definition. Sorry if I bothered anyone with this question. :) –  Ahmed Ali Sep 8 '13 at 23:26