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poochie_d
Sep15-11, 11:36 AM
I have read that a function f: A -> B can be defined as an ordered triple of sets (A,B,X), where X is the set of all ordered pairs X = \{(a,f(a)) \in A \times B\}. But ordered tuples are really functions from \{1, ..., n\} to (whatever set under consideration), right? So isn't this a circular definition? Or is there a more basic definition of functions that does not involve tuples? Any help would be greatly appreciated.

micromass
Sep15-11, 12:12 PM
But ordered tuples are really functions from \{1, ..., n\} to (whatever set under consideration), right?

No, this is not true. The ordered tuple (a,b) is defined as {{a},{a,b}}. It's not defined as a function.

poochie_d
Sep15-11, 12:27 PM
But aren't tuples other than the ordered pair defined as functions, so that the definition of functions as triples would still be circular?

micromass
Sep15-11, 12:28 PM
But aren't tuples other than the ordered pair defined as functions, so that the definition of functions as triples would still be circular?

No, triples can be defined as

(a,b,c)=((a,b),c)

And the definition of a function only uses ordered pairs and triples. So there is nothing circular.

poochie_d
Sep15-11, 12:59 PM
Oh I think I get it now. Thanks micromass!