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0ary functions 
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#1
Mar2805, 11:18 AM

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From my book:
"We agree that there is exactly one 0tuple in [set] X, and we designate it by ( ). A 0ary function from X to Y is then completely determined by its value for the argument ( ). We shall identify the function with this value. This means that a 0ary function from X to Y is simply an element of Y." I don't understand the reasoning behind the second sentence. How does a function even work on an empty sequence? Edit: I may as well tack this on here. "universe" and "individuals" are undefined terms, right? I've never seen them defined, not formally anyway, and they seem to be basic concepts. 


#2
Mar2805, 11:42 AM

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It just arbitrarily assigns a member of Y to the empty sequence. Since it is "completely determined by its value for the argument ( )", any such function can be identified with that particular value. In that sense, a "0ary function from X to Y is simply an element of Y."



#3
Mar2805, 11:47 AM

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It isn't a function on X. This just a convention just like 0!=1, I shouldn't worry about it. In lots of bits of mathematics we deal with maps more generally than just the ones you think of with inputs and outputs.



#4
Mar2805, 11:57 AM

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0ary functions
Okay, but how does it behave like a variable or constant or something else? If it behaves as a variable, what does it vary through?



#5
Mar2805, 01:10 PM

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Variable and constant aren't particulalry meaningful terms, really, if you think about it  this is set theory, not physics. Or let me put it this way: functionf f:X \to Y behave as a constant or a variable?
The space of 0ary functions is equivalent to Y, the space of 1ary functions is equivalent to Y^X, the space of nary functions is Y^{Xx..xX} with n copies of X in the product for n >0. Seems reasonable, doesn't it? 


#6
Mar2805, 02:14 PM

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I hate to frustrate you, but I don't know what f:X \to Y means, and I haven't learned about metric spaces yet. This is a mathematical logic book and only the second chapter at that. Perhaps I won't be able to learn logic and foundational material first, but I'm not ready to give up yet. If you feel like helping, what I have in mind is if I let y be a/the value of the 0ary function from X to Y, is y an arbitrary element of Y, so that I can generalize from y to every element of Y, or is y a particular element of Y, so that I cannot generalize from it?



#7
Mar2805, 02:49 PM

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f:X \to Y is some fucntion from X to Y, it's written in pseudotex, and I'm not doing anything with metric spaces.
The set of 0ary functions frmo X to Y is naturally isomorphic as a set to Y. Thus there is an 0ary function for each element of Y ande each element of Y determines an 0ary fucntion. Incidentally, I've never met an 0ary function before reading this thread, i'm just telling you what the definition you wrote in the first post says in almost exactly the same words. An 0ary function is a map from the set of 0tuples, of which there is just one the empty string with no elements  to Y. 


#8
Mar2805, 03:35 PM

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Okay, great. I think I'll need to broaden my concept of a function, but I understand it well enough for now isomorphism was the key. Thank you.



#9
Mar2805, 04:08 PM

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well, it is a function in the ordinary sense. The domain is the set of 0tuples of which there is exactly one denoted (), or since we're only thinking about sets, we may as well denote it @, or anything else we may care to use. Thus we're looking at the space of all functions in the proper sense from the set {@} to the set Y. Any function from a set with one element to any set S takes a unique value,so we can identify the set of all functions with Y. Just as we can always identify maps(X,Y) with Y^X.
The more general concept of maps are called morphisms, btu now i come to look at the question more closely I realize it was completely unnecessary to even allude to them. 


#10
Mar2905, 12:02 AM

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Wow, I don't know how in the world I did this, but I completely missed the set of 0tuples, and was thinking of the (0) objects in the 0tuple as the domain.! It wasn't actually having a function defined on an empty domain that confused me (I don't really know if that in itself would fail) it was how the values were assigned that baffled me. Anyway, I understand what the actual definition says now. Thanks again.



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