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0-ary functions

by honestrosewater
Tags: 0ary, functions
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honestrosewater
#1
Mar28-05, 11:18 AM
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From my book:
"We agree that there is exactly one 0-tuple in [set] X, and we designate it by ( ). A 0-ary 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 0-ary 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.
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HallsofIvy
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Mar28-05, 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 "0-ary function from X to Y is simply an element of Y."
matt grime
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Mar28-05, 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.

honestrosewater
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Mar28-05, 11:57 AM
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0-ary 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?
matt grime
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Mar28-05, 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 0-ary functions is equivalent to Y, the space of 1-ary functions is equivalent to Y^X, the space of n-ary functions is Y^{Xx..xX} with n copies of X in the product for n >0. Seems reasonable, doesn't it?
honestrosewater
#6
Mar28-05, 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 0-ary 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?
matt grime
#7
Mar28-05, 02:49 PM
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f:X \to Y is some fucntion from X to Y, it's written in pseudo-tex, and I'm not doing anything with metric spaces.

The set of 0-ary functions frmo X to Y is naturally isomorphic as a set to Y. Thus there is an 0-ary function for each element of Y ande each element of Y determines an 0-ary fucntion.

Incidentally, I've never met an 0-ary 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 0-ary function is a map from the set of 0-tuples, of which there is just one -the empty string with no elements - to Y.
honestrosewater
#8
Mar28-05, 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.
matt grime
#9
Mar28-05, 04:08 PM
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well, it is a function in the ordinary sense. The domain is the set of 0-tuples 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.
honestrosewater
#10
Mar29-05, 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 0-tuples, and was thinking of the (0) objects in the 0-tuple 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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