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February 26th, 2018, 09:49 PM   #1
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Finite mapped on infinite set

Can a finite set ever be mapped onto an infinite set?
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February 26th, 2018, 10:08 PM   #2
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No.
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February 27th, 2018, 12:29 AM   #3
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Couldn't, e.g., each member of a finite set be mapped countless times into an infinite set?
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February 27th, 2018, 12:34 AM   #4
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Then, contrary to the definition of a function, f(n) would have more than one value. How could you avoid that? Can you be more explicit as to exactly how each member of N would be mapped to S?
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February 27th, 2018, 01:42 AM   #5
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In set theory, we have the notion of a relation and of a function.
A function is a special relation, but most of the function-language makes sense for relations really.

So you mean, to take the finite set {*} and the infinite set N={0,1,2,3,...},
we want to define *R0, *R1, *R2, etc (meaning that * is related to 0, * is related to 1, etc.) This is a perfectly fine relation. A function, however, is a relation R such that for each x, there is EXACTLY one y such that xRy.

If we have *R0 and *R1, this condition is violated since there are multiple numbers y such that *Ry.

Last edited by skipjack; February 27th, 2018 at 01:55 AM.
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February 27th, 2018, 02:19 AM   #6
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Quote:
Originally Posted by Micrm@ss View Post
A function, however, is a relation R such that for each x, there is EXACTLY one y such that xRy.
"At most" rather than "exactly".
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February 27th, 2018, 04:55 AM   #7
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Quote:
Originally Posted by v8archie View Post
"At most" rather than "exactly".
No, exactly is the correct word. This IS the most common usage of the term function by mathematicians, regardless of what high school teachers say.
If you want "at most", it is nowadays called a partial function, among other names.
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February 27th, 2018, 05:55 AM   #8
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Consider f to be the finite set $\displaystyle f = \{-1, 0, 1 \}$. Then define $\displaystyle p: f \to \mathbb Z$ be the map defined by p(-1) = all negative integers, p(0) = 0, and p(1) = all positive integers.

Isn't this a surjective function? Or do I need more coffee this morning?

-Dan
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February 27th, 2018, 09:41 AM   #9
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Quote:
Originally Posted by topsquark View Post
Consider f to be the finite set $\displaystyle f = \{-1, 0, 1 \}$. Then define $\displaystyle p: f \to \mathbb Z$ be the map defined by p(-1) = all negative integers, p(0) = 0, and p(1) = all positive integers.

Isn't this a surjective function? Or do I need more coffee this morning?

-Dan
But then, for example, p(1) equals both 1 and 2, which is not allowed.
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February 27th, 2018, 11:36 AM   #10
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Quote:
Originally Posted by topsquark View Post
Consider f to be the finite set $\displaystyle f = \{-1, 0, 1 \}$. Then define $\displaystyle p: f \to \mathbb Z$ be the map defined by p(-1) = all negative integers, p(0) = 0, and p(1) = all positive integers.

Isn't this a surjective function? Or do I need more coffee this morning?

-Dan
topsquark seems to have defined most closely to what I was getting at -- say p(-1) representing all negative integers, perhaps a surjective function. I think my main concern is whether one may repeat beyond limit identical elements of the domain (below). Would "exactly" allow that?

Microm@ss, is there a way around the strict definition of a function in this case, like topsquark's example? What if f={1} such that p=(1, 1, 1...)? Possibly countless elements from one, but can they all be the same?

Please forgive my atrocious notation.
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