Re: Axiom of choice in proof for surjectivity
From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 06/14/04
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Date: Mon, 14 Jun 2004 22:13:35 +0000 (UTC)
On 14 Jun 2004, Adam wrote:
>"Arturo Magidin" <magidin@math.berkeley.edu> wrote in message
>news://cakna5$13pe$1@agate.berkeley.edu...
>>> Non-empty is not enough; I assume that you mean he made them ->finite<-
>sets
>
> The author did not specify that they be finite sets, so perhaps that
>condition wasn't needed for the proof he made.
I'm sorry, but this comment makes no sense to me in light of your
previous comments (which here you removed). You had said:
"That the Axiom of Choice is complicated for infinite sets is
probably the reason why the author let the domain and codomain
in the theorem be non-empty sets."
First, the Axiom of Choice is necessary when the sets involved in the
proof are INFINITE. You are correct that the Axiom of Choice is
"complicated", but restricting the proof to "non-empty sets" does
not, in any way, affect the use of the Axiom of Choice. So if you
meant "non-empty", then your previous comment is a phenomenal
non-sequitur. I assumed that your comment was not a non-sequitur,
and so guessed that you meant to say "finite" (if both A and B are
finite, then you DO avoid using the Axiom of Choice, and thus
avoid the 'complications' inherent in them).
In fact, there is no need to specify "non-empty" for the theorem to
hold in general, but I suspect that the reason that the author
specified 'non-empty' is that his definition of "function" probably
requires that the sets be nonempty. But if we do not require that,
note that for every set B there is only one possible function
from the empty set to B, namely the empty function; and only one
possible function whose codomain is the empty set, namely the
function whose domain is also the empty set.
Assume f:A->B is surjective. Let us consider what happens if either
A or B are empty.
If B is nonempty, then A must be nonempty by surjectivity. So the
only way that one of A or B is empty is if both are empty. In that
case, the empty function from the empty set to itself is in fact
bijective (not just surjective), and has itself as an inverse.
The reason one might want to avoid the empty function is that
the condition for injectivity (a function is injective if and only
if it has a left inverse) is no longer true for functions whose
domain is the empty set but whose codomain is not empty.
Which is why I suspect the author excluded the empty sets, to be
able to have all three conditions (left inverse<->injective; right inverse<-> surjective; inverse <-> bijective) stated similarly,
and not necessarily in their most general form.
But this is neither here nor there with regards to the Axiom of
Choice. My point is that "non-empty" does not affect whether or
not one uses the Axiom of Choice to define a right inverse to a
surjective map. On the other hand, if we specify "finite" for the
sets (or even just for B), then one can avoid using the Axiom of
Choice.
-- Arturo Magidin, sans .sig
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