Re: bijective mapping of sets
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Fri, 24 Mar 2006 02:03:19 +0000 (UTC)
In article <1143161498.846509.112430@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<sikari.h@xxxxxxxx> wrote:
Is this reasoning generally valid to show that
two sets are equal and also that f is injective;
I mean is the following true:
\forall A,B \subset X such that
A \subset B and f(X-A)=f(X-B) => B \subset A (i.e. B=A) and f is
injective
You are not being very careful with your notation; specifically, it is
not clear what the scope of the quantification is, and whether the
implication includes the "and" or not. If you mean:
If f:X->Y if a function, such that
for all subsets A and B of X, if A is contained in B and
f(X-A) = f(X-B) then B is contained in A
then f is injective
then the answer is yes: given any a,b in X such that f(a)=f(b), then
letting A={a}, B={a,b}, we have A contained in B, f(X-A)=f(X-B), so we
conclude that A=B, and hence that a=b, which proves f is injective.
On the other hand, you could mean:
Let f:X->Y be a function. Then for all subsets A and B of X, [if A is
contained in B and f(X-A)=f(X-B), then A=B]; in particular f is
injective.
Then this is not true: a constant function will f(X-A)=f(X-B) in all
cases except when one of A and B is all of X and the other one is
not.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
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