Hi: Given sets A = { 0, 1, 2, ... } and B = { 1, 2, ... }, the
following can be proved:
1. Exists f:B -> A defined by f(x)=x-1 is into and onto A, thus |A|=|
B|.
2. Exists g:B -> A defined by f(x)=x is into but not onto A, thus |A|>|
B|
Given there exists a procedure to convert all (2) to (1) and reversing
the procedure all (1) to (2), why do we say if any bijection exists
between two sets they are equal rather than saying if any injection
but not surjection exists they cannot be equal?

Scott wrote:

Hi: Given sets A = { 0, 1, 2, ... } and B = { 1, 2, ... }, the
following can be proved:
1. Exists f:B -> A defined by f(x)=x-1 is into and onto A, thus |A|=|
B|.
2. Exists g:B -> A defined by f(x)=x is into but not onto A, thus |A|>|
B|
Given there exists a procedure to convert all (2) to (1) and reversing
the procedure all (1) to (2), why do we say if any bijection exists
between two sets they are equal rather than saying if any injection
but not surjection exists they cannot be equal?

A bijection is defined as a mapping that is both injective (into)
and surjective (onto).

If there exists no bijection between two sets, then there is no
mapping that is both surjective and injective.

If equipollence between two sets is defined as the existence
of a bijection between the two sets, then the sets are not
equipollent if there is no mapping that is both surjective
and injective.

If equipollence between two sets is defined as the existence
of a bijection between the two sets, then the sets are not
equipollent if there is no mapping that is both surjective
and injective.

If a function F from A to B is injective, then B dominates A (by
definition).

If a function G from A to B is surjective, then A dominates B (but we
need the Axiom of Choice to prove this in general - Lemma 10.20 of Kunen
p. 30).

By the Schroeder-Bernstein theorem, these two conditions together imply
that A and B are equipollent.

On May 13, 7:35 am, Norman Megill <n... at (no spam) see.signature> wrote:
If a function F from A to B is injective, then B dominates A (by
definition).

If a function G from A to B is surjective, then A dominates B (but we
need the Axiom of Choice to prove this in general - Lemma 10.20 of Kunen
p. 30).

By the Schroeder-Bernstein theorem, these two conditions together imply
that A and B are equipollent.

Yes, but we can reverse the procedure in Schroeder-Bernstein theorem
to derive an injective but not surjective function to show A and B are
not equipollent. My question is why (mathematically) if any bijection
exists A and B are equipollent but any injection that is not
surjective does not prove A and B are not equipollent.

Yes, but we can reverse the procedure in Schroeder-Bernstein theorem
to derive an injective but not surjective function to show A and B are
not equipollent.

My question is why (mathematically) if any bijection
exists A and B are equipollent but any injection that is not
surjective does not prove A and B are not equipollent.

On May 13, 1:27 pm, magi... at (no spam) math.berkeley.edu (Arturo Magidin) wrote:
The ->definition<- of "equipollent" is that there exists at least one
bijection between A and B. So "not equipollent" must be the negation
of this; hence, there does not exist a bijection between A and B.

Okay. Thanks for clearing that up.

One definition of "infinite" (Dedekind-infinite) is that there exists
a non-surjective injection from A to itself. If an injection that is
not surjective were sufficient to establish that A and B are not
equipollent, it would serve to establish that no Dedekind-infinite set
is equipollent to itself, and so if there are any Dedekind infinite
sets, then "equipollence" would not be an equivalence relation.

Follow up question: Isn't there a bit of circular logic here? Isn't
the definition of equipollent creating the property of Dedekind-
infiniteness?

If an injection that is not surjective were sufficient
to establish that A and B are not equipollent,

then many, if not all,
sets we consider Dedekind-infinite would not be Dedekind-infinite

and
equipollence would still be an equivalence relation; e.g, A is not
Dedekind-infinite. Changing the definition of equipollent would change
a few theorems,

but not render the theory inconsistent or incomplete.
(Kind of like the parallel line postulate of geometry.)

No, say strict-equipollent is:
S and T are strict-equipollent <-
Ef [f is a bijection from S to T]
^
~Eg[ g is S into T ^ ~(g is S onto T)].

On May 14, 3:04 pm, MoeBlee <jazzm... at (no spam) hotmail.com> wrote:

say strict-equipollent is:
S and T are strict-equipollent <-> Ef [f is a bijection from S to T] ^
~Eg[ g is S into T ^ ~(g is S onto T)].

So let's take the case of S = T = { 0, 1 } and f is a bijection and g
is an injection but not surjection. By equipollent, S and T have the
same cardinality. By strict-equipollent, S and T do not have the same
cardinality. My question is: why do we use equipollent rather than
strict-equipollent?

But that does NOT entail that there are not Dedekind infinite sets, so

I did not say there are no Dedekind infinite sets, but that some sets,
like A, that are Dedekind infinite under equipollent are not Dedekind
infinite under strict-equipollent.

it does NOT entail that the strict-equipollence is reflexive, so it

Why not? Strict-equipollence guarantees that every every element of S/
T is related to itself.