```Proof that ZFC is inconsistent
Posted: Jan 28, 2008 4:19 AM
Update: 2013 August version
by: Han de Bruijn

More precise: proof that the axiom of Infinity in ZFC is inconsistent.

WAS: Re: 1-1/2+1/3-1/4+1/5-1/6+1/7 (: main thread in 'sci.math' where
Victor Meldrew has been trying repeatedly to spoil the party with his

http://mathforum.org/kb/message.jspa?messageID=6063213

The standard definition of the naturals is considered once again:

0 = {}
1 = {0} = {{}}
2 = {0,1} = { {} , {{}} }
3 = {0,1,2} = { {} , {{}} , { {} , {{}} } }
.
.
n+ = (0,1,2,3,4, .. ,n} = [ .. oh, well .. ]
.
.
The axiom of Infinity in ZFC is as follows. There exists a set which
contains with each member n also its successor n+, starting with 0.
So the natural numbers are defined by a successor function, called s,
where:
s(n) = n+ = n + 1
We find:
0
1 = s(0) = {0}
2 = s(1) = s(s(0)) = {0,1}
3 = s(2) = s(s(1)) = s(s(s(0))) = {0,1,2}
.
.
n+ = s(n) = s(s(s(s( .. s(s(0)) .. ))) [ 2(n+1) parentheses ]
.
.
Theorem: the mapping S(n), where S(n) = s(s(s(s( .. s(s(n)) .. ))) and
the numerosity of successor functions in the composed function S equal
to the first transfinite cardinal is the set of all natural numbers N.
Proof: S(n) and {0,1,2,3,4,5, .. oo } are identical. And the latter is
precisely the set of all naturals N with cardinality Aleph_0.

Theorem: the mapping S(N) does exist and is equal to {N}: S(N) = {N}.
Proof: this follows by applying the previous theorem to all naturals
in the set {0,1,2,3,4,5, .. oo } and: S(N) = {S(0),S(1),S(2), .. } =
= {N,N,N, .. } = {N} . Mind the subtlety: the infinite composition of
successors is _not_ the naturals, but the singleton with THE naturals
as the only element in it.

However, we also have the following sequences:

N = {0,1,2,3,4,5,6,7,8,9, .. ,n, .. }
s(N) = {1,2,3,4,5,6,7,8,9, .. ,n, .. }
s(s(N)) = {2,3,4,5,6,7,8,9, .. ,n, .. }
s(s(s(N))) = {3,4,5,6,7,8,9, .. ,n, .. }
.
.
Lemma: all naturals from 0 up and including n are _not_ present in the
range of the composite function consisting of n+ successor functions s.
Proof: as seen from the above, more formally by mathematical induction.

Theorem: the range of the mapping S(N) = s(s(s(s( .. s(s(N)) .. ))) is
a set of naturals with not a single natural left in it.
Proof: due to the above lemma for finite compositions, NO naturals can
be present in this infinite composition.
Corollary: S(N) is the empty set = {}.

But we have also proved that S(N) = {N}. Consequently {N} = {} .

Take a good look at the latter formula. It says exactly _this_: the set
containing only the set of all naturals is empty. This can mean nothing
else than: THE set of ALL naturals does not exist.

This completes the paradox, and the proof that the axiom of Infinity in
ZFC is inconsistent. The good news is that Infinity is the _only_ axiom
which is inconsistent in ZFC, as has been established previously by HdB
in the 'sci.math' thread "Implementable Set Theory" and an accompanying
article. The other axioms form a consistent system as long as the axiom
of Infinity and hence Infinity itself is considered as: not even known.

It is noted that, with such a theory about heredetary finite sets, many
axioms become theorems and hence are redundant. Set theory has become a
PART of mathematics, instead of its purported foundation.

http://mathforum.org/kb/message.jspa?messageID=5946892
../jaar2007/set_theory.pdf

The paradox essentially is caused by the idea that a natural should be
identified with an initial segment of (previously "defined") naturals.

Disclaimer: the above does not mean that common mathematical reasoning
as demonstrated here is supported by this author in any respect. Quite
obviously, the outcome of that resoning deprecates any reasoning with
infinitary objects itself; it has been invalid from the start.

Han de Bruijn

P.S.
The argument has been formalized an refined with a great deal of effort
by Jesse F. Hughes and G. Frege in the abovementioned sci.math thread:

http://mathforum.org/kb/message.jspa?messageID=6063213

This is what Jesse says, in the end:

> So, Han's notation is actually acceptable, whether by accident or
> not.  But of course, this shouldn't be a surprise and I'm a bit
> embarrassed I didn't think about it before.

Thus supporting the form but not the content; the latter remains my own
achievement, quite unfortunately.

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