The seven deadly sins of set theory
From
WM@21:1/5 to
All on Wed Dec 13 19:51:52 2023
1. Scrooge McDuck's bankrupt
Scrooge Mc Duck earns 1000 $ daily and spends only 1 $ per day. As a cartoon-figure he will live forever and his wealth will increase without
bound. But according to set theory he will get bankrupt if he spends the dollars in the same order as he receives them. Only if he always spends
them in another order, for instance every day the second dollar
received, he will get rich. These different results prove set theory to
be useless for all practical purposes.
The above story is only the story of Tristram Shandy in simplified
terms, which has been narrated by Fraenkel, one of the fathers of ZF set theory.
"Well known is the story of Tristram Shandy who undertakes to write his biography, in fact so pedantically, that the description of each day
takes him a full year. Of course he will never get ready if continuing
that way. But if he lived infinitely long (for instance a 'countable
infinity' of years [...]), then his biography would get 'ready',
because, expressed more precisely, every day of his life, how late ever, finally would get its description because the year scheduled for this
work would some time appear in his life." [A. Fraenkel: "Einleitung in
die Mengenlehre", 3rd ed., Springer, Berlin (1928) p. 24] "If he is
mortal he can never terminate; but did he live forever then no part of
his biography would remain unwritten, for to each day of his life a year devoted to that day's description would correspond." [A.A. Fraenkel, A.
Levy: "Abstract set theory", 4th ed., North Holland, Amsterdam (1976) p. 30]
2. Failed enumeration of the fractions
All natural numbers are said to be enough to index all positive
fractions. This can be disproved when the natural numbers are taken from
the first column of the matrix of all positive fractions
1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
... .
To cover the whole matrix by the integer fractions amounts to the idea
that the letters X in
XOOO...
XOOO...
XOOO...
XOOO...
...
can be redistributed to cover all positions by exchanging them with the
letters O. (X and O must be exchanged because where an index has left,
there is no index remaining.) But where should the O remain if not
within the matrix at positions not covered by X?
3. Violation of translation invariance
Translation invariance is fundamental to every scientific theory. With n
m ∈ ℕ and q ∈ {ℚ ∩ (0, 1]} there is precisely the same number of rational points n + q in (n, n+1] as of rational points m + q in (m,
m+1] . However, half of all positive rational numbers of Cantor's
enumeration
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, ...
are of the form 0 + q and lie in the first unit interval between 0 and
1. There are less rational points in (1, 2] but more than in (2, 3] and
so on.
4. Violation of inclusion monotony
Every endsegment E(n) = {n, n+1, n+2, ...} of natural numbers has an
infinite intersection with all other infinite endsegments.
∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ .
Set theory however comes to the conclusion that there are only infinite endsegments and that their intersection is empty. This violates the
inclusion monotony of the endegments according to which, as long as only non-empty endsegments are concerned, their intersection is non-empty.
5. Actual infinity implies a smallest unit fraction
All unit fractions 1/n have finite distances from each other
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Therefore the function Number of Unit Fractions between 0 and x, NUF(x),
cannot be infinite for all x > 0. The claim of set theory
∀x ∈ (0, 1]: NUF(x) = ℵo
is wrong. If every positive point has ℵo unit fractions at its left-hand side, then there is no positive point with less than ℵo unit fractions
at its left-hand side, then all positive points have ℵo unit fractions
at their left-hand side, then the interval (0, 1] has ℵo unit fractions
at its left-hand side, then ℵo unit fractions are negative. Contradiction.
6. There are more path than nodes in the infinite Binary Tree
Since each of n paths in the complete infinite Binary Tree contains at
least one node differing from all other paths, there are not less nodes
than paths possible. Everything else would amount to having more houses
than bricks.
7. The diagonal does not define a number
An endless digit sequence without finite definition of the digits cannot
define a real number. After every known digit almost all digits will follow.
Regards, WM
--- SoupGate-Win32 v1.05
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