What is this supposed to mean? — Michael
What do you mean by an "imperfect" triangle? — Michael
You should see non-Euclidean geometry where the angles in a triangle can be more or less than 180 degrees. — Michael
where there are an infinite number of elements (including fractions and irrationals) between 0 and 1 — punos
The answer to your problem is quite simple. In mathematics things are done by axiom. If you want to count to infinity and beyond, simply produce an axiom which allows you to do that, and bingo the infinite is countable, and you're ready to go beyond. Look closely at the following: — Metaphysician Undercover
But having a different concept and definition of infinitude doesn't thereby entail that there is a contradiction in set theory or mathematics. — TonesInDeepFreeze
Again, yes, there may be a contradiction between set theory and certain other formulations. But that does not entail that there is a contradiction within set theory. — TonesInDeepFreeze
set theory does not refer to an object named 'infinity' but rather to the property of being infinite, which is a crucial distinction. — TonesInDeepFreeze
Your posting is an absurd loop — TonesInDeepFreeze
But {0 1 2 3 ...} is not notation that for every natural number there is a greater natural number, but rather it is an informal notation to stand for the set of all and only the natural numbers. — TonesInDeepFreeze
But {0 1 2 3 ...} is not notation that for every natural number there is a greater natural number, but rather it is an informal notation to stand for the set of all and only the natural numbers. — TonesInDeepFreeze
There is least infinite cardinal, which is the cardinality of the set of the infinite set of natural numbers. And there are cardinals greater than the least infinite cardinal. Moreover, for each cardinal, whether it is a finite cardinal or infinite cardinal, there is a greater cardinal. — TonesInDeepFreeze
The only reason something like a sequence of numbers can go on forever, is because of Infinity. It is not because the sequence of numbers are Infinite. — Philosopher19
A quick look will tell you that there are twice as many feet as there are people. You do not need to count the number of people to know this to be true; just check for amputees... — Banno
Perhaps one might argue that there is no count involved with regards to the latter and that it's just a fact that Infinity encompasses an infinite number of natural numbers. But if that's the case, then Infinity also encompasses an infinite number of possible real numbers and possible letters or possible x. But where there is no counting involved, all infinites are of the same size/quantity (or rather, infinity is one quantity as opposed to different quantities). — Philosopher19
for any infinite cardinality there is a greater infinite cardinality — TonesInDeepFreeze
Time well spent would be to learn some mathematics rather than claiming untrue things about it. — TonesInDeepFreeze
Is it that Philosopher19 has a picture of infinity such that, since one cannot count to infinity, one cannot have a grasp of infinity? — Banno
By bijection. See Open Logic Ch.4. — Banno
"Counting", and ill-defined notion, is not involved in bijection, although "enumeration", a well-defined notion, is. — Banno
See Cantor's diagonal argument. — Banno
Because the choice is between the whole of the remainder of that project being founded on an error unnoticed by more than a century of study by logicians world wide; and your being mistaken. — Banno
And this shows that you have not understood R = {x : x ∉ x}: — Banno
Proof. If R = {x : x ∉ x} exists, then R ∈ R iff R ∉ R, which is a contradiction. — Open Logic: Complete build
I can't give an opinion but you have put some work into it to your credit. — Mark Nyquist
Yes, we can establish set X as being "bigger" than set Y without counting the number of items in X and Y. We can establish this by using Cantor's diagonal argument. If you were a mathematician you would understand it. — Michael
I've seen Cantor's diagonal argument. It does not answer the questions I asked you in my last post to you. — Philosopher19
Russell proved that these two axioms entail that there is a set that only contains all sets that are not members of themselves (the Russell set). — Michael
So when you say:
There exists a set B whose members are precisely those objects that satisfy the predicate
— Michael
are you essentially saying "there is a set that contains all sets that are not members of themselves"? If not, can you clarify? — Philosopher19
1) There exists a set whose members are sets that are not members of themselves
2) There exists a set that contains all sets that are not members of themselves — Philosopher19
You've already admitted that you're not a mathematician, so it's strange that you think you know mathematics better than Cantor (and Russell). — Michael
↪Philosopher19 Yes. Given the axiom schema of unrestricted comprehension there exists a set B whose members are sets that are not members of themselves. This leads to a contadiction. If B is not a member of itself then it should be a member of itself. — Michael
In Russell's paradox, φ is "sets that are not members of themselves". — Michael
are you essentially saying "there is a set that contains all sets that are not members of themselves"? If not, can you clarify?There exists a set B whose members are precisely those objects that satisfy the predicate — Michael
It says that A and B are equal if every member of A is a member of B and every member of B is a member of A. — Michael
There exists a set B whose members are precisely those objects that satisfy the predicate — Michael
Given any set A and any set B, if for every set X, X is a member of A if and only if X is a member of B, then A is equal to B (the of axiom of extensionality) — Michael