No.However, we still have to consider fact 2 by which we can determine inequality of cardinality of sets
The set of natural numbers N = {1, 2, 3, 4,...}
The set of even numbers E = {0, 2, 4, 6,...} — TheMadFool
Again no!It's clear that N can be separated into two proper subsets viz. the set of even numbers V = {0, 2, 4, 6,...} and the set of odd numbers, D = {1, 3, 5, 7,...}
Notice how set E has a bijection with set V and set V is a proper subset of set N. If so, then in accordance with fact 2 above, n(E) < n(N) i.e. the set of even numbers is less than the set of natural numbers.
This presents a problem doesn't it? — TheMadFool
2. A set G has a cardinality greater than a set H if and only if the there's a bijection between set H and a proper subset of G — TheMadFool
What is changed? What does not change? You really are confused about usage. Imagine I have a pile of sand, big as you like. One grain of sand is added. Does that mean that "pile of sand" is meaningless because even though it's not the same pile, still it's called the same thing? Or if a family has a new baby, it's not a family any more?Transfinite maths and reality are greatly in opposition. ∞+1=∞ tells us that there exists something, that when it is changed, it does not change. — Devans99
From what you write and how you ague, I'd say you actually are not interested in maths at all. If you were, you'd have paid some attention to what people who actually do know have said in replying to you many times over. All these "maths" you balk at are really a lot more interesting, the more you actually know about them.My interest in maths stops when maths stops telling me about the nature of reality. — Devans99
However, it is classically true that A<BA<B iff there is an injection from AA to BB but no surjection. — quickly
Again no! — ssu
If there exists a bijection between two sets, even a single one, even if there are plenty of functions that aren't bijections, then we DEFINE the two sets as being cardinally equivalent. It's a definition, not a proof. — fishfry
I'm not sure what he meant, but a bijection is both an injection and a surjection.I think fishfry said something to the effect that bijection has precedence of injection. Why? — TheMadFool
I'm not sure what he meant, but a bijection is both an injection and a surjection.
But to your question: a set is infinite if and only if it is equivalent to one of its proper subsets. And this counters your argument. You are making the mistake of thinking about an infinite set as finite.
The proof of this actually refutes your idea: See for example here, Proof 1
You see with a finite set it isn't so: a finite set can not be in one-to-one correspondence with one of its proper subsets. — ssu
Yep, you got it! :up:I think I get it now. The proper subset of the infinite set itself has to be infinite.The number of elements in a proper subset B of a finite set A is necessarily less than the number of elements in the set A. n(B) < n(A) so long as A and B are finite. - In other words the fact that the set of even numbers don't contain odd numbers and that the set of natural numbers is the union of odd and even numbers didn't translate into a numerical difference like it does with finite sets. — TheMadFool
I think fishfry said something to the effect that bijection has precedence of injection. Why? — TheMadFool
n(set of natural numbers) - n(set of even numbers) = infinity - infinity = zero because both are, well, infinite. — TheMadFool
Thanks. I'll just stick to the simple.injective, then there may be some elements of BB that are not mapped to by any element of AA;
surjective, then there may be some elements of AA that map to the same element of BB; and
bijective, then there is a unique pairing between elements of AA and elements of B — quickly
Thanks. I'll just stick to the simple. — TheMadFool
Can you have a look at what I said below. — TheMadFool
The proper subset of the infinite set itself has to be infinite — TheMadFool
Really? Consider N={1,2,3,...} and S={1,9} No 1:1 correspondence. You are missing "equivalence"
"A" proper subset? Not "The", I think. — John Gill
Can you have a look at what I said below. It seems wrong and right.
n(set of natural numbers) - n(set of even numbers) = infinity - infinity = zero because both are, well, infinite.
— TheMadFool
Consider two infinite sets A and B of equal cardinality i.e. n(A) = n(B) = infinity
Shouldn't n(A) - n(B) = 0?
Yet n(set of natural numbers, infinite) - n(set of even numbers, infinite) = n(set of odd numbers, infinite) which is infinite indicating that infinity - infinity = infinity — TheMadFool
Consider two infinite sets A and B of equal cardinality i.e. n(A) = n(B) = infinity
Shouldn't n(A) - n(B) = 0? — TheMadFool
Beyond finite instances, cardinalities are not numbers. They are equivalence classes. — John Gill
If 'the cardinality of N and E are equal', means there are as many elements in
N as E, I can accept that. — sandman
Infinity is not an integer, or any type of number, and definitely not a quantifier.They think aleph-naught is, or acts like, just another integer, — tim
I accept. I ought to because I've delivered a few and felt justified doing so, so I ought to take if I've earned. — tim wood
I think fishfry said something to the effect that bijection has precedence of injection. Why? — TheMadFool
Beyond finite instances, cardinalities are not numbers. They are equivalence classes. — jgill
The point that you were hung up on before is that the definition of a bijection says that there exists at least one bijection between two sets. — fishfry
The main point is that this is how you define cardinal numbers these days. They're no longer equivalence classes of sets that themselves aren't sets. That was a problem so it got fixed at the expense of needing to do some technical work. — fishfry
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