In terms of cardinality or largeness or magnitude N = E. — TheMadFool
we pair the members of E with the even numbers in N. — TheMadFool
What now of the odd numbers in N? They have no matching counterpart in E.
Doesn't this mean N > E? — TheMadFool
My math is at high school level so bear with me. — TheMadFool
I have a straight stick with one end. Can anyone tell me its length? — sandman
No, your not doing that.My argument is like Cantor's diagonal argument regarding the absence of bijection between the real numbers and the natural numbers — TheMadFool
Do understand what you are making with a bijection: every member of E has a pair in N and vice versa. That is a bijection, one to one correspondence. You are now somehow assuming it wouldn't be a bijection, but either a surjection or an injection.So we pair the members of E with the even numbers in N. We can do that perfectly and with each member of E in bijection with the even number members of N. What now of the odd numbers in N? — TheMadFool
My math is at high school level so bear with me. — TheMadFool
What's wrong with my argument? — TheMadFool
This is contradicted by:
1. Random sampling of integers results in an average of 50% even E, 50% odd D.
Statistics can be verified in the real world, and is useful in applications of probability.
2. In the above example, removing E from N leaves D, removing E from E leaves nothing, so where is the logic? An odd feature of this example is the appearance of the same integers in both sets.
The 'bijection' for example 1 defines y=2x, as a mapping from N to E. The results are not about the size of sets, but the definition used for mapping. — sandman
A set without limit (infinite) is not measurable, since boundaries enable measurement. — sandman
In mathematical analysis, a measure on a set is a systematic way to assign a number to each suitable subset of that set, intuitively interpreted as its size. In this sense, a measure is a generalization of the concepts of length, area, and volume. A particularly important example is the Lebesgue measure on a Euclidean space, which assigns the conventional length, area, and volume of Euclidean geometry to suitable subsets of the n-dimensional Euclidean space Rn. For instance, the Lebesgue measure of the interval [0, 1] in the real numbers is its length in the everyday sense of the word, specifically, 1. — Wiki
Sure.
Starting from the fact that we don't know the answer to the Continuum Hypothesis. Which tells us quite plainly that we still don't understand everything about mathematical infinity. — ssu
Basically two different bijections are possible. — TheMadFool
N = {2, 4, 6, 8,...,1, 3, 5,...} — TheMadFool
As you can I see can form a pairing (1-to-1 correspondence) between E and the even numbers in set N like so: (2,2), (4,4), etc. and that leaves the odd numbers without a corresponding pair in the set E. — TheMadFool
This then is used to "prove" that the set E is equivalent to set N — TheMadFool
This then is used to "prove" that the set E is equivalent to set N — TheMadFool
Basically two different bijections are possible. One agrees with Cantor's "proof" but the other contradicts Cantor. You'll have to show that Cantor's bijection is the correct one and the alternative is nonsensical. — TheMadFool
No, that's a great source of confusion. 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
That's where the problem is isn't it?
The definition is inadequate for the reason that, on one hand, Cantor's "preferred" bijection leads to an equivalence between the set of even numbers and natural numbers but on the other hand there exists another bijection that shows that the set of natural numbers is not equivalent to the set of natural numbers. — TheMadFool
there exists another bijection that shows that the set of natural numbers is not equivalent to the set of natural numbers. — TheMadFool
It's a definition. It can't be right or wrong. — fishfry
If a definition leads to a contradiction? — TheMadFool
That's not technically Cantor's diagonal argument. That proof comes into use only with the reals R.I understand Cantor's argument well enough to see that there's a pair (1-to-1 correspondence) between the natural numbers and even numbers. — TheMadFool
Please understand you are not talking of a bijection! It isn't a bijection if one group has more members.Basically two different bijections are possible. — TheMadFool
It doesn't.
Do you agree that there exists at least one bijection from E to N?
If you agree, then you must agree that E and N have the same cardinality, because the definition says that there must be at least one bijection between them, and this is manifestly the case.
ps -- It's like a guy who cheats on his wife. She says to him, "You're a cheater." He says no! Think about all the times I DIDN"T cheat on you.
But that's not the point. If you cheat once, that's the definition of a cheater. If you cheated Monday but not on Tuesday or Wednesday, you can't say you're not a cheater because you didn't cheat on Wednesday. Right? The definition is doing it once.
Likewise the definition of cardinal equivalence is that there's at least one bijection. It doesn't matter that some other function isn't a bijection. — fishfry
Do you agree that there exists at least one bijection from E to N? — fishfry
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