While appealing to dictionary definitions of mathematical concepts is pretty weak in itself, it also ignores the fact that plenty of words have different meanings in different contexts. In the context of cardinality, ‘+’ is taken to mean the disjoint union of two sets. A disjoint union of two countably infinite sets is itself countably infinite, so it has the same cardinality as each of the sets.
This is why you were insitent on using ‘size’ and other ambiguous concepts: it’s only through that ambiguity that a contradiction can be produced.
Have you read the page on Hilbert’s Grand Hotel Paradox? You really should.
Let A be the set InfA, and An be the nth element of A. Let B be the set InfA1 + InfA2, and B(s,n) be the nth element from set s (either InfA1 or InfA2, which I’ll just call 1 and 2 for simplicity)
Let f(x) be a function from A to B. The following definition of f(x) is a bijection:
For x=0,
f(x) = B(1,0)
For x=1,
f(x)= B(2,0)
For all other x, if x is even,
f(x) = B(1,x/2)
If x is odd,
f(x) = B(2,(x-1)/2)
Now, the challenge for you, if you still want to claim that this is not a bijection: name one element from either set A or set B that is not uniquely mapped to an element in set B or A (respectively) by this function.
Since I just provided a bijection between InfA and InfA1+InfA2, and you aren’t able to point to an element that does not map uniquely, you must acknowledge that they have the same cardinality.
That is obviously not true because the very “last” (pseudo-last) term in that list is:
1.0 - 0.9 = 0.1
1.0 - 0.99 = 0.01
1.0 - 0.999 = 0.001
1.0 - 0.9999 = 0.0001
.
.
. 1.0 - 0.999… =
…the very exact term you are saying isn’t there.
Using your very same reasoning: “The difference between 1.0 and .999… is 0.000… with a theoretical 0 at the end. Except, there is no such thing. By your own argument…”
I have a reason put forth for believing that there is a “1” (as stated above). What is your reason for believing that there is a “0” (which must exist as such for the difference to be zero?
A) We are not restricted to “common math”, but rather to consistent logic.
B) It isn’t “obvious” either way.
What it breaks down to is that:
1/3 leads to 0.333…, with no equivalency resolution.
3 * 0.333… can only lead to another unresolved, endless string, in this case 0.999…, still no equivalency.
I can understand how someone who doesn’t think or doesn’t understand math and logic would mindlessly accept that 0.999… = 1.0 (despite the obviousness of the loss of logic). The art of persuasion doesn’t require logical reasoning.
That is exactly the point right there.
BY DEFINITION, there is always a difference between 1.0 (the limit) and 0.999… (the limitless string).
1.0 is a “bounded decimal”
0.999… is an “unbounded decimal”.
The same number cannot be both bounded and also unbounded. I think that constitutes the FOURTH Proof that I have put forth without flaw being pointed out.
Firstly, your effort to appeal to a higher definition would be far more effective is you could actually present one as I requested. Calling the addition a “disjointed union” doesn’t change anything at all (a semantic appeal at best). The definitions that I have provided from multiple sources are the very ones used to create what you now call “math”. If you were to come up with anything different (and your new one isn’t different), it would have to be someone’s odd fantasy definition that is new or unrelated to math.
Secondly,
That is your hypothesis. Care to prove it? Realize that I have already disproven it.
That’s just your fantasy BS. I haven’t used “size” to mean anything any different than every mathematician since day one. You trying to claim ambiguity is nonsense and irrelevant.
I looked at that years ago and resolved it to be just another word game as are all “paradoxes”.
That is exactly what I was expecting you to try to pull. You are attempting to hide the ignore-ance. You first hide the fact that you are ignoring that disappearing difference and then you try to hide an entire set within that disappearing difference (an unfortunate political trick played on populations for thousands of years).
The problem is (and I don’t have to point to a single mis-bijection) that we must stick to what we know in order to both come to a conclusion and also avoid deception. So let’s see what we know:
We know that infA and the infA1 are equinumerous? A) Agree B) Disagree?
or Forfeit
Of course what that will mean is the there is a bijection between them and that one is exactly countable by the other. Which will in turn mean that nothing else can be additionally counted because ALL of the countability is already used up.
That is something that we KNOW even before we add anything at all.
So, two points there: one, that it isn’t just word games, it shows something that is counter-intuitive about the way infinity works.
Two, the lack of equivalence between the statements means that the statement, “one is exactly countable by the other”, doesn’t entail that “nothing else can be additionally counted”. Those are not equivalent, the latter does not follow from the former.
You aren’t playing fair, James. I’ve answered your series of true-false questions, and now you’re avoiding my question to show one element that is not uniquely mapped by the bijective function I described. I’m happy to continue answering your questions after you either point to such an element or acknowledge that you can’t.
I pointed to an entire infinite set that was not uniquely mapped. But you propose a false dichotomy. You propose that unless I stand on one foot and explain all of reality, then I must accept that 2+2=2. I proved that neither is the case.
If you would like merely a single one out of the infinite set that was not uniquely mapped (and not mapped at all), that is easy enough. Let’s take one of your “even” number designations:
For all other x, if x is even,
f(x) = B(1,x/2)
B is the set of infA1+infA2. The bijection function “f{x}” applies only to x from 0 to infA1 or from 0 to infA2 (since those are the same range), but not infA1+infA2. Thus you can do your bijection function up to:
f(infA1) = B(1,infA1/2)
But you cannot and did not include:
f(infA1+infA2) = B(1, (infA1+infA2)/2)
(infA1+infA2)/2 is a valid element of set 1 in union B. But (infA1+infA2) is not a valid number for x. Since infA1 is the entire countable set of integers, infA1+1 (or more) is not countable. You left out the entire range of x from infA1 through infA1+infA2 because those are not countable integers for your x.
Thus your claim:
…has been proven wrong by two methods.
The real question is why you bothered to claim it in the first place.
To repeat, f(x) if a function from A (a.k.a. InfA) to B (a.k.a. InfA1+InfA2, the disjuctive union of InfA1 and InfA2). B(s,n) specifies the set s that an element comes from (either InfA1 or InfA2, which I had represented by 1 or 2 for succinctness, though perhaps this shorthand is what’s throwing you off) and the element n, the nth element of the originating set. The function acts on elements, not sets, so “f(InfA1+InfA2)” is not a proper application of the function, and needn’t be for f(x) to remain a bijective function from A to B.
You’re still trying to provide a general response, when I’ve asked you to point to a single, specific element of the set A or B that is not mapped uniquely from one to the other.
To respond your general point, that A is mapped by f(x) to either InfA1 or InfA2, but not both, let’s take two elements of A, 4 and 5, and see where they are mapped by f(x):
Here we see that there are elements in A that map to elements of B coming from both InfA1 and InfA2. The 4th element of A maps to the 2nd element of InfA1, and the 5th element of A maps to the 2nd element of InfA2.
So A does indeed map to the elements of the disjoint union of InfA1 and InfA2. f(x) picks out the elements coming from both sets.
So again, can you point to a single element that fails to be so mapped by f(x)? Or will you concede that f(x) is indeed a bijective function, and that A and B have the same cardinality?
Which is exactly what I said. There can be no “f(infA1+infA2)”, yet there must be in order to complete the bijection of elements in the union B of infA1+infA2 with anything else.
That is your flaw.
Since B is the union of an amount1 with an amount2 (disjointed) any counter involved must count from 0 through amount1 and on up through amount2. So the counter (in this case “x”) must be an integer in the range from 0 through infA1+infA2. The problem is that “x” can’t actually do that.
Not so. I gave you one specific sample from that infinite list:
That would be one particular bijection, the one at infA1+infA2, that you inherently excluded simply because x cannot count to infA1+infA2.
Now that was kind of silly. I gave you a specific which you claimed to be a general. Then you say that you are responding to the general by giving me a specific.
And your sample is not actually relevant.
Look at the max value for x:
f{infA} = B {infA1, infA1/2)
That maps the half way point of the set infA1 (which is infA1/2) to f{infA}. And that is fine. But what about the rest of infA1? infA1 has half of its elements beyond infA1/2. Those are not mapped, the max of which I specifically pointed out as the element B{infA1, (infA1+infA2)/2}, which happens to be the max element in infA1 (and was not mapped).
Any number extended to infinity approaching another number is going to define as that number it approaches. It may never “technically” reach it, but since we’re talking about literally an infinitely small distance between two numbers (0.999… and 1) it makes perfect sense to equate them.
InfA1+InfA2 is not counterexample to the claim that f(x) is a bijection from A to B. In f(x), x is an element of A, and f(x) is an element of B. InfA1+InfA2 is not an element of A, it is a disjunctive union of sets. Since A = InfA = the set of integers, that disjunctive union of sets is not an element of A.
Surely you understand the distinction between a set and and an element of a set? And that a disjunctive union of sets is not an integer?
Set A isn’t an element of set A.
Set B isn’t an element of set A.
Set InfA is not a number, it’s a set, and it’s not an element of set A nor set B.
InfA1 and InfA2 are sets, not numbers, and neither is an element of either set A or set B.
The disjoint union of InfA1 and InfA2 is not an element of set A nor set B.
Set A’s elements are the integers.
Set B’s elements are number pairs (s,n) where s is a set and n is an element of the set, in this case an integer since set B is a disjoint union of two sets of all integers.
My claim is that f(x), a function that operates on the elements of set A, maps each element of set A to a unique element of set B, and that the mapping fully covers set B. If f(x) fails to work on anything that isn’t an element of set A, it does not affect my claim. If x is undefined for some f(x) that is not an element of set B, it does not affect my claim.
So point to an element of A (not a set, there is no element of A that is a set) that does not map uniquely via f(x) to an element of B. To give you a little assistance, all the elements of A are integers, so your response, if you believe there is one, should be in the form of an integer (also appropriate would be to provide an element of set B (a number pair (s,n) where s is either InfA1 or InfA2, and n is an integer) that is not mapped from an element of set A, or is mapped by multiple elements of set A).
The truth of it is:
[10.000…:0R] * [0.999…:9R] = [9.999…:90R]
[9.999…:90R] - [0.999…:9R] = [9.000…:81R]
[10.000…:0R] - [9.000…:81R] = [0.999…:19R] [1.000…:0R] - [0.999…:19R] = [0.000…:81R]
An infinitesimal is easy to overlook and ignore (much like socialist governments overlooking the needs of peons). But one shouldn’t overlook 81 of them.
Overlooking the peons leads to unnecessary disharmony bringing the collapse of very large empires.
We know that infA1+infA2 is not an element of A. But we also know that it IS an element of B.
That one fact alone tells you that you cannot form a bijection between them.
Surely you understand that when a set is merely a count of numbers, the set must contain ALL of the numbers involved. Set A has infA elements, numbered 0 through infA. Thus infA is one of its elements.
Your set B is two of set A, labeled: infA1 + infA2.
Thus set B has elements 0 through infA1+infA2.
Your bijection f(x) cannot mate 0–infA to 0–(infA1+infA2).
The above “10.000…” is incorrect. There is no “.000…” after the number 10. Certainly not in the sense there is a .999… This is because 0 isn’t a number like 9 is a number; 0 is only a placeholder for orders of magnitude of other numbers.
In our base 10 system to multiply anything by 10 you simply move the decimal one space to the right.
I put the 0s there merely for clarity. They don’t change the math at all.
No. That is just a shortcut method for simple minded issues.
0.999… is a long division issue that turns out to be endless, indicating that it cannot ever resolve to become the ratio. The same with 1/3 and a great many ratios.
What you’re claiming is tantamount to saying that the set of all integers is itself an integer. It isn’t. The set of all integers just does not include the set of all integers as an element.
Again, I will suggest, this time without derision, that you don’t understand the different between a set and an element of a set, and that distinction really is a prerequisite for making strong claims about the nature of infinite sets.
