zero divided by zero - one, or undefined?

[Hitman928]

No, the reason I can't (easily) create a mapping from [0,1] to (1,100] is that one set is a closed subset of the reals and the other is an open subset of the reals. But I can really easily inject the set [0,100] into the set [0,1], thereby proving that |[0,1]| >= |[0,100]|. In fact, given any two real numbers x and y with x != y, I can inject the entire real number line into [x,y].

This leads directly to the counter-intuitive conclusion that any continuous subset of the real numbers has the same cardinality as the entire real number line.

One is actually a half-closed set and I was being very sloppy with my notation, my apologies. However, I would actually love to see this proof as this would imply a proof of the continuum hypothesis (specifically there exists no set S such that N0 < |S| < 2^N0, where N0 is aleph-naught). which has never been proven (or disproven) after over 100 years. In fact, the latest developments have leaned toward it being disproven but nothing in the current axioms of set theory can be used to prove or disprove it.

On a more general note, as I've already said more than once, I was not trying to show a rigorous proof of anything, I was merely trying to show a concept the best way I thought it would be easy to understand. I should have probably put a big disclaimer saying that a proof of what I was saying is not possible and that only the concept of bigger and smaller infinite sets was the point. Perhaps I should have provided a better example. Either way, I just hoped people would understand the concept, I didn't think anyone would take that and then try to use it in any actual number theory. Consider it equal to a bad car analogy that are made so often to computers but don't ever really hold up as true on either end. If you can provide an easily understandable, yet fully verifiable treatment of how 2^N0 > N0, you are more than welcome to pick up where I failed.

Don't front, kid. If you have a rigorous argument to show us, then show us. Plenty of people on this forum can understand formal math.

Presumption and/or condescension I believe are universally frowned upon, though I have no proof (pun intended).

 

[Slammy1]

Problem is 0 is not something that exists in nature, but is instead a simplification to help us model reality. If you start to break from basic assumptions, 0/0 could be just about anything (1, 0, infinity).

 

[DeathRayLoveMachine]

One is actually a half-closed set and I was being very sloppy with my notation, my apologies. However, I would actually love to see this proof as this would imply a proof of the continuum hypothesis (specifically there exists no set S such that N0 < |S| < 2^N0, where N0 is aleph-naught).

No, it doesn't. All it implies is that any continuous infinite subset of R has the same cardinality as R itself. Notice that only continuous subsets have this property; that's why Cantor named it the continuum hypothesis. Cantor was fully aware of these issues, having been the one who discovered them.

Here's something even crazier than that: |R²| = |R|
That means there are as many points in a plane as there are in a line.

And by extension: |R| = |R²| = |R³| = ...

You can even go all the way, and say: |R| = |R^&#1488;|

And for that matter: |&#1488;| = |&#1488;^&#1488;|

(The aleph character messes up text direction and I can't create subscripts, so just imagine I put a little zero after each aleph.)

If you can provide an easily understandable, yet fully verifiable treatment of how 2^N0 > N0, you are more than welcome to pick up where I failed.

The proof you're looking for is Cantor's diagonality argument. I won't repeat it here because you can just google it for yourself.

 

[Hitman928]

The proof you're looking for is Cantor's diagonality argument. I won't repeat it here because you can just google it for yourself.

I am familiar with the diagonality argument, but a mathematical proof is not what I was attempting as I didn't think recreating or linking to this would have helped anyone without the math background to understand it, and those that are able to understand it I'm sure know about it already, so what would be the point? Anyway,

No, it doesn't. All it implies is that any continuous infinite subset of R has the same cardinality as R itself. Notice that only continuous subsets have this property; that's why Cantor named it the continuum hypothesis. Cantor was fully aware of these issues, having been the one who discovered them.

I honestly don't see what you are trying to argue here except the validity of the continuum hypothesis itself. If set S is not continuous and infinite, then it doesn't involve the cardinality of an infinite set. If S is an infinite subset of the real numbers, then the continuum hypothesis says that the infinite subset has to have either a cardinality equal to N0, or a cardinality equal to the cardinality of the entire infinite set of real numbers. You are essentially saying that this is true, which is fine if that's what you believe, but it is unproven and the math "community" is leaning toward the opposite. Unless I'm not reading what your point is correctly?

Here's something even crazier than that: |R²| = |R|
That means there are as many points in a plane as there are in a line.

And by extension: |R| = |R²| = |R³| = ...

You can even go all the way, and say: |R| = |R^&#1488;|

And for that matter: |&#1488;| = |&#1488;^&#1488;|

(The aleph character messes up text direction and I can't create subscripts, so just imagine I put a little zero after each aleph.)

Yep, cardinal arithmetic, although your last two are wrong, e.g. N0^N0 = 2^N0, see here.

 

[Wizlem]

People were just pointing out that your initial example of infinity being larger than another infinity was incorrect. While that is a true statement your example was of two sets that have the same cardinality.

I'm not sure why, if you indeed understand all of this other stuff, haven't accepted that and moved on. You just keep going on about with more incorrect statements which you continue to argue about when people correct you.

Your logic is flawed when you say that a proof of the existance of an injective function from the entire real numbers to an infinite subset of the real numbers proves the continuum hypothesis. That is incorrect.

 

[_Rick_]

In fact, if you were to find a non-finite subset of R that does not have the same cardinality as R (or N), you would have immediately disproven the discontinuity hypothesis.

All your examples had identical cardinality, which can be proven by defining a bijective relation on the two sets.

If you are arguing there are other values for set size than cardinality - then I doubt you'll find many people to agree, as a bijective relation is currently the standard way of determining set size equality.

 

[DeathRayLoveMachine]

Yep, cardinal arithmetic, although your last two are wrong, e.g. N0^N0 = 2^N0, see here.

|&#1488;^&#1488;| is something even a bit weirder than the power set of aleph. It's a shorthand for the Cartesian product of a countably infinite number of countably infinite sets - which is still countably infinite, surprisingly.

Sometimes mathematical notation is weird.

 

[havecold]

Change zero to the word nothing. The answer should be referred to what is gained or lost, not what you started out with. Simple enough. So 5/0 does not equal what you are left with which is 5. Answer(or zero in this case) is always defined. Terminology to describe what is happening is what's missing.