Comment by powertower | Hacker News Reader

Comment by powertower | Hacker News Reader

Infinity can be relative and can be defined unlike a division by zero which in undefined (as in ... it can't be relative to anything else, and can't be used in a formula).

And hence infinity can be used in a formula and can cancel out with another relative infinity...

Example:

1) There are an infinite amount of real numbers between 1 and 2.

2) The amount of real numbers between 2 and 4 is twice the amount of real numbers between 1 and 2.

I would guess that if numbers are defined in terms of relativity/relationship, then infinity is a number.

But it seems that people wrongly define numbers in absolute terms, as if they exist outside the mind, and are separate from one another. Like the Universe cares about 1.24545434 and 7656.45433477.

But that's just my guess.

Correct me if I'm wrong, but I believe the cardinality of the two sets you describe are equal, as there exists a bijective mapping between them, meaning there are an equal "number" of real numbers in both.

Not only does there exist a bijective mapping between [1,2] and [1,4], there exists infinite different bijective mappings between a subset of [1,2] and [1,4].

i.e.: One could map bijectively from [1,1.5] to [1,4] and map bijectively from [1.5,2] to [1,4] (1)

To talk about there being "twice as much" in one uncountable infinity than in another uncountable infinity is nonsense, since you can't apply words like "twice", since the infinities can't be counted.

(1) https://imgur.com/NkKEI

For every real number between 1 and 2...

1) I can find that number also in the set between 1 and 4.

2) But I can't find that number in the set between 2 and 4.

To myself numbers are always relative to one another, don't exist outside the mind, and the set between 2 and 4 is twice the set between 1 and 2.

But like I said, it's a personal spin on it.

Where 1 <= x <= 4:

f(x) = (x-1)/9 + 1, maps bijectively from [1,4] to [1,4/3]

g(x) = (x-1)/9 + 4/3, maps bijectively from [1,4] to [4/3,5/3]

h(x) = (x-1)/9 + 5/3 maps bijectively from [1,4] to [5/3,2]

Therefore, [1,2] must contain three times as many numbers as [1,4], right?

It doesn't work like that.

I'm not sure what you are doing with the above.

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