infinity?

Dadeyemi

infinity + 1 = infinity so induction infinity - (- infinity) = infinity + infinity = infinity.

You can't induct to infinity.

yusufu

Yes, but the string is also continuous. :yep:

Ah, but it most certainly isn't, not if it's real! It would only be continuous if between any two atoms we can find another atom, but this is obviously not true if you have two adjacent atoms. The string is discrete!

And what if space itself is discrete? Hehe.

tan 90

Simba

It's interesting to note that you can have infinitely many different sizes of infinity. We usually donate the size of N (the natural set) as

\displaystyle \mathbb{N}

and proceed from there. This is known as a 'countably infinite' set, as is any set A if we can find a bijection f:

\displaystyle \mathbb{N}

-> A.

As an example, consider the set A of all even, positive integers, i.e. A = {2, 4, ...} Then we can construct a bijection f:

\displaystyle \mathbb{N}

-> A; let f(x) = 2x. This implies that the set of all even, positive integers is the same size as the set of all positive integers.

As for there being infinitely many different sizes of infinity, here is a proof (and indeed a test of Simba's study of set theory :p:

...)

"Consider the set A. If A is a finite set, then we know that the power set, P(A) (the set of all subsets of the set) is larger in magnitude than A. Indeed, |P(A)| = 2^|A|. This is quite easy to prove if you consider each element of A and whether or not it is included in each subset of A.

Now we move onto the pressing matter - proving that we can create a set bigger than any existing set; even infinite sets. Let our infinite set be B.

For two sets to be of the same size, we require f: B -> P(B) to be a bijective function. If this is not the case, the sets are not the same size. Our aim is to prove that f can not be a bijection.

Define

\displaystyle C =\lbrace x \in B | x \notin f(b) \rbrace\ where \ C \subseteq B.

Now suppose f is a bijective function. Then

\displaystyle \exists x \in B

such that

\displaystyle f(x) = C

.

Then

\displaystyle x \in C \iff x \notin f(x) \iff x \notin C

.

A clear contradiction, so f is not a bijection. Hence the power set is larger than the infinite set we started with."

It's a really interesting field to look into if you are interested in set theory and such :smile:

.

Here's an interesting tidbit: you can also prove

X \neq P(X)

using diagonalization.

For the purposes of explanation, I'm going to first show the idea with sets that can be listed. Of course, none of the uncountable ones are this way, but doing them is the same idea.

Let

X = \{x_1,x_2,x_3,...\}

.

Then, we can represent any subset of

X

by the "sequence"

(0,1,0,1,1,...)

, where the value is 1 if the element is in the subset, and 0 otherwise. For that particular subset,

x_1,x_3

are not in it, while

x_2,x_4,x_5

are.

So assume there is a bijection. It would look something like this:

x_1 \to (0,1,0,0,1,0,...)

x_2 \to (0,1,1,0,1,1,...)

x_3 \to (1,0,0,0,1,0,...)

...

So define the sequence

x

as: the nth element if x is the negation of the nth element of

x_n

.

Clearly,

x

differs from all

x_n

(at the nth element), so it cannot be in this bijection. This contradicts that the bijection is onto.

------

To do this with uncountable sets, we note that the index of a sequence doesn't necessarily have to be the natural numbers, they can be the elements of

X

itself.

So we represent each subset of

X

by a sequence:

S=\{s_\alpha\}_{\alpha \in X}

, with

s_\alpha=1

if

\alpha

is in the subset, and 0 otherwise.

So assume there is a bijection. It would look something like

\alpha \to S_\alpha

\beta \to S_\beta

\gamma \to S_\gamma

etc. In fact, we could just write

S_\alpha

as the subset being mapped to by

\alpha \in X

, and there'd be no ambiguity.

So we define the "bad" subset

x

similarly: the

\alpha

th element of x is the negation of the

\alpha

th element of

S_\alpha

.

Clearly,

x

differs from all

S_\alpha

at the

\alpha

th element. This contradicts that the bijection is onto.

Therefore, P(X) is bigger than X.

Reply 83

The Bachelor

13

Oh, and the definition of the set C in your post is slightly wrong.

Reply 84

The Bachelor

13

Bead

and while we're on it, there is another question regarding infinity. if we have 0.9 recurring, when do we actually reach 1? is 1 just an approximation of 0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999?

The archemedian principle: given a real number

x \geq 0

,

If

x < \frac{1}{n}

for all

n \in \mathbb{N}

, then

x =0

.

You can prove it by assuming the real numbers are a valid field (I.e if you multiply or divide 2 reals, with the exception of 0, the answer is a real. Hint: would you agree that, for any real number y, there exists natural number N greater than y?)

zero

(read the whole thread)

wow thanks for explaining stuff for me, guys :smile:

although i don't understand all, mind 0.0 (i will be looking up on that set theory). but it's weird that infinity, which is not a number, should be treated like a number in maths :rolleyes:

Reply 87

jamesair99

Kolya

Don't forget that:
INFINITY IS NOT A REAL NUMBER

surely it's not a complex number :p:

Reply 88

deadgenius

2

Hmmm, the answer to this has something to with Cantor's levels of infinity or something like that. basically, infinity is NOT a number, it is a set of numbers, all infinitely bigger than the last. Maybe.

Reply 89

Infinity is not a number, it has NO position on the umber line (complex or real), and can be thought of as the process of counting without end.

Reply 90

yusufu

16

DeanK2

Infinity is not a number, it has NO position on the umber line (complex or real), and can be thought of as the process of counting without end.

What about the ends of the real line?

EDIT: And there's no such thing as the complex number line. :smile:

Reply 91

RyanT

15

DeanK2

Infinity is not a number, it has NO position on the umber line (complex or real), and can be thought of as the process of counting without end.

Oh, i thought infinity was a number - thanks for helping me out :smile:

Wouldn't it just be an infinity bigger then the other two?
They can be bigger then one another y'know :s-smilie:

Reply 92

20083

2

DeanK2

I've spent months (literally months) wandering about infinity.

Did you get there?

Reply 93

I'm vaguely amused by the fact that the real number line is a closed set, yet its boundary is empty. It's somewhat mind-boggling.

Reply 94

There is no actual end to the number line. However, there does seem to be some coonfusion (slighlty off topic) as to why a circle's circumference is not exact.
A rational number can be expressed in the form (p/q). So, taking a rational number (such as 2) and splitting it half. And then one of those pieces in half again. And so on... you can create an infinite series that will CONVERGE to 2.

The difference with pi is that it is IRRATIONAL. A series can be used to converge to pi, but you cannot terminate the series, or add up infintely many terms. So, the circle's circumference can be 'trapped' between values, yet will not have an exact value.

Reply 95

It has an exact value, defined by the limit of the series. It just so happens that value is not rational.

Reply 96

DFranklin

18

DeanK2

There is no actual end to the number line.Loosely speaking,

\infty

lies at both ends of the number line following one-point compactification.

A rational number can be expressed in the form (p/q). So, taking a rational number (such as 2) and splitting it half. And then one of those pieces in half again. And so on... you can create an infinite series that will CONVERGE to 2.

I don't know what you think this is proving; it's not exactly hard to produce an infinite series that will CONVERGE (sic) to pi, y'know.

Reply 97

It isn't hard to produce one at all. When using pi in calculations, you will ALWAYS obtain (unless dividing pi by pi etc) an irrational number hence circumference can only be trapped between two values on the number line, but can never be thought of as a point.

Reply 98

But the definition of convergence means that the limit is trapped between two values that can be arbitrarily close, therefore, it can be considered a point on the real number line.

Reply 99