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There are infinite number of real numbers between 0 and 1,i.e in the interval [0,1]. So definitely there should be more numbers in the interval [0,10] because it includes the numbers in the first case and also the numbers in the interval (1,10]. But we can map(one to one) every number in [0,1] to every number in [0,10]. This means that there are equal numbers in [0,1] and [0,10]. What does this really mean though??

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"more" is a bad term. If you mean one set is a subset of the other, then yes. If you mean this in terms of some measure of "size" for a set, it depends on how you measure it. The standard way is cardinality, where we denote the "size" of a set, $A$, by $|A|$. Then if there is an injective function $f:A\to B$ we say $|A|\le B$ and if there is a bijective function, then we say $|A|=|B|$. But there is a bijection here, $f(x)=10x$ is one and its inverse is $f(x)=x/10$. – Adam Hughes Jul 4 '14 at 7:52
I'm confused. I was showing the map from $[0,1]\to[0,10]$. Maybe I should answer this way: "should" is a dangerous phrase which is incorrect because it relies on too naïve a notion of "size" for sets which is purely subset related. – Adam Hughes Jul 4 '14 at 7:57
Cardinality is a fair way to compare which set is "bigger" in a set theory sense. If you want "bigger" in a geometric sense, use the length of the interval. It depends on what kind of question you're asking, there is no one golden "true" measure of "size." You have to specify what kind of size you mean. – Adam Hughes Jul 4 '14 at 8:02
There were about zillion questions like that on the site before. Have you made any effort to find them, read them and understand their answers? – Asaf Karagila Jul 4 '14 at 8:02
@AsafKaragila yes, I have. I went through the following question I found a similar argument there, but couldn't find a convincing answer to that – Maximus Jul 4 '14 at 8:05

The fact that the set $A$ is a proper subset of a set $B$ does NOT mean that the cardinality of the set $A$ is smaller than the set $B$. In fact, a set is infinite if and only if it contains a proper subset with the same cardinality as itself.

For example, the set $\{2,3,4,\dots\}$ is a proper subset of $\mathbb N$, but has the same cardinality as $\mathbb N$, as the bijection $n\mapsto n+1$ clearly shows.

I admit, this can, at first, be slightly confuzing, but eventually, you get used to it. The main point is this: the ONLY proper way to determine whether sets have the same cardinality is to show whether there exists a bijection from one to the other. A bijection can be viewed as a pairing. By that I mean that a bijection $f:A\to B$ produces a set of pairs $\{(a,f(a)|a\in A\}$ where you provided each element of $A$ with an element of $B$ to pair up with. Since all the elements are now paired up, it makes sense to say that yes, indeed, there is the same "amount" of them in both sets.

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So, are you saying that,all infinities that have a one to one mapping are equal. Dont take me wrong here, but I think that it fails the axioms that "Whole is always greater that the part". – Maximus Jul 4 '14 at 8:19
@Maximus: Axioms don't exist "just like that". Modern mathematics have axioms as formal statements which are part of specific theories. "The whole is always greater than the part" sounds like something that Plato or Aristotle would said, or something that Kant might argue. It does not sound like mathematics, by a modern standard. – Asaf Karagila Jul 4 '14 at 8:21
@Maximus There is no such axiom in mathematics. That is exactly why I said you should only consider cardinalities through the eyes of bijections. If there exists a bijection, the sets have equal cardinality (notice I am avoiding the use of common phrase terms like "greater" or "bigger". For example, the set $\{2,3,4\dots\}$ has the same cardinality as $\mathbb N$, $\mathbb N^2$, $\mathbb Q$ and many more sets, but it has a smaller cardinality as $\mathbb R$ (which, again, has the same cardinality as both $(0,1)$ and $[0,1]$) – 5xum Jul 4 '14 at 8:22
The proof of the equality of cardinality of $\mathbb Q$ and $\mathbb N$ is a beautiful example of this. – tpb261 Jul 4 '14 at 10:59
@tpb261 Strictly speaking, the beautiful proof is the one proving the equal cardinality of $\mathbb N^2$ and $\mathbb N$, since proving it with $\mathbb Q$ requires leaving out some elements (like $4/2$, because you already have $2/1$). But yes, it is a perfect example. – 5xum Jul 4 '14 at 12:04

People are used to work with finite numbers. In case of infinity, our intuition fails. In set theory, the 'measure' of the set is its cardinal number. It tells us how many elements does the set have. In the finite set it is easy to find its cardinal number: count the elements. Counting means finding a bijection between the given set and the set $N_k=\{1,2,...,k\}$ for any $k \in N$.

For sets with infinitely many elements there are two problems with our intuition:

i) all sets seem to have same number of elements: infinitely, so their cardinals are equal;

ii) as in your problem, the 'bigger' set must have more elements.

But non of above is correct.

What we do here, is just the same: take a bijection between two sets. Because we've used to work mostly with natural or real numbers, in the infinite case we search for bijections between that set and natural or real numbers. First person to handled these problems was Georg Cantor. He proved that we cannot find any bijection between natural and real numbers, so it means that cardinals of natural and real numbers are not equal!

But, we can always find a bijection between any interval [a,b] and real numbers, and a bijection between any two intervals, so we can also find a bijection between [0,1] and [0,10].

So, the explanation why it is so is related to the nature of infinity (or infinities). Imagine if you add a finite elements to a set with infinitely many elements. How many elements will the set have? What if we add infinitely many elements to a set with infinitely many elements?!

Also remember that 'counting the number of elements' means finding a bijection!

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Looking at the comments to the other answer, it seems that the OP has a problem based on a philosophical principle that "Whole is greater than the part". – Asaf Karagila Jul 4 '14 at 12:59
Well, it depends related to what we compare the sets. If we compare the mass of the given sets, [0,1] has mass 1 and [0,10] has mass 10. That philosophical axiom is related to mas, but not to the number of elements. – Emin Jul 4 '14 at 13:04
No. It is just that the philosophical principle is based on finiteness of "the whole", rather than an abstract principle. And because we naturally associate an interval with its length, which is finite, it seems to us that measure and cardinality needs to somehow coincide and whatnot. – Asaf Karagila Jul 4 '14 at 13:07
You are right!! – Emin Jul 4 '14 at 13:11

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