Does any interval of $\mathbb R$ have the same number of elements as $\mathbb R$?

I think I understand Cantor's diagonalization argument, but I'm trying to wrap my head around this consequence of it.

Let's suppose I pick an arbitrary interval, say, $[5, 6]$. Is it true that the number of reals in this interval is the same as the number of reals in $\mathbb R$? How can that be when $\mathbb R$ is clearly bigger than that interval?

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It just means there is some one-to-one correspondence between the interval and all of $\mathbb{R}$. Yes, it is quite bizarre, but you've actually encountered something like it before. The function $\frac{1}{x}$ on $[-1,1] \setminus \{0\}$ shows a correspondence between the domain and the vast majority of $\mathbb{R}$. – Austin Mohr Nov 3 '11 at 19:58
@AustinMohr Thank you for the clarification, I edited accordingly. – Evan Riley Nov 3 '11 at 19:58
"the same number of elements" is loose speaking, infinite sets do not really have a "number of elements". If by that you mean the cardinality, yes, that's true. And that (an infinite set can have the same cardinality of a proper subset of itself) is a basic "paradoxical" property of infinite sets, rather more basic than then diagonalization argument (it applies also to countable infinite sets). See the Hotel paradox en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel – leonbloy Nov 3 '11 at 20:03
@Austin, you mean $(-\frac\pi2,\frac\pi2)$, right? – Henning Makholm Nov 3 '11 at 20:05
You have to reconcile yourself with a different notion of "bigger." For example, under Cantor's definition of "same size," any infinite set of natural numbers is the same size as the natural numbers. So there are as many perfect squares, for example. – Thomas Andrews Nov 3 '11 at 20:08

You are mingling two different notions of size. Indeed there is a sense in which $\mathbb R$ is "bigger" than a finite interval of $\mathbb R$. The branch of mathematics dealing with this notion of size is called measure theory. For standard measures such as the Lebesgue measure, the measure of $\mathbb R$ is infinite and the measure of $[5,6]$ is $1$.

However, when we talk about the "number" of reals, this refers to a different notion of "size", the one formalized as cardinality. This is purely concerned with the possibility of mapping sets to each other, not with any notion of size in the physical sense of extent. There's no contradiction in the fact that these two notions of size lead to different comparisons between $\mathbb R$ and $[5,6]$.

The situation is somewhat similar to the difference between comparing subsets of natural numbers using cardinality or using natural density. For instance, the set of natural numbers and the set of even numbers have the same cardinality and are in that sense equally "big". However, the set of natural numbers has natural density $1$ and the set of even numbers has natural density $1/2$, so there is also a well-defined sense in which there are twice as many natural numbers as even numbers.

The case of natural numbers is perhaps somewhat easier to visualize than the case of real numbers.

To visualize the comparison using cardinality, imagine an infinite row of elves extending towards the West. Each elf has a natural number on the front of his or her garment, beginning with $1$ and increasing by $1$ with each elf along the row. Now all the elves simultaneously turn around, and on the back of their garment each elf has twice the number on the front. They're still the same elves with the same garments, but now when you look along that row, you only see the even numbers. Clearly there are as many even numbers as natural numbers.

To visualize the comparison using natural density, imagine an infinite row of lights, each associated with one of the natural numbers and being turned on to represent that a number is in the subset being represented. When we represent the naturals numbers, all the lights are on. You can take a closer look at any section of the row and see that all the lights are on in that section. Now we represent the even numbers by turning off all the odd-numbered lights and leaving only the even-numbered ones on. Now you can take a closer look at any section of the row and see that half the lights in that section are on and half the lights are off. Only half the lights are on, so clearly there are only half as many even numbers as natural numbers.

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Because the finitary notion of "bigger" (in terms of set containment) doesn't react well with infinite cardinalities. Perhaps more shocking is that the set of all continuous maps $\mathbb{R}\to\mathbb{R}$, $\mathbb{R}^n$ for any $n\in\mathbb{N}$, $\mathbb{C}^n$, and $(a,b)$ for any $a,b\in \mathbb{R}$ all have the same number of elements, if we define "the same number of elements" to mean there exists a bijection between the two sets. To get a visually appealing reason why $(5,6)$, say, is equipotent to $\mathbb{R}$ (in the sense that I have said) imagine bending $(5,6)$ into a $U$-shape, drawing a vertical line through the center of the $U$ and projecting from that line, through the $U$, to the real line. You should see that this is the required bijection. For more information akin to what I have told you, see here http://en.wikipedia.org/wiki/Stereographic_projection

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The function $f(x) = 5+\frac{1}{1+e^x}$ is a bijection $\mathbb R \to (5,6)$. We can make it hit the endpoints too by shuffling countably many points a la Hilbert's Hotel.

This obviously works for every proper interval, so every such interval are equinumerous with $\mathbb R$ itself.

It is clear that $\mathbb R$ is a proper superset of $[5,6]$, but that isn't the same as being "clearly bigger" as regards cardinality. This is not more (or less) mysterious than the fact that the set of perfect squares is equinumerous with $\mathbb Z$, despite being a proper subset.

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