Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My math teacher said that. I disagreed, but he said that I was wrong. But I'm not convinced - is it really right? Please notice that I'm not talking about $\mathbb{R}$ $⊂$ $\mathbb{C}$, but $\mathbb{R}$ $<$ $\mathbb{C}$.

share|improve this question
4  
My answer to a closely related old question (about even integers and all integers) also answers this question. –  Noah Snyder Oct 16 '11 at 22:24
    
@NoahSnyder I think this kind of "agrupament" is almost "velocity" in Physics. It's like "who grows faster", but I don't know if you understand what I mean. –  Ian Mateus Oct 16 '11 at 22:36
4  
Ask what his notion of size is. $\mathbb{C}$ has more elements in its basis (as a rea vector space) for example. Often these misunderstandings come up because two people mean different things. –  AnonymousCoward Oct 16 '11 at 23:15
    
@GottfriedLeibniz I think it's more a linguistic question than a mathematical one :) –  Ian Mateus Oct 17 '11 at 0:51

4 Answers 4

up vote 16 down vote accepted

It really depends on what one means by "fewer elements".

The usual notion that most mathematicians use for this is "cardinality". Two sets $X$ and $Y$ have the same cardinality if and only if there is a bijection between them (a function $f\colon X\to Y$ that "pairs them up", so that every element of $Y$ corresponds, under $f$, to one and only one element of $X$). In the case of finite sets, this corresponds to our usual notion of "size", in that two finite sets have the same "cardinality" if and only if they have the same number of elements.

But for infinite sets, the notion introduces many counterintuitive consequences: the set of all even integers has "the same cardinality" as the set of all integers, even though it is a proper subset of it. In fact, one way to define "infinite set" is to say that a set is infinite if and only if it has the same cardinality as a proper subset.

We say a set $X$ has "strictly smaller cardinality" than a set $Y$ if and only if there is a one-to-one function $f\colon X\to Y$, but there is no onto function $g\colon X\to Y$. In particular, the fact that there is a one-to-one but not onto surjection from $\mathbb{R}$ to $\mathbb{C}$ shows that the cardinality of $\mathbb{R}$ is less than or equal to that of $\mathbb{C}$, but it does not show that they are not equal (just because this particular function is not onto does not mean there are no onto functions).

In that sense, $\mathbb{R}$ and $\mathbb{C}$ both have "the same size" because they both have the same cardinality: there is in fact a bijection between $\mathbb{R}$ and $\mathbb{R}^2$, and of course there is a bijection between $\mathbb{R}^2$ and $\mathbb{C}$.

The arithmetic of cardinalities is interesting, and not without its dangers (the answers may depend on exactly what set-theoretic axioms you accept or not accept, for example).

On the other hand, you can try to measure "size" in other ways. There is a natural notion of "size" for subsets of the complex numbers: their area, or at least their outer measure, when viewed as subsets of the plane; if you use that notion of size, then the real numbers have "size" $0$, while the complex numbers have infinite size, so the complex numbers are "larger". Added. Or as t.b. suggests in the comments, you could try to compare them in terms of their dimension as an $\mathbb{R}$-vector space ($\dim_{\mathbb{R}}(\mathbb{R}) = 1\lt 2 = \dim_{\mathbb{R}}(\mathbb{C})$. Or any other of several ways in which you could try to compare their "size".

I would argue, however, that it would not be very felicitous to call these other ways of comparing their size, "the reals have fewer elements than the complex numbers", because "fewer elements" seems to refer to a "count" of elements, and that leads you straight back to the notion of cardinality.

share|improve this answer
7  
Maybe measuring the sizes in terms of dimension as $\mathbb{R}$-vector spaces would be another way to grasp their apparent difference? Interesting that as $\mathbb{Q}$-vector spaces their "size" becomes the same again thanks to cardinal arithmetic... –  t.b. Oct 16 '11 at 23:22
    
@t.b. Good point. –  Arturo Magidin Oct 16 '11 at 23:23
1  
@ArturoMagidin oh, and if I've understood this, so quaternions would have the same cardinality than complex numbers, right? –  Ian Mateus Oct 17 '11 at 0:30
2  
@IanMateus: Careful: "countable" has a specific meaning (it means that there is a surjection from $\mathbb{N}$ onto the set). As to your second question, yes, the quaternions (with real coefficients) have the same cardinality as the complex numbers. –  Arturo Magidin Oct 17 '11 at 1:37
1  
@PatrickDaSilva: The argument is due to Cantor, who also gave other ways of doing the bijection. Worse case scenario, it's easy to come up with injections $\mathbb{R}\hookrightarrow \mathbb{R}\times\mathbb{R}\times\mathbb{R}$ and $\mathbb{R}\times\mathbb{R}\hookrightarrow\mathbb{R}$ (e.g., using the interweaving digits), and then apply Cantor-Bernstein to get a bijection out of them, which will be as explicit and constructive as the original functions were (CB is constructive, not merely existential). –  Arturo Magidin Nov 11 '11 at 5:49

$\mathbb R$ is smaller than $\mathbb C$ ... ??? In the sense of cardinality, NO. (As explained in other answers.) In the sense of Hausdorff dimension, YES. What sense did your teacher mean?

I'm sorry, but we often find when a student reports what a teacher says, some details may get distorted or even omitted. Unless the teacher comes here and makes a defense, we cannot know what is going on. Or, even better, if the student talks to the teacher (what an idea!) and finds out what was intended.

share|improve this answer
    
I think that's a good idea. Next class, I'll talk to him and ask what was intended. Thank you, man. –  Ian Mateus Oct 17 '11 at 0:17

No, it is not correct. Both the real numbers $\mathbb{R}$ and the complex numbers $\mathbb{C}$ have the same cardinality, $\mathfrak{c}$, which is the cardinality of the continuum. See this wikipedia page.

As a general rule, if $A\subseteq B$ then you can say that $|A|\leq|B|$, but if $A\subset B$ (that's a strict subset relation) you cannot conclude that $|A| < |B|$.

share|improve this answer
    
I recommend reading Arturo's answer for a more complete discussion. –  Chris Taylor Oct 17 '11 at 0:29

There are two senses of size, which I think any mathematician has thought in terms of (here, both strict comparisons):

  1. Subsets: $A \subset B$: $A$ is a smaller set than $B$ in that $B$ contains elements in $A$, and more.
  2. Cardinality: $A \prec B$: for some bijection $f$, $f[A] \subset B$

In this context, both $\mathbb{R} \subset \mathbb{C}$ and $\mathbb{R} \prec \mathbb{C}$. One way to see the second fact is that $\mathbb{C} = \mathbb{R}\times\mathbb{R}$ as sets and that the product (or sum, or max) of two infinite cardinalities is the maximum of those two values.

Anyways, I think it would be hard to find a mathematician who hasn't thought in terms of 'bigger', 'smaller', and 'size' in both the comparative senses of set membership (1) and set cardinality (2). The notion of subset is obviously useful, while the notion of cardinality shows up everywhere when you don't care about the properties of the elements of the set, you just care about what you can construct out of a set. Blind to any properties the elements of a set may have, the only thing left over is cardinality.

I think the only trouble here is not the mathematics, but semantics and terminology, which may be a point of contention. For example, I've known a few people who object to any association between 'size' in the intuitive sense and the mathematical definition of cardinality. I haven't seen this happen within the mathematical community. Perhaps it's because mathematicians are wired differently or that we often find the exotic objects in the mathematical world not more unnatural than the exotic species in the biological world... and that in fact the mathematical universe wouldn't work right without them.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.