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 question concerns the cardinality of two subsets of $\mathbb{R}$. It feels very naive, but, isn't $\left[\dfrac{1}{7,000,000,000},1\right]$ "smaller" than $[0,1]$? I think that this is dependent upon one's set theory axioms, but I'd like an answer from the most common axiomatic models.

After consulting Wikipedia, I've seen that you must show there is only an injective function from $B$ to $A$ to prove that $|A|>|B|$. Such a function is obvious: $f:A\to B$ with $x\mapsto x$. The hard part seems showing there is not a bijective function.

I don't know how to do that, but I think I might be able to figure out why I don't know how to do that. It seems like the continuum hypothesis is at work here: I am trying to find some sort of cardinality, $X$, such that $\aleph_0 < |X| < 2^{\aleph_0}$. This is because it would be intuitive that any subset of the real number system which is an interval would have a larger cardinality than $\aleph_0$. Similarly, I'm trying to find a similar cardinality $Y$ such that $X<Y$.

What's going on here?

share|improve this question
7  
For any $a < b$ finite reals, $f:[a,b] \rightarrow [0,1]$ sends $x \mapsto \frac{x-a}{b-a}$ is a homeomorphism of the usual topologies, and in particular a bijection of sets. I can't imagine many axiomatic models wouldn't accomodate such a map. –  A Blumenthal Feb 2 '13 at 21:52
    
So, @ABlumenthal, $f$ is a bijection from $[a,b]$ to $[0,1]$ with $a<b$ and $a,b\in \mathbb{R}$? Hence, the two sets are of equivalent cardinality? –  000 Feb 2 '13 at 21:55
1  
This function is a continuous bijection with a continuous inverse when both intervals are equipped with the usual topology, i.e. subspace Euclidean topology from the reals, so the function is in fact a homeomorphism. –  Thomas E. Feb 2 '13 at 21:58
1  
In this context you may interpret it as saying simply that $f$ is a continuous bijection whose inverse is also continuous. –  Brian M. Scott Feb 2 '13 at 21:58
1  
Just to clarify something: 'size' in the intuitive sense has no bearing on cardinality, in a catastrophic way. An example is the cantor set (see en.wikipedia.org/wiki/Cantor_set), which has cardinality of the continuum and yet has "size zero" in the most natural sense (that of Lebesgue measure, which is a kind of rigorous model of the notion of 'size' for (most) subsets of the real line). –  A Blumenthal Feb 2 '13 at 22:03
show 7 more comments

2 Answers

up vote 5 down vote accepted

All intervals have the same cardinality as the reals $2^{\aleph_0}$. A Blumenthal has given you a bijection between any closed interval and $[0,1]$. Then you can find a bijection between $[0,1]$ and $(0,1)$ (how can two points matter?) and between $(0,1)$ and $\mathbb R$. It is true that the length of the intervals is different, but the cardinality is not.

According to the continuum hypothesis, known to be consistent with ZFC but not required, there are no sets $X$ such that $\aleph_0 < |X| < 2^{\aleph_0}$, so don't look too hard for them.

share|improve this answer
    
Actually, how do you find a bijection between $[0, 1]$ and $(0, 1)$? –  Joe Z. Feb 13 '13 at 13:38
1  
@JoeZeng: You find two chains of points and shift them down the line. I'll just do one end: $f(0)=1/3, f(1/3^n)=1/3^{(n+1)}, f(x)=x$ for $x \ne 1/3^k$. The other end is similar. –  Ross Millikan Feb 13 '13 at 13:51
    
Didn't think of that. Thanks. –  Joe Z. Feb 13 '13 at 13:55
add comment

Just to supplement Ross’ already-very-clear answer. In what follows, I shall assume that you know some basic measure theory.

Let $ \mu $ be the standard Borel measure defined on the Borel $ \sigma $-algebra $ \mathcal{B} $ of the closed interval $ [0,1] $. Define a quasi-ordering $ \preceq $ on $ \mathcal{B} $ as follows: $$ \forall B_{1},B_{2} \in \mathcal{B}: \quad B_{1} \preceq B_{2} \stackrel{\text{def}}{\iff} \mu(B_{1}) \leq \mu(B_{2}). $$ If $ B_{1} $ and $ B_{2} $ are sub-intervals of $ [0,1] $, then $ \preceq $ is simply a comparison of their lengths.

Define also a quasi-ordering $ \leqslant $ on $ \mathcal{B} $ as follows: \begin{align} \forall B_{1},B_{2} \in \mathcal{B}: \quad B_{1} \leqslant B_{2} &\stackrel{\text{def}}{\iff} |B_{1}| \leq |B_{2}| \\ &\iff (\exists f)(\text{$ f: B_{1} \to B_{2} $ is injective}). \end{align}

The point to note is:

The quasi-ordered set $ (\mathcal{B},\preceq) $ is not isomorphic to the quasi-ordered set $ (\mathcal{B},\leqslant) $ in the category of quasi-ordered sets.

share|improve this answer
    
Is this a very precise way of mathematically qualifying the statement that 'length' and 'cardinality' are two very different concepts? –  000 Feb 3 '13 at 0:16
    
Yes. Although ‘length’ and ‘cardinality’ offer ways of comparing Borel subsets of $ [0,1] $, they are mutually incompatible. –  Haskell Curry Feb 3 '13 at 0:28
    
Your skill of exposition is beautiful. –  000 Feb 3 '13 at 1:16
add comment

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.