What is the cardinality of $[a,b] $?

Question

It is a well-known fact that any open interval $(a,b)$ has the same cardinality as $\mathbb R$; that is, there exists a bijection $f\colon(a,b)\mapsto \mathbb R$. What about the closed interval $[a,b]$?

I figured since $[a,b]=\{a\}\cup(a,b)\cup\{b\}$, and since $\{a\}\cup(a,b)\cup\{b\}$ are all disjoint, $$\text{card}\,[a,b] = \text{card}\,\{a\} + \text{card}\,(a,b) + \text{card}\,\{b\}=\text{card}\,(a,b) + 2$$ But if $\text{card}\,(a,b)=\text{card}\,\mathbb R=\mathfrak c$, then $\text{card}\,(a,b) + 2=\mathfrak c+2=\mathfrak c$. So then that means $\text{card}\,[a,b]=\text{card}\,\mathbb R$. A contradiction. (Because there is no possible bijection $g\colon[a,b]\mapsto \mathbb R$.)

Which of my assumptions is faulty?

You've simply asserted there is no bijection between $[a,b]$ and $\mathbb R$. It's true there is no continuous bijection between the two sets, but it isn't true there is no bijection. — Thomas Andrews, Aug 05 '13 at 02:27
The appendix of this answer describes explicit bijections $[a,b]\leftrightarrow\Bbb R$ and $[a,b]\leftrightarrow(a,b)$, among other things. — MJD, Aug 05 '13 at 02:31

score 5 · Accepted Answer · edited Aug 05 '13 at 05:18

5

There is a bijection $g:[0,1]\rightarrow \Bbb R$. Let $h:(0,1) \rightarrow \Bbb R$ be a bijection. Let $f:[0,1] \rightarrow (0,1)$ be the bijection defined as follows. Map $f(0)=1/2$, $f(1)=1/3$, $f(1/2)=1/4$, and in general, $f(1/n)=1/(n+2)$. For all other values, map $f(x)=x$. Then $g=h(f(x)):[0,1]\rightarrow \Bbb R$ is a bijection.

$$f(x)= \begin{cases} 1/2 &x=0\\ 1/(n+2) & \exists n\in\mathbb N : x=1/n\\ x & \text{else} \end{cases} $$

Why is $f:[0,1]\rightarrow (0,1)$ a bijection? First note that for $U = \{1/2,1/3,\dots \}$ we have $f:\{0,1\} \cup U \rightarrow U$ and $f:(0,1) \setminus U \rightarrow (0,1) \setminus U$. This partitions $[0,1]$ into two disjoint sets and we only need to show $f$ is a bijection to the corresponding set in $(0,1)$ for each one. In the later case, $f$ is just the identity map. In the first case, we just shift $0 \to 1/2$, $1 \to 1/3$, $1/2 \to 1/4$, and so on.

edited Aug 05 '13 at 05:18

learner

6,726

answered Aug 05 '13 at 02:27

abnry

14,664

I like your example. What could make it even better is if you could prove (even informally) that it is indeed a bijection. – chharvey Aug 05 '13 at 04:07
I added a quick explanation. – abnry Aug 05 '13 at 04:23
Thanks. It really is a beautiful function. I added the notation for easy reading. – chharvey Aug 05 '13 at 04:46
http://math.stackexchange.com/a/160741/ – Did Aug 06 '13 at 20:38

What is the cardinality of $[a,b] $?

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