Cardinality of real numbers

Question

Given an open interval, say $(a,b]$, how do we show that it has the same cardinality as the set of real numbers? Or is there a bijective mapping in the 1st place?

Many related questions. This one is almost a duplicate. http://math.stackexchange.com/questions/28568/bijection-between-an-open-and-a-closed-interval — Ethan Bolker, Apr 08 '17 at 16:58
Is there a way to show using bijective functions? I am thinking of tangent functions but not too sure how to construct — Jus, Apr 08 '17 at 17:05
@Jus: There does not exist a continuous bijection function between $(a,b]$ and the reals. Any explicit bijection is going to have to get dirty and dig into doing some of the discrete set manipulations. The real numbers offer no insight into how this works, so it's better to just do it in the abstract to show that $\alpha + 1 = \alpha$ whenever $\alpha$ is an infinite cardinal. — , Apr 08 '17 at 17:05
Is it possible to find a function that is bijective if tangent doesn't work? There should be such a function that exist since there is indeed a bijective mapping. — Jus, Apr 08 '17 at 17:08
Hint what doesn $b$ map to? Call it $b_1$ what does $b_1$ map to? Call it $b_2$ etc. Can we do that a countably many times and leave all those not in $b_i$ alone? — fleablood, Apr 08 '17 at 17:13

score 0 · Answer 1 · answered Apr 08 '17 at 16:59

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The simplest proof sketch, in my opinion is:

$|(a,b)| = \mathfrak{c}$
$|(a,b]| = |(a,b)| + 1$
$\mathfrak{c} + 1 = \mathfrak{c}$

where $\mathfrak{c}$ is the cardinality of the reals.

answered Apr 08 '17 at 16:59

... assuming the OP is comfortable with that much cardinal arithmetic. Unlikely for a beginner. – Ethan Bolker Apr 08 '17 at 17:01
I think one must prove $\mathfrak c+ 1 = \mathfrak c$ first. – fleablood Apr 08 '17 at 17:03

fleablood · Answer 2 · 2017-04-08T17:34:45.080

Take a countable subset of $(a,b)$. and index the elements as $K=\{c_i\}$.

Take the map $f:(a,b] \rightarrow (a,c)$ as:

$f(b) = c_1$

$f(c_i) = c_{i+1}$

$f(x)_{x\ne b; x\not \in K} = x$.

That is easily shown to be a bijection so the cardinality of $(a,b]$ is the same cardinality as $(a,b)$.

If you are like me, the blase command "Take a countable subset of $(a,b)$ " will leave you cold. But it is legitimate via the inductive property of the natural numbers. I don't think (I could be wrong) that it would be valid to say "take an uncountable subset" but a countable subset is fine.

Oh, and $(a,b)$ has the same cardinality as $\mathbb R$ is a standard exercise. Your tangent function with proper shifting and scaling will do.

score 0 · Answer 3 · answered Apr 08 '17 at 17:14

You can use the tangent function to construct a bijection $f$ from $(a,b)$ to $\mathbb{R}$. (That's an idea from your comment.)

Now you have to find a way to map the extra point $b$. The trick is to give up on looking for a nice formula and reuse a trick you may have seen when "counting" the natural numbers. Use your bijection $f$ to find the points $x_n$ such that $f(x_n) = n$ for $n= 0, 1, \ldots$. Now redefine $f$ so that $f(x_{n+1}) = n$. Then $f$ is is injective on $(a,b)$ and its range is all of $\mathbb{R}$ except $0$. So define $f(b) = 0$ and you're done.

Cardinality of real numbers

3 Answers3