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Lets say there is an open interval (a,b) where a,b are real numbers. Is it true that (a,b) has cardinality equal to real numbers?

Jus
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2 Answers2

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Yes. $f(x)=\tan (\frac {\pi}{2}\cdot \frac {x-m}{d}),$ where $m=(b+a)/2$ and $d=(b-a)/2,$ is a bijection from $(a,b)$ to $\mathbb R.$

DanielWainfleet
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  • How do you arrive at such a function? Is there any method? – Jus Apr 06 '17 at 09:42
  • @Jus The map $x \mapsto x - (b+a)/2$ sends the open interval $(a, b)$ to $(-(b-a)/2, (b-a)/2)$. The idea is that this gives an interval of the form $(-d, d)$, so that the subsequent map $x \mapsto x/d$ will send this new interval to $(-1, 1)$. Finally, as in my answer about bijecting $(0, 1)$ with $\mathbb{R}$, there is the frequent question of bijecting $(-1, 1)$ with $\mathbb{R}$. A common example is a modification of the tangent function, since $x \mapsto \tan(x)$ bijects the open interval $(-\pi/2, \pi/2)$ with $\mathbb{R}$. So, the suggestion here is a composition of several functions. – Benjamin Dickman Apr 06 '17 at 09:57
  • There are many examples of continuous 1-to-1 functions that map an open bounded interval onto $\mathbb R.$.... It is well-known than $\tan (x) $ maps $(-\pi /2,\pi /2)$ bijectively to $\mathbb R,$ ..... The $f$ of my answer is $\tan g(x)$ where $g$ is a linear map from $(a,b)$ onto $(-\pi /2,\pi /2)$. – DanielWainfleet Apr 06 '17 at 09:59
  • (In fact, my answer at the earlier MSE link similarly notes that the inverse hyperbolic tangent, $\tanh^{-1}$, bijects $(-1, 1)$ with $\mathbb{R}$; so, one could also use here: $\tanh^{-1}(\frac{x-m}{d})$...) – Benjamin Dickman Apr 06 '17 at 10:01
  • Does it work if the interval is closed? Say [a,b] – Jus Apr 06 '17 at 10:03
  • There is a bijection from $[a,b]$ to $\mathbb R$ but not a continuous one. The usual method is to apply the Cantor-Berstein (a.k.a. Schroeder-Bernstein, a.k.a. Cantor-Schroeder-Bernstein) theorem: If there is an injection of $A$ into $B$ and an injection of $B$ into $A$ then there is a bijection from $A$ to $B.$ ... The identity function injects $[a,b]$ into $\mathbb R$, while a bijection from $\mathbb R$ to $(a,b)$ is an injection of $\mathbb R$ into $[a,b].$ So there exists a bijection from $[a,b]$ to $\mathbb R.$ – DanielWainfleet Apr 06 '17 at 10:25
  • Hi sorry I couldn't really see your message as in it doesn't show the symbols. – Jus Apr 06 '17 at 10:35
  • Must be a network error . Try again. – DanielWainfleet Apr 06 '17 at 10:51
  • Can I say the same function works for cases (a,b], [a,b) too? – Jus Apr 06 '17 at 17:05
  • You need different functions, but the C-S-B method works for half-open intervals. I have seen 2 proofs of the C-S-B theorem, one long and complicated, the other short and simple. – DanielWainfleet Apr 06 '17 at 19:12
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This question likely exists on MSE in some form or another; for example, one often encounters a question around finding a bijection between a particular open interval, such as $(0,1)$, and $\mathbb{R}$.

If you check MSE 2203081, then you will find that there is a 1-1 onto map from $(0,1)$ to $\mathbb{R}$ that is defined by $x \mapsto (2x-1)/(x^2 - x)$.

Since you are starting instead with the open interval $(a, b)$, you could use $x \mapsto (x-a)/(b-a)$ in order to create a bijection from $(a, b)$ to $(0, 1)$, and then you could compose this with the function above (or any of the linked examples) that map $(0,1)$ to $\mathbb{R}$ to find a bijection from $(a, b)$ to $\mathbb{R}$.

So: Yes, given any open interval $(a, b)$, it has the same cardinality as the real numbers, as can be demonstrated by a map that puts it into a 1-1 correspondence with $\mathbb{R}$.

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Benjamin Dickman
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