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I want to prove that any open interval $(a,b)$ has the same cardinality of the real numbers: $|(a,b)| = |\Bbb R|$.

Do I have to find an function to prove it? Or is there a theorem to prove it easier? or any idea?

Hanul Jeon
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Alex Turner
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    It's easy in this case to find a bijective function between the two. In fact, you can choose a continuous function that works – Ben Grossmann Aug 30 '15 at 01:48
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    You know $( - \frac{\pi}{2}, \frac{\pi}{2} )$ is mapped to $\mathbb R$ under the function $\tan x$ – ThePortakal Aug 30 '15 at 01:48
  • Google "cardinality open interval" and the first result I got was http://math.stackexchange.com/questions/300815/show-that-open-segment-a-b-close-segment-a-b-have-the-same-cardinality, so while Google personalizes search results, I find it hard to believe this woul dhave been very far down the search for you either. – Asaf Karagila Aug 30 '15 at 05:41

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The function $y = \tan(x)$ is bijective on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. We would like to shift/stretch it so that it's bijective on $(a,b)$. The period should be $b-a$, so we would at least have $y = \tan\left(\frac{\pi}{b-a}x \right)$. Now translate it.

user217285
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Hint: First find an easy bijection between any two given open intervals (you can even find a continuous one), and then use a function and certain interval that you can show a bijection to ${\mathbb R}$ with (for example, tangent function).

user2566092
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There's a great visual proof for this, have you seen it? It's simple enough that I can just describe it, and perhaps you can draw it yourself:

Take a circle $C$ of circumference $|a-b|$ (this will be our interval $(a,b)$, after we remove a single point), and place it above a straight line $L$ of infinite length having no endpoints (this is our stand-in for $\mathbb{R}$). Mark the point $v$ on $C$ that is furthest from $L$.

The bijection between the points on $C$ (excluding $v$) and the points on $L$ is simply: the point $u$ on $C$ maps to the point $p$ on $L$ if both lie on a line segment from $v$ to $L$.

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