I'm trying to prove any interval has the same cardinality of the reals numbers $\mathbb R$. In order to prove this I define two functions $f:(s,t)\to (u,v),f(x)=\frac{v-u}{t-s}(x-s)+u$ and $g:\mathbb R\to (-1,1),g(x)=\frac{x}{1+|x|}$.

My question is are these functions bijections and from that can I conclude any interval has the same cardinality of $\mathbb R$?

Thanks in advance

  • $\begingroup$ $g$ is not defined at $x=-1$. $\endgroup$ – T. Eskin May 6 '14 at 3:48
  • $\begingroup$ @ThomasE. I changed the function, is it ok now? thanks for the remark. $\endgroup$ – user42912 May 6 '14 at 3:50
  • $\begingroup$ The problem is a little different for open intervals $(a,b)$ than for closed intervals $[a,b]$, or half-open intervals. $\endgroup$ – André Nicolas May 6 '14 at 3:51
  • $\begingroup$ The range of your new $g$ is not $(0,1)$. No big problem, you can work with $(-1,1)$. $\endgroup$ – André Nicolas May 6 '14 at 3:52
  • $\begingroup$ @AndréNicolas is it of now? $\endgroup$ – user42912 May 6 '14 at 3:53

Your first function is a bijection and proves that any two finite intervals have the same cardinality. Your second fails because $g(-\frac 12) \not \in (0,1)$ but you have the right idea. There are many bijections between $\Bbb R$ and some finite interval. One of the simplest is $\arctan(x)\to (\frac {-\pi}2,\frac \pi 2)$ This solves the open intervals. For closed intervals, you need to "swallow" the endpoints somehow.

  • $\begingroup$ I changed the function, do you think it's a bijection now? thanks for the answer. $\endgroup$ – user42912 May 6 '14 at 3:55
  • $\begingroup$ Yes, it is. Can you prove it? $\endgroup$ – Ross Millikan May 6 '14 at 3:58
  • $\begingroup$ I'm trying to do this right now. $\endgroup$ – user42912 May 6 '14 at 3:59
  • $\begingroup$ $f(x)$ is a bijection because it is a linear function, and all linear functions are bijections (why?). For $g$ consider $\frac{x}{1+|x|}=\frac{x}{1+x}$ if $x \geq 0$. Its derivative is $\frac{1}{(1+x)^2}>0$. This means $g(x)$ is increasing when $x \geq 0$. Note the same is true about $g'(x)$ in the case $x<0$ and notice that $g(x)$ is continuous on $\mathbb{R}$. Conclude. $\endgroup$ – Darrin May 6 '14 at 4:03
  • $\begingroup$ I defined two functions $g_1:\mathbb R^+\to (0,1), g_1(x)=\frac{x}{1+x}$ and $g_2:\mathbb R^-\to (-1,0], g_2(x)=\frac{x}{1-x}$ $\endgroup$ – user42912 May 6 '14 at 4:04

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