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i can't find a bijective, continuous map from $\mathbb{R}$ to the closed interval $[0,1]$. Give an example.

If not bijective then what is the difference between cardinal no of $(0,1)$ and $[0,1]$ ?


marked as duplicate by Najib Idrissi, Tomás, hardmath, user296602, user147263 Jan 3 '16 at 19:34

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  • 1
    $\begingroup$ Welcome to MSE! What have you tried? Can you think of any reason why there should/shouldn't be such a map? Also without some more context, it's hard to help. What context did this problem appear, and what theorems do you know about this kind of thing? $\endgroup$ – Zach Stone Jan 3 '16 at 18:42
  • $\begingroup$ If not bijective then what is the difference between cardinal no of (0,1) and [0,1] ? $\endgroup$ – Rajkrishna Mondal Jan 3 '16 at 19:23

Since $f$ is continuous, $f^{-1}((0,1))$ is open, and it is $\mathbb{R}/\{a,b\}$ with $a,b \in \mathbb{R}$ and $a \not = b$. But now $f([a,b]) = [0,1]$ so the map can't be bijective.

  • $\begingroup$ if not bijective then what is the difference between cardinal no of (0,1) and [0,1] ? $\endgroup$ – Rajkrishna Mondal Jan 3 '16 at 19:02
  • $\begingroup$ Same cardinality. $\endgroup$ – Lubin Jan 3 '16 at 19:03
  • $\begingroup$ there does not exit any bijection ,how it will be possible? $\endgroup$ – Rajkrishna Mondal Jan 3 '16 at 19:04
  • $\begingroup$ There is a bijective map, but it is not continuous! $\endgroup$ – Maffred Jan 3 '16 at 19:06
  • $\begingroup$ proof it with example. $\endgroup$ – Rajkrishna Mondal Jan 3 '16 at 19:09

Here is my try:$$\\$$ Since you are trying to find a bijective, continuous mapping, and that is a homeomorphism from the view of topology, we need to verify that if $R$ is momeomorphic to $[0,1]$. We know that $R$ is homeomorphic to $(0,1)$ by $f=\ln(\frac{x}{1-x})$, so the problem is whether $(0,1)$ is homeomorphic to $[0,1]$ or not. However, citing what Tsemo Aristide wrote above, $[0,1]$ is compact, and $(0,1)$ fails to be so. Hence, you can't find a bijective, continuous mapping between $R$ and $[0,1]$. And here's my try to prove that the inverse function of a bijective, countinuous function is also continuous.

$$ \forall y \in f(D) $$ where D is the domain of the original function $f$ $$\forall \epsilon>0 , \exists \delta >0$$ that for all $y^{'}\in \left| y-y^{'} \right|<\epsilon, \left| f^{-1}(y)-f^{-1}(y^{'}) \right|<\epsilon$ holds. Because $\left| f^{-1}(y)-f^{-1}(y^{'}) \right|= \left| x-x^{'} \right|$, and $ \left| y-y^{'} \right| = \left| f(x)-f(x^{'}) \right|$

$$ \forall x \in D, \forall \delta>0 , \exists \epsilon >0$$ that for all $x^{'}\in\left| x-x^{'} \right|<\epsilon, \left| f(x)-f(x^{'}) \right|=\left| y- y^{'}\right|<\delta$ since $f$ is continuous.

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    $\begingroup$ There are bijective continuous functions which are not homeomorphisms (because their inverses are not continuous). The usual example is $f : [0,2 \pi) \to S^1,f(t)=(\cos(t),\sin(t))$. Here the inverse of a point in the first quadrant near $(1,0)$ is far away from the inverse of a point in the fourth quadrant which is also near $(1,0)$. $\endgroup$ – Ian Jan 3 '16 at 19:34
  • $\begingroup$ Thank you for pointing it out! That makes a lot of senses! I really appreciate it! :) $\endgroup$ – sang747 Jan 3 '16 at 19:40

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