0
$\begingroup$

This question already has an answer here:

We know that there are bijections between $[0,1]$, $(0,1)$ and $\mathbb{R}$. But my question is can we obtain a continuous bijection between $[0,1]$ and $(0,1)$, and between $[0,1]$ and $\mathbb{R}$? I think there will not exist but I am not sure.

$\endgroup$

marked as duplicate by Nosrati, YuiTo Cheng, Thomas Shelby, postmortes, Leucippus Jun 26 at 6:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I do not know how to use mathjax,so please someone edit my question. $\endgroup$ – user684834 Jun 26 at 3:43
  • 2
    $\begingroup$ What is IR?${}{}$ $\endgroup$ – Clayton Jun 26 at 3:43
  • $\begingroup$ Put them between dollars. $\mathbb{R}$ gives $\mathbb{R}$. $\endgroup$ – Nosrati Jun 26 at 3:45
  • 3
    $\begingroup$ Note: $[0,1]$ is compact and $(0,1)$ is not $\endgroup$ – J. W. Tanner Jun 26 at 3:46
  • 1
    $\begingroup$ @Clayton I think it's interval $\mathbb R$. $\endgroup$ – Michael Rozenberg Jun 26 at 3:48
2
$\begingroup$

The image of a continuous map of a compact metric space is compact. In particular, the image of a continuous map of a compact metric space into $\mathbb R$ is closed and bounded. Therefore, don't expect to find a continuous bijection between $[0,1], $ which is compact, and $(0,1)$, which is open. Other explanations can be found here.

$\endgroup$
  • $\begingroup$ assuming the standard topology for $\Bbb R$ $\endgroup$ – J. W. Tanner Jun 26 at 4:06
  • $\begingroup$ same argument applies for $[0,1]$ and $\Bbb R$, but there are continuous bijections between $(0,1)$ and $\Bbb R$ $\endgroup$ – J. W. Tanner Jun 26 at 4:23