It is said that the one point compactification of R is a circle. But how do i show it? I know it suffices to show R is homeomorphic to a punctured circle but how can i prove it?

  • $\begingroup$ You can search: Alexandroff compactification: en.wikipedia.org/wiki/Alexandroff_extension $\endgroup$ Jun 16, 2016 at 15:04
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    $\begingroup$ SImply consider $\Bbb{R} \to S^1$ defined by $t \mapsto e^{2i \arctan t}$. $\endgroup$
    – Crostul
    Jun 16, 2016 at 15:14
  • $\begingroup$ @Novati: Thanks i will check it up $\endgroup$
    – Mathcho
    Jun 16, 2016 at 15:44
  • $\begingroup$ @Crostul: That is not even a surjection $\endgroup$
    – Mathcho
    Jun 16, 2016 at 15:55
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    $\begingroup$ It's a surjection onto $S^1 \setminus \{ -1 \}$. $\endgroup$
    – Crostul
    Jun 16, 2016 at 16:43

1 Answer 1


Pictorially, you can imagine $\mathbb R$ to be a tangent of the circle at point $x \in S^1$. Now for each point $y \in \mathbb R$ imagine a segment joining $y$ and $-x$; $z$ be the point of intersection of this segment and the circle. The correspondence $y \mapsto z$ is a homeomorphism between $\mathbb R$ and $S^1 - \{-x\}$.

The nice thing about this construction is that it is generalizable!

homeomorphism between real line and punctured circle

  • $\begingroup$ this is a quite old answer, but nonetheless; shouldn't it read, for any point $y\in\mathbb R$ instead of $y\in S^1$? $\endgroup$
    – Chaos
    Mar 11, 2020 at 9:50
  • $\begingroup$ @RScrlli Yes. Fixed it. $\endgroup$
    – hrkrshnn
    Mar 12, 2020 at 9:45

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