In the book Topology by Munkres, there is an example of a bijective continuous function that is not a homeomorphism.

I don't think that I completely understood the explanation. I understand that the function is a bijection and continuous.

Firstly, what is the topology specified on $[0, 1)$ so that $[0, \frac{1}{4})$ is open? And why does $f(0) = (1, 0)$ not lie in any open set $V$ such that $V \cap S^{1} \subset f(U)$? Is it similar to the reason that every open set containing $(6, 7]$ in standard topology on $\mathbb R$ contains some point greater than $7$ also?

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    $\begingroup$ The topology on $[0,1)$ is the induced topology from $\mathbb{R}$. A subset $U$ of $[0,1)$ is open if and only if there is an open subset $\mathcal{O}$ of $\mathbb{R}$ such that $U=\mathcal{O}\cap[0,1)$. Yes: any open subset of $S^1$ that contains $(1,0)$ will necessarily include points corresponding to small negative angles. (Alternatively: you can disconnect $[0,1)$ by removing a single point, but if you remove a single point from $[0,1)$, the result is never disconnected, so the two cannot be homeomorphic). $\endgroup$ – Arturo Magidin Feb 4 at 19:21
  • $\begingroup$ @ArturoMagidin Thanks for the explanation and the alternative approach. Now I get it $\endgroup$ – ab123 Feb 4 at 19:23
  • $\begingroup$ (that should be "if you remove a single point from $\mathbf{S}^1$, the result is never disconnected"...) $\endgroup$ – Arturo Magidin Feb 4 at 19:30
  • $\begingroup$ @ArturoMagidin oh yeah, I got the gist of it $\endgroup$ – ab123 Feb 4 at 19:31

The function $\varphi\colon\mathbb{R}\to S^1$ defined by $\varphi(t)=\bigl(\cos(2\pi t),\sin(2\pi t)\bigr)$ is continuous and surjective.

Therefore its restriction $f$ to the interval $[0,1)$ is continuous as well. It is also injective and surjective.

It cannot be a homeomorphism, because $C=S^1\setminus\{(-1,0)\}$ is connected, being equal to $\varphi\bigl((-1/2,1/2)\bigr)$, but $f^{-1}(C)=[0,1)\setminus\{1/2\}$ is not connected.

The proof in the book uses a different method. Here $U=[0,1/4)$ is open in $[0,1)$, because it is equal to $(-1/4,1/4)\cap[0,1/4)$.

However, $f\bigl([0,1/4)\bigr)$ is not open, because any open disk around $(1,0)$ contains points of $S^1$ that are not in $f\bigl([0,1/4)\bigr)$.

  • $\begingroup$ +1 @egreg sir.. $\endgroup$ – jasmine Aug 15 at 13:12

Here is a simple example.
The identity map from the reals given the discrete topology, to the reals with the usual topology generated by open intervals.


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