1
$\begingroup$

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?


enter image description here

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $f\bigl([0,1/4)\bigr)$.

$\endgroup$
  • $\begingroup$ +1 @egreg sir.. $\endgroup$ – jasmine Aug 15 at 13:12
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.