(1) $X= (0,1]$ and $Y = [0,1]$;

(2) $X = [0,1]$ and $Y$ is the topologist's sine curve

(3) $X = [0,1] \cup [2,3]$ and $Y = X \times X$

I believe there is a continuous function for the first one.

I know there doesn't exist a continuous function for the second one, because $X$ is path connected and the topologist's sine curve is not. As well as, $f(0)$ doesn't exist in $Y$.

I feel like there is no continuous function for (3) because of the disconnection; however, I could be wrong.

  • 3
    $\begingroup$ Count connected components. $\endgroup$ – Daniel Fischer Apr 25 '14 at 18:44
  • $\begingroup$ What topology do the sets have? A function $\mathrm{f} : X \to Y$ is called continuous if the pre-image of each open set in $Y$ is an open set in $X$. To know if this is true, or not, you need to know which sets are open. You need to state the topologies that you are using on each choice of $X$ and $Y$. For example, if $X$ is given the discreet topology then every function is continuous. $\endgroup$ – Fly by Night Apr 25 '14 at 19:00
  • $\begingroup$ Well, you can take $f(x)=|2x-1|$ (or some similar piecewise linear function) to get a surjective continuous map from $(0,1]$ onto $[0,1]$. $\endgroup$ – Martin Sleziak Apr 25 '14 at 20:38

These problems all use variations on the idea of connectivity. Let me give you some hints.

  1. $Y$ has the property that if you remove two certain points (namely $0$ and $1$), it is still connected. Does $X$ has this property?

  2. Are $X$ and $Y$ path-connected?

  3. How many connected components does $X$ have? What about $Y$?

  • $\begingroup$ (2) I know that X is path-connected, but Y is not. $\endgroup$ – user114634 Apr 26 '14 at 18:32
  • $\begingroup$ That's right. So if there were a surjective continuous function $X\to Y$ we could take any path in $X$ and map it over to get a path in $Y$. Do you see a contradiction? $\endgroup$ – user134824 Apr 26 '14 at 18:36
  • $\begingroup$ There's no f(0) in Y. $\endgroup$ – user114634 Apr 26 '14 at 18:39
  • $\begingroup$ (3) X obviously have 2 connect components, and I believe that Y has 3 connected components. So there can't be a continuous function. $\endgroup$ – user114634 Apr 26 '14 at 19:30

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.