Is there a continuous surjection from the closed unit square $[0,1]\times[0,1]$ to $\mathbb R ^2$?

If yes, please give examples. I'm a little stuck on this. What if I replace the closed unit square with the open one?

  • 2
    $\begingroup$ For the second part, do you know an answer for an open interval and the real line? $\endgroup$ – Mark Bennet Sep 9 '14 at 17:00
  • $\begingroup$ Please make the body of your question self-contained. The title is important, but should be separate from the body. $\endgroup$ – Asaf Karagila Sep 9 '14 at 17:02

No. The continuous images of unit square are compact so they are bounded in plane.

  • $\begingroup$ If they replace the closed unit square with the open one, it is possible however (using some form of tangent function). $\endgroup$ – icurays1 Sep 9 '14 at 17:00
  • $\begingroup$ I get the argument for the closed square. But what about the open case? $\endgroup$ – adrija Sep 9 '14 at 17:05
  • $\begingroup$ @user173185 It is a basic fact in topology that a continuous image of a compact set is compact. The unit square is compact, the plane is not, therefore the plane is never the continuous image of the square. $\endgroup$ – Slade Sep 9 '14 at 19:19
  • $\begingroup$ @you-sir the closed unit square is compact. The open unit square is in fact homeomorphic to the plane (hint: $\tan$). $\endgroup$ – Najib Idrissi Sep 9 '14 at 19:55

For the open case: take $f(x,y)=(\ln(-\ln x),\ln(-\ln y))$.


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