1
$\begingroup$

I was recently taught that the Peano curve is an example of a continuous bijection from the closed unit interval to the closed unit square. However, if we take a point in the square and take it's preimage in the interval and delete both points, the interval gets disconnected while the square remains connected. How can such a function be continuous? As continuous maps preserve connectedness, there should be no continuous bijection between the unit interval and the unit square, yet the Peano curve is supposedly such an example. Where am I going wrong?

$\endgroup$
  • 2
    $\begingroup$ A space-filling curve cannot be a bijection. $\endgroup$ – J. J. Sep 11 '14 at 9:06
  • $\begingroup$ ..with that argument u just convinced yourself that it is not a bijection. $\endgroup$ – FWE Sep 11 '14 at 9:17
1
$\begingroup$

Peano curve is a surjection, not bijection from an interval onto its cartesian square.

| cite | improve this answer | |
$\endgroup$

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.