Brouwer fixed-point theorem: Difference between revisions

Revision as of 03:25, 26 July 2022

Revision as of 08:39, 1 September 2022

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Convexity is not strictly necessary for BFPT. Because the properties involved (continuity, being a fixed point) are invariant under [[homeomorphism]]s, BFPT is equivalent to forms in which the ___domain is required to be a closed unit ball <math>D^n</math>. For the same reason it holds for every set that is homeomorphic to a closed ball (and therefore also [[closed set|closed]], bounded, [[connected space|connected]], [[simply connected|without holes]], etc.).

The following example shows that BFPT doesn'tdoes not work for domains with holes. Consider the function <math>f(x)=-x</math>,

which is a continuous function from the unit circle to itself. Since ''-x≠x'' holds for any point of the unit circle, ''f'' has no fixed point. The analogous example works for the ''n''-dimensional sphere (or any symmetric ___domain that does not contain the origin). The unit circle is closed and bounded, but it has a hole (and so it is not convex) . The function ''f'' {{em|does}} have a fixed point for the unit disc, since it takes the origin to itself.