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Brouwer fixed-point theorem: Difference between revisions

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→Closedness: Fixed some language issues, and removed a paragraph that was wrong and misleading. (The problem with the example is indeed the closedness of the ___domain - the theorem never uses that the *range* of the function is equal to the ___domain, only that the *codomain* is. Indeed, if it did it would be a much weaker theorem. It is true that the *range* of f is (0,1), but we can easily extend the *codomain* to (-1,1).)
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Consider the function
:<math>f(x) = \frac{x+1}{2},</math>
which is a continuous function from the open interval <math>(−1-1,1)</math> to itself. Since xthe =point <math>x=1</math> is not part of the interval, there is notno apoint fixedin pointthe of___domain such that <math>f(x) = x</math>. The spaceset <math>(−1-1,1)</math> is convex and bounded, but not closed. On the other hand, the function ''<math>f''</math> {{em|does}} have a fixed point forin the ''closed'' interval <math>[−1-1,1]</math>, namely ''f''<math>x=1</math>. The closed interval <math>[-1,1]</math> is compact, the open interval <math>(-1,1)</math> =is 1not.
 
(The ___domain of f is (-1,1) but the range is (0,1), which are not the same. Earlier, one of the conditions for functions satisfying the theorem is that the ___domain and range were the same, not that one be a subset of the other. Thus the reason for f failing is not closure.)
 
===Convexity===
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