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:<math>f(x) = \frac{x+1}{2},</math>
which is a continuous function from the open interval (−1,1) to itself. Since x = 1 is not part of the interval, there is not a fixed point of f(x) = x. The space (−1,1) is convex and bounded, but not closed. On the other hand, the function ''f'' {{em|does}} have a fixed point for the closed interval [−1,1], namely ''f''(1) = 1.
(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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