Problem: Every differentiable function f: (0, 1) --> [0, 1] is uniformly continuous.
Discussion;
False
Note that every differentiable function f: [0,1] --> (0, 1) is uniformly continuous by virtue of uniform continuity theorem which says every continuous map from closed bounded interval to R is uniformly continuous. However in this case the domain is an open interval.
We can easily find counter example such as $latex f(x) = \sin ( \frac {1}{x} ) $. Intuitively speaking the function oscillates (between -1 and 1) faster and faster as we get close to x = 0. Hence we can get two arbitrarily close values of x such that their functional value's difference equals a particular number (say 1) therefore exceeding any $latex \epsilon < 1 $
An interesting discussion:
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