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Continuity PropertyIn mathematics, the continuity property may be presented as follows. THEOREM: Suppose that f : b -> R is continuous on the closed bounded interval b &sube R . Then the image f(b) is also a closed bounded interval. The theorem acts as a stronger form of the intermediate value theorem, containing the three assertions:
(i) The image set is an interval. This is the intermediate value theorem.
(ii) This interval is bounded, so that f is bounded.
(iii) This bounded interval is closed, so that f attains both its bounds. Proof (i) This is dealt with elsewhere (ii) We proceed by contradiction Suppose is unbounded on any interval a',b', and continuous Then f is unbounded on a',c' or b', for a'< c' < b' As if not f cannot be bounded on a',b' This allows us to find an interval where f is unbounded, for some y, such that delta is arbitrarily small. For this, take a,b split into and , then into — and so on. However, this can be shown to contradict the continuity of f, using the definition of continuity for the real line. Let be a closed interval of width We now define recursively to be a closed interval of width This can always be found by the above argument Now by another theorem the intersection is non-empty. So define to be a point in Now this point is in each and any other point in is at most from So letting So f is unbounded on However 2/n can be made arbitrarily small so for all we have a such that, there exists an This is more than enough to disprove continuity, that is show that there exists an x_0 such that for some epsilon, for all delta there exists a point x in the open interval or width delta around x_0, such that the absolute value of x_0 - x is greater than epsilon, which we have done. (In fact we have shown that this is the case for all epsilon) This gives us a contradiction So f a,b must be bounded (iii) This comes from the least upper bound property of the real line By the least upper bound axiom, we know that there is a minimum M such that for Let Now A will also have least upper bound m, this must be in a,b, as and a,b contains all its limit points. This point will then be the maximum as Further as we can find a member of A, b such that f(b) is arbitrarily close to M otherwise M is not minimal, so if f(m)< M,M cannot be an upper bound. Thus Similarly considering and noting that ,we see f obtains its minimum Thus f obtains its minimum and maximum at at least one point so, by the intermediate value theorem, it obtains all values in between Caveats It is important to note that this theorem only applies to continuous real functions. It does not apply to the rationals, as these do not satisfy the least upper bound axiom; they are not complete. To illustrate this consider f would obtain its maximum value at but this is not in the set. If f is not continuous consider as a counterexample This is unbounded, but 0,1 is bounded. Further one should reiterate that the set must be closed, otherwise the maximum and minimum values maybe not be obtained. Someone might like to add some thoughts about more general functions from any complete set.
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