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An extreme example: if a set X is given the discrete topology (in which every subset is open), all functions : to any topological space T are continuous. On the other hand, if X is equipped with the indiscrete topology (in which the only open subsets are the empty set and X) and the space T set is at least T 0, then the only continuous ...
Given two metric spaces (X, d X) and (Y, d Y), where d X denotes the metric on the set X and d Y is the metric on set Y, a function f : X → Y is called Lipschitz continuous if there exists a real constant K ≥ 0 such that, for all x 1 and x 2 in X,
A continuous function fails to be absolutely continuous if it fails to be uniformly continuous, which can happen if the domain of the function is not compact – examples are tan(x) over [0, π/2), x 2 over the entire real line, and sin(1/x) over (0, 1]. But a continuous function f can
The converse does not hold, since the function :, is, as seen above, not uniformly continuous, but it is continuous and thus Cauchy continuous. In general, for functions defined on unbounded spaces like R {\displaystyle R} , uniform continuity is a rather strong condition.
The function = {, < is continuous, but not differentiable at x = 0, so it is of class C 0, but not of class C 1. Example: finitely-times differentiable ( C k ) [ edit ]
A continuous function () on the closed interval [,] showing the absolute max (red) and the absolute min (blue).. In calculus, the extreme value theorem states that if a real-valued function is continuous on the closed and bounded interval [,], then must attain a maximum and a minimum, each at least once.
is continuous at every irrational number, so its points of continuity are dense within the real numbers. Proof of continuity at irrational arguments Since f {\displaystyle f} is periodic with period 1 {\displaystyle 1} and 0 ∈ Q , {\displaystyle 0\in \mathbb {Q} ,} it suffices to check all irrational points in I = ( 0 , 1 ) . {\displaystyle I ...
Then f (−1) = f (1), but there is no c between −1 and 1 for which the f ′(c) is zero. This is because that function, although continuous, is not differentiable at x = 0. The derivative of f changes its sign at x = 0, but without attaining the value 0.