# Closedness of the Set of Continuous and Increasing functions

Let $$A=\left\{ f\in C\left[ 0,1\right] |\text{ }f\text{ is strictly increasing and }f\left( 0\right) =0\text{ and }f\left( 1\right) =1\right\}$$, where $$C\left[ 0,1\right]$$ is the set of continuous functions over $$\left[ 0,1\right]$$. Is $$A$$ closed with respect to sup metric (or, some other nontrivial metric)?

• What did you try? What is a nontrivial metric? – José Carlos Santos Apr 17 at 13:24

No, it isn't. Think of a continuous (non-strictly) increasing function $$f$$ with $$f(0) = 0$$ and $$f(1) = 1$$ that has a "plateau" around $$x = 1/2$$. That means, $$f(x) = 1/2$$ for all $$x$$ in a neighborhood of $$x=1/2$$. It oculd even by piecewise linear. You can now easily think of a sequence of functions in $$A$$ that converge uniformly to $$f$$, can you?