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I’m currently working on the following problem:

Let $X$ denote the set of nondecreasing functions $f:[0,1]\to\mathbb{R}$. We endow $X$ with the sup metric. Prove that $X$ is complete.

I notice that if we are working with continuous functions, then we can use the nice property of $C([0,1])$ that it is complete under sup metric and prove that any closed subset is a complete subspace. However, since continuity is not specified, I wonder what nice properties for general functions from $[0,1]$ to $\mathbb{R}$ that we can use. Thank you!

Here is the question from its original source: https://ww3.math.ucla.edu/wp-content/uploads/2021/09/basic-19F.pdf (Q11).

Michael Wang-Wakamatsu
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The space of bounded functions with the sub-norm is complete (see Space of bounded functions is complete). Thus, all you need to prove is that your subspace is closed.

Pick $(f_n)_n$ a converging sequence (in the sup-norm) of monotonically increasing functions functions. This means, for all $n\in \mathbb{N}$ and all $0\leq x\leq y \leq 1$ we have $f_n(x) \leq f_n(y)$. However, if $f(x)=\lim_{n\rightarrow \infty} f_n(x)$ and $f(y)=\lim_{n\rightarrow \infty} f_n(y)$, then we also get $f(x)\leq f(y)$ as pointwise convergence preserves $\leq$ (see Suppose that $(s_n)$ converges to $s$, $(t_n)$ converges to $t$, and $s_n \leq t_n \: \forall \: n$. Prove that $s \leq t$.).

Severin Schraven
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  • Thanks for the answer. My professor edited the question to be functions from [0,1] to R instead of [0,1] to [0,1] as in the original version. Is there anything we can say about the unbounded real valued functions on [0,1] in general? – Michael Wang-Wakamatsu Feb 24 '24 at 06:03
  • @MichaelWang-Wakamatsu It does not matter. If you look at the first link, you'll see that you can pick any complete metric space (in particular $\mathbb{R}$ with the standard metric is fine) for the target of your functions. – Severin Schraven Feb 24 '24 at 08:58