Is the set of uniformly bounded non-decreasing functions a compact set with the metric $(,)=\sup|−|$?

Question

Fix $M>0$. Let $\Phi = \{f|f:[a, b] \to [-M, M] \, \text{is an non-decreasing function} \}$. Define a metric $d: \Phi \times \Phi \to [0, \infty)$ by $(,)=\sup_{x \in [a, b]}|(x)−(x)|$. Is the topology induced by compact?

Answer 1

As written, no, since the set is unbounded.

Answer 2

The answer is NO. Counter-example: Consider the case that $M=10$, $a=0$, $b=1$. Recall that for a metric space, the metric topology is compact iff it is sequentially compact (i.e., Every sequence has a convergent subsequence). Prove by contradiction. Suppose the contrary that $(\Phi,d)$ is compact. For each $n\in\mathbb{N}$, define $f_{n}:[0,1]\rightarrow[-10,10]$ by $f_{n}(x)=x^{n}$. Clearly $f_{n}$ is increasing, so $f_{n}\in\Phi$. Consider the sequence $(f_{n})$. By the compactness assumption, there exists $f\in\Phi$ and a subsequence $(f_{n_{k}})$ such that $d(f_{n_{k}},f)\rightarrow0$ as $k\rightarrow\infty$. In particular, for each $x\in[0,1]$, $|f_{n_{k}}(x)-f(x)|\rightarrow0$ as $k\rightarrow\infty$. It follows that $$ f(x)=\begin{cases} 0, & \mbox{ if }x\in[0,1)\\ 1, & \mbox{ if }x=1 \end{cases}. $$ Let $\varepsilon=\frac{1}{100}$. Then there exists $K\in\mathbb{K}$ such that $d(f_{n_{k}},f)<\varepsilon$ whenenver $k\geq K$. Hence, for any $x\in[0,1)$, we have \begin{eqnarray*} & & |x^{n_{K}}|\\ & = & |f_{n_{K}}(x)-f(x)|\\ & \leq & d(f_{n_{K}},f)\\ & < & \frac{1}{100}. \end{eqnarray*} Letting $x\rightarrow1-$ and observing that $x^{n_{K}}\rightarrow1$, we have $1<\frac{1}{100}$, which is a contradiction.