Is the set of uniformly bounded non-decreasing functions a compact set with the metric $(,)=\sup|−|$? 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?
 A: 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.
A: As written, no, since the set is unbounded.
