Let $K$ be a compact Hausdorff space and let $C(K)$ be the space of all scalar-valued continuous functions on $K$. Let $(f\_{n})\_{n}$ be a sequence in $C(K)$ satisfying $\sup\limits\_{n}\sup\limits\_{t\in K}|f\_{n}(t)|<\infty$. We define an equivalence relation $R$ on $K$ by $$t\_{1}Rt\_{2}\Leftrightarrow f\_{n}(t\_{1})=f\_{n}(t\_{2}), \forall n.$$
Let $K\_{1}:=K/R$ be the quotient space with the quotient topology $\tau$. That is, $\tau=\{V\subseteq K\_{1}:Q^{-1}(V)$ is open in $K\}$, where $Q:K\rightarrow K\_{1}$ is the quotient mapping. It is easy to see that $(K\_{1},\tau)$ is compact. My question is the following:


Question. Is $(K\_{1},\tau)$ metrizable?



Metrization of quotient spaces defined by sequences of continuous functions
Yes, it is metrizable. Assume the scalar field is real; if it is complex, replace the functions $f\_n$ with their real and imaginary parts. WLOG each $f\_n$ maps $K$ into $[0,1]$. Amalgamate the $f\_n$ into a single function $f: K \to [0,1]^{\omega}$. Then $K\_1$ is homeomorphic to $f(K)$ with topology induced from $[0,1]^{\omega}$. Metrizability of $K\_1$ now follows from metrizability of the cube.


