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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?

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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.

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