Compactness of Special Metric on Maps from Countable Space to Compact Metric Space Let $X$ be a countable set and $(Y, d) $ be a compact metric space. I'm trying to define a metric $b $ on $Y ^ X $ so that it characterizes the point-wise convergence and that $(Y ^ X, b)$ is a compact space.
I already knew that $\delta_t: (f,g)\mapsto |f(t)-g(t)|$ ($\forall t\in X$) are a group of quasi-metrics that characterizes the point-wise convergence. So I've tried $b: (f,g) \mapsto \sum_{x\in X}|f(x)-g(x)|$ but it seems that it actually characterizes the uniform convergence.
I also noticed that I can use the sequential compact property of the space $Y$, and also use this property to prove the compactness of the space $Y^X$. But although I managed to form a map $f$ from a sequence of maps $(f_n)$, I failed to prove that $f$ is a limit point, since I cannot specify a specific subsequence and I have to say it should undertake countable processes of taking subsequences.
 A: 
给定$X$上一个序列$X=\{x_1,x_2,\cdots\}$. $\forall f,g\in Y^X$, 定义度量$b$:
$$(f,g)\mapsto b(f,g):=\sum_{i\geqslant 1}{1\over
 2^i}d(f(x_i),g(x_i)),$$        首先, 它是良定义的, 因为在$Y$是紧度量空间的情形,
连续函数$d$在其上有最大值$M$, 由此级数可和. 下面验证$b$是一个度量:
(1) 由$d$的对称性, $b(f,g)=b(g,f)$;
(2) 由$d$的规范性, $b(f,f)=0$;
(3) 由$d$的三角不等式, $b(f,h)=\sum_{x\in X}{1\over
 2^i}d(f(x),h(x))\leqslant \sum_{x\in X}{1\over
 2^i}\Big[d(f(x),g(x))+d(g(x),h(x))\Big]$ $=b(f,g)+b(g,h) $;
(4) 由$d$的分离性, $b(f,g)=\sum_{x\in X}{1\over 2^i}d(f(x),g(x))=0$
$\Rightarrow$ $d(f(x),g(x))=0$ ($\forall x\in X$) $\Rightarrow$
$f(x)=g(x)$ ($\forall x\in X$) $\Rightarrow$ $f=g$.
其次, 度量$b$刻画了点态收敛. 事实上, 由P1L1, $\{\delta_x: (f,g)\mapsto d(f(x),g(x));x\in X\}$伪度量族刻画了点态收敛, 只需证明按$b$收敛与按诸$\delta_x$收敛等价. 一方面,
当映射序列$(f_n)_{n\geqslant 1}$按$b$收敛于$f$时, $\forall\varepsilon>0$,
$\exists N$, $\forall n>N$, $b(f_n,f)<\varepsilon$, 从而$\forall
 i\geqslant 1$, $d(f_n(x_i),f(x_i))<2^i\varepsilon$,
亦即序列$(f_n)_{n\geqslant 1}$按诸$\delta_x$也收敛于$f$. 另一方面,
当序列$(f_n)_{n\geqslant 1}$按诸$\delta_x$收敛于$f$时, $\forall\varepsilon>0$,
$\exists N_1$, $$    \sum_{i\geqslant N_1}{1\over
 2^i}d(f_n(x_i),f(x_i))\leqslant\sum_{i\geqslant N_1}{M\over
 2^i}={M\over 2^{N_1}}<{\varepsilon\over 2};   $$ $\forall i<N_1$,
$\exists N_{0,i}$, $\forall n>N_{0,i}$, $$    {1\over
 2^i}d(f_n(x_i),f(x_i))<{\varepsilon\over 2N_1};   $$         从而,
令$N_0:=\max_{i<N_1}\{N_{0,i}\}$, $\forall n>N_0$, $$
    b(f_n,f)=\left(\sum_{i<N_1}+\sum_{i\geqslant N_1}\right){1\over
 2^i}d(f_n(x_i),f(x_i))<\varepsilon,   $$       亦即序列$(f_n)_{n\geqslant
 1}$按$b$也收敛于$f$. 从而得证$b$刻画了点态收敛.
于是, $(Y^X,b)$形成度量空间, 下面验证它是紧的. 任取其中序列$(f_n)_{n\geqslant 1}$,
则对给定的$i\in\mathbb{Z}$, $\{f_n(x_i)\}_{n\geqslant 1}$ ($\forall i$)
均是$Y$中的序列, 由$Y$的紧性, 它们均有子列收敛. 先取$\{f_n(x_1)\}_{n\geqslant
 1}$的收敛子列$\{f_{n_{1,k_1}}(x_1)\}_{k_1\geqslant 1}$, 极限设为$y_1\in Y$,
并设$f\in Y^X$: $f(x_1)=y_1$. 再取$\{f_{n_{1,k_1}}(x_2)\}_{k_1\geqslant
 1}$的收敛子列$\{f_{n_{2,k_2}}(x_2)\}_{k_2\geqslant 1}$, 极限设为$y_2\in Y$,
并设$f(x_2)=y_2$. 如此进行下去, 便得到了$f\in Y^X$.
$$    \begin{matrix}
     f_{n_{11}}(x_1) 
      & f_{n_{21}}(x_2)
      & f_{n_{31}}(x_3)
      & \cdots\\
     f_{n_{12}}(x_1) 
      & f_{n_{22}}(x_2)
      & f_{n_{32}}(x_3)
      & \cdots\\
    f_{n_{13}}(x_1) 
      & f_{n_{23}}(x_2)
      & f_{n_{33}}(x_3)
      & \cdots\\
     \vdots&\vdots&\vdots&\ddots\\
     f(x_1)&f(x_2)&f(x_3)&\cdots    \end{matrix}   $$        其中$(f_{n_{ii}})_{i\geqslant 1}$是$(f_n)_{n\geqslant 1}$的子列;
对于给定的$j\in\mathbb{Z}$, $\{f_{n_{ii}}(x_j)\}_{i\geqslant
 j}$也是$\{f_{n_{j,k_j}}(x_j)\}_{k_j\geqslant 1}$的子列, 收敛于$f(x_j)$;
因而$(f_{n_{ii}})_{i\geqslant 1}$点态收敛于$f$, 由上面的刻画, 也按$b$收敛于$f$, 得证.

