Showing Completeness of a Space 
*

*Given $s\geq1$, $X$ is nonempty set and $d: X\times X \rightarrow [0,\infty)$ is $b_s$-pseudo metric on $X$ if for $x,y,z\in X$
i. $d(x,x)=0$
ii. $d(x,y)=d(y,x)$
iii. $d(x,y)\leq s[d(x,z)+d(z,y)]$


*$F=\{d_n:n\in\mathbb{N}\}$ is a family of $b_s$-pseudo metric.


*i. a sequence $\{x_n\}$ in $X$ is convergent-$F$ to $x\in X$ if for all positive $\varepsilon$ there exist $N_0\in\mathbb{N}$ such that $$d_k(x_n,x)<\varepsilon$$
for all $n\geq N_0$ and $d_k\in F$.
ii. a sequence $\{x_n\}$ in $X$ is Cauchy-$F$ if for all positive $\varepsilon$ there exist $N_0\in\mathbb{N}$ such that $$d_k(x_n,x_m)<\varepsilon$$
for all $n,m\geq N_0$ and $d_k\in F$.
iii. $X$ is complete-$F$ if each of its Cauchy-$F$ sequence is convergent-$F$.
I have $X =C([0,\infty),\mathbb{R})$ i.e. a set of all continuous function from $[0,\infty)$ to $\mathbb{R}$ and I also have $F=\{d_k:k\in\mathbb{N}\}$ where $$d_k(x,y)=\max_{t\in[0,k]}(x(t)-y(t))^2$$
for all $x,y\in X$
How can I show that $X$ is complete-$F$? I can't find $N_0$ such that its Cauchy-$F$ sequence is convergent-$F$.
 A: $d_k$ are indeed $b_s$-pseudo metrics on $X$. $(i),(ii)$ being obvious while for $(iii)$ we have
\begin{align}
d_k(x,y)&=\max_{t\in [0,k]}(x(t)-y(t))^2\\
&=\max_{t\in [0,k]}(x(t)-z(t)+z(t)-y(t))^2\\
&\leq 2\max_{t\in [0,k]}(x(t)-z(t))^2+2\max_{t\in [0,k]}(z(t)-y(t))^2\\
&=2d_k(x,z)+2d_k(z,y)
\end{align}
where in the third step we used the inequality $2\alpha \beta \leq \alpha^2+\beta^2$. Hence, $d_k$ are indeed $\beta_s$-pseudo metrics with $s=2$. Now, suppose we have an $F-$Cauchy sequence $(x_n)$ on $X$. Then, for $\epsilon>0$ we may find $N$ such that
$$\tag{1} d_k(x_n,x_m)=\max_{t\in [0,k]}(x_n(t)-x_m(t))^2<\epsilon$$
for every $n,m\geq N$. Fix some $t\geq 0$ and find $k\geq t$. Then, by $(1)$ we have
$$\tag{2}|x_n(t)-x_m(t)|\leq \sqrt{\epsilon}$$
for every $n,m\geq N$. This shows that $(x_n(t))$ is Cauchy sequence on $\mathbb{R}$. Hence, it converges to some limit denoted by $x(t)$. I.e. $\lim_{n\to \infty}x_n(t)=x(t)$ for every $t\geq 0$. Letting $n\to \infty$ in $(2)$ and squaring we obtain
$$|x(t)-x_m(t)|^2\leq \epsilon$$
for all $m\geq N$ and every $t\in [0,k]$. Taking $\sup$ to the above inequality we obtain
$$\tag{3} \sup_{t\in [0,k]}(x(t)-x_m(t))^2\leq \epsilon$$
for every $m\geq N$ and $k$. Now using $(3)$ we  can show that $f$ is continuous. Let $t_0\geq 0$ and $\epsilon>0$. By $(3)$ we may find $m$ such that
$$\tag{4} \sup_{t\in [0,k]}(x(t)-x_m(t))^2<\epsilon^2$$
for every $k$. By continuity of $x_m$ at $t_0$ there is and $\delta>0$ such that $|x_m(t)-x_m(t_0)|<\epsilon$ for every $|t-t_0|<\delta$. Pick $k>t_0+\delta$. Then, using $(4)$ and the continuity of $x_m$, for every $|t-t_0|<\delta$ we have
\begin{align}
|x(t)-x(t_0)|\leq |x(t)-x_m(t)|+|x_m(t)-x_m(t_0)|+|x_m(t_0)-x(t)|\leq 3\epsilon
\end{align}
Hence, $x$ is continuous. Now, returning to $(3)$ by the continuity of $x$ we can substitute $\sup$ with $\max$. Hence, rewriting $(3)$ we've shown that for $\epsilon>0$ there is an $N$ such that $d_k(x,x_m)\leq \epsilon$ for every $k$ and $m\geq N$. This shows that $x_m$ is convergent-$F$ to $x$.
