The Completeness of A Space

Question

Before I ask my question, let me list some definitions that's used in the question.

1. 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. $d(x,y)\leq s[d(x,z)+d(z,y)]$

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

3. i. a sequence $\{x_n\}$ in $X$ is $F$-convergent to $x\in X$ if for all $d_k \in F$ and positive $\varepsilon$ there exist $N\in\mathbb{N}$ such that $$d_k(x_n,x)<\varepsilon$$ for all $n\geq N_0$.

   ii. a sequence $\{x_n\}$ in $X$ is $F$-Cauchy if for all $d_k \in F$ and positive $\varepsilon$ there exist $N\in\mathbb{N}$ such that $$d_k(x_n,x_m)<\varepsilon$$ for all $n,m\geq N_0$

   iii. $X$ is $F$-complete if each of its $F$-Cauchy sequence is $F$-convergent.

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Now let's say I have:

1. $X =C([0,\infty),\mathbb{R})$ i.e. a set of all continuous function from $[0,\infty)$ to $\mathbb{R}$, AND
2. $F=\{d_k:k\in\mathbb{N}\}$ where $$d_k(x,y)=\max_{t\in[0,k]}(x(t)-y(t))^2$$ for all $x,y\in X$ Every element of $F$ is $b_2$-pseudo metric on $X$.

Here, I want to show that $X$ is $F$-complete. This is the answer I get from my previous question here (Showing Completeness of a Space). But unfortunately, I wrote the wrong definition for $F$-convergent and $F$-Cauchy sequence there. So I tried to work on it again.

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Suppose we have an arbitrary $F$-Cauchy sequence $\{x_n\}$ on $X$. Then, for any $k\in \mathbb{N}$ and $\epsilon>0$ 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 $n,m\geq N$.

By $(1)$, we have for every $t\in [0,\infty)$, $\{x_n(t)\}$ is a Cauchy-real number sequence. Hence, it's convergent to some point $y_t\in\mathbb{R}$, i.e. $y_t = \lim_{n\rightarrow\infty} x_n(t)$. The limit of this real number sequence is unique, so I can define a function $x : [0,\infty)\rightarrow\mathbb{R}$ where $$x(t) = \lim_{n\rightarrow\infty} x_n(t)$$ for all $t\in[0,\infty)$.

We should show that $x(t)$ is continuous and finally show that $\{x_n\}$ is $F$-converges to $x$. This is how I show that $x$ is continuous.

For any $k\in \mathbb{N}$ and $\varepsilon>0$, we can pick any $t \in [0,k]$. Then, we have $N(t)\in\mathbb{N}$ such that $$|x_n(t) - x(t)|<\varepsilon$$ for all $n\in\mathbb{N}$ where $n\geq N(t)$

Now, we must choose a fix $n_*\in\mathbb{N}$ (I chose $n_*=\max_{t\in[0,k]}N(t)$) such that the inequality below holds for every $n\in\mathbb{N}$ where $n\geq n_*$ and for all $t\in[0,k]$. $$\tag{2} |x_n(t)-x(t)|<\varepsilon$$ From $(2)$ we got $x_n$ is uniformly convergent to $x$ on $[0,k]$ for all $k\in\mathbb{N}$. Because $x_n$ is continuous on $[0,k]$ for all $n\in\mathbb{N}$, thus $x$ is continuous on $[0,k]$. Hence, $x$ is continuous on $[0,\infty)$.

I looked again at the $n_*$ I've chosen and soon realized that $[0,k]$ has an infinite number of elements and $[0,k]$ is bounded, but the function $N(t)$ is not necessarily continuous. So I can't choose $n_*$ as maximum value of $N(t)$ a.k.a the way I chose $n_*$ is invalid.

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Now, here's my question. Is there a way to find a fix $n_*$ for $(2)$?

Answer 1

On the right side of (1) change $\epsilon$ to $\epsilon^{2}$. You will then have $|x_n(t)-x_m(t) |<\epsilon$ for all $n, m \geq N$ for all $t \in [0,k]$. Let $m \to \infty$. You get $|x_n(t)-x(t) |\leq \epsilon$ for all $n \geq N$ for all $t \in [0,k]$. Hence you can take $n_{*}$ to be $N$.