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I am currently having some doubts about some proofs that $C_{[a, b]}|$ with metric $\rho(x(t), y(t)) = \max\limits_{a \le t \le b} |x(t) - y(t)|$ is complete.

All the proofs I have seen were divided into three parts.

Given a Cauchy sequence $\{x_n\} \subset C_{[a, b]}$.

  1. Prove that $\{x_n(t_0)\} \subset \mathbb{R}$ is Cauchy for all $t_0 \in [a, b]$. And then define $x: x(t) = \lim_{n\to\infty}x_n(t)$
  2. Prove that $x_n \to x$ when $n \to \infty$.
  3. Prove that $x$ is continuous.

I am confused about the second part. Some proofs are so simple, yet some is so complex that I wonder if there is anything wrong about the simple proof.

  • First proof:

Since $\{x_n\}$ is Cauchy, $\forall \varepsilon > 0, \exists N_0 \in \mathbb{N}$ such that $\forall n, m > N_0$ we have $\rho(x_n, x_m) < \frac{\varepsilon}{2}$.

That means $\forall t_0 \in [a, b]: |x_n(t_0) - x_m(t_0)| \le \rho(x_n, x_m) < \frac{\varepsilon}{2}$

And so we have $|x_n(t_0) - x(t_0)| = |x_n(t_0) - \lim\limits_{m\to\infty} x_m(t_0)| = \lim\limits_{m\to\infty} |x_n(t_0) - x_m(t_0)| \le \frac{\varepsilon}{2}$

Because that is true for all $t_0 \in [a, b]$, we then have: $\rho(x_n, x) = \max\limits_{a \le t \le b} |x_n(t_0) - x(t_0)| \le \frac{\varepsilon}{2} < \varepsilon$

Thus, $x_n \to x$ when $n \to \infty$.

  • Second proof

For all $t_0 \in [a, b]$, since $\{x_n(t_0)\} \subset \mathbb{R}$ converges to $x(t_0)$, $\forall \varepsilon > 0, \exists N_{t_0} \in \mathbb{N}$ such that $\forall n > N_{t_0}$ we have $|x_n(t_0) - x(t_0)| < \varepsilon$.

Consider $N = \max\limits_{a \le t_0 \le b} N_{t_0}$, for all $n > N$ we have $|x_n(t) - x(t)| < \varepsilon, \forall t \in [a, b] \Rightarrow \rho(x_n, x) = \max\limits_{a\le t \le b} |x_n(t) - x(t)| < \varepsilon$

And so we conclude that $x_n \to x$ when $n \to \infty$.

  • Third proof

This one is from this answer.

Since $\{x_n\}$ is Cauchy, $\forall \varepsilon > 0, \exists N \in \mathbb{N}$ such that $\forall n, m \ge N$ we have $\rho(x_n, x_m) < \frac{\varepsilon}{3}$.

Now we have that $\rho(x_n, x) \le \rho(x_n, x_N) + \rho(x_N, x) < \frac{\varepsilon}{3} + \rho(x_N, x)$.

For any $t_0 \in [a, b]$ and any $m \ge N$, $|x_N(t_0) - x(t_0)| \le |x_N(t_0) - x_m(t_0)| + |x_m(t_0) - x(t_0)|$ $(1)$.

We have that $|x_N(t_0) - x_m(t_0)| \le \rho(x_N, x_m) < \frac{\varepsilon}{3}$.

Since $x(t_0) = \lim_{m\to\infty}x_m(t_0)$, and $(1)$ holds for any $m \ge N$, we can find large enough $m$ such that $|x_m(t_0) - x(t_0)| < \frac{\varepsilon}{3}$

And so, $|x_N(t_0) - x(t_0)| \le |x_N(t_0) - x_m(t_0)| + |x_m(t_0) - x(t_0)| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} < \frac{2\varepsilon}{3}$. This holds for all $t_0 \in [a, b]$, thus, $\rho(x_N, x) = \max\limits_{a\le t \le b} |x_N(t) - x(t)| < \frac{2\varepsilon}{3}$.

Now we have that, $\rho(x_n, x) < \frac{\varepsilon}{3} + \rho(x_N, x) < \frac{\varepsilon}{3} + \frac{2\varepsilon}{3} = \varepsilon$.

We can conclude that $x_n \to x$ when $n \to \infty$.

I feel like all of them is correct. But then if that so, why bother going with such a long approach like the third proof. The second one is so straight-forward that it feels kinda deceptive. Hope that someone can explain which proof is correct. I appreciate your help a lot.

sleeper161
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1 Answers1

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The first proof is fine. The third one is correct but uselessly complicated. The second one is flawed because "$N = \max\limits_{a \le t_0 \le b} N_{t_0}$" might not exist ($\{N_t\mid a\le t\le b\}$ might even be unbounded).

Anne Bauval
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  • It seems against intuition for me that $N = \max_{a \le t_0 \le b} N_{t_0}$ might not exist even though the individual $N_{t_0}$ exists and is finite. Are there any counter-examples for this? – sleeper161 Dec 25 '24 at 23:23
  • For each $t$ there is a smallest $N_t$ which does the job, let us call it $SN_t$. You can then easily construct a counterexample: define any unbounded map $[a,b]\to\Bbb N,t\mapsto N_t$ such that $\forall t, N_t\ge SN_t$. On the other hand, the map $t\mapsto SN_t$ itself is bounded (hence takes only finitely many values, hence has a max): this is what the first and third method prove. – Anne Bauval Dec 26 '24 at 00:02
  • So if I change the proof to $N = \max_{a \le t \le b} SN_t$ would that make the second proof correct? – sleeper161 Dec 26 '24 at 00:16
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    It would need the first or third proof as a lemma (see my previous comment), and would then become redundant. – Anne Bauval Dec 26 '24 at 04:50