$\ell^\infty$空间中的全有界性与紧性相关问题求解

Alright, since you've already wrapped up part (a) showing $S$ is closed, let's walk through part (b) using the given hint—this is a classic totally boundedness argument for coordinate-decaying sets in $\ell^\infty$.

Step 1: Define Key Sets and Pick $N_\epsilon$ #

First, let's formalize the hint's components for any arbitrary $\epsilon > 0$:

• Let $B_\epsilon = \left(\frac{\epsilon}{2}\mathbb{Z}\right) \cap [-1,1]$. This set is finite: we're taking integer multiples of $\frac{\epsilon}{2}$ that fit within $[-1,1]$, so the total number of elements is at most $\left\lfloor \frac{2}{\epsilon} \right\rfloor - \left\lceil -\frac{2}{\epsilon} \right\rceil + 1$, which is clearly a finite number.
• Choose $N_\epsilon$ as any integer greater than $\frac{2}{\epsilon}$ (for example, $N_\epsilon = \left\lfloor \frac{2}{\epsilon} \right\rfloor + 1$). For all $n > N_\epsilon$, this guarantees $\frac{1}{n} < \frac{\epsilon}{2}$—a crucial detail because every $x = (x_n) \in S$ satisfies $|x_n| \leq \frac{1}{n}$.

Step 2: Build the Finite $\epsilon$-Net $S_\epsilon$ #

Now define the set:
$$S_\epsilon = \left{ (y_n) : y_n \in B_\epsilon \text{ for } n \leq N_\epsilon, , y_n = 0 \text{ for } n > N_\epsilon, , |y_n| \leq \frac{1}{n} \text{ for all } n \right}$$
(This is exactly the hint's $S_\epsilon = {(y_n): y_n\in B_{\epsilon} \text{ 且 } y_n=0 \text{ } \forall n>N_{\epsilon}}\cap S$, just with the intersection with $S$ spelled out explicitly.)

$S_\epsilon$ is finite: For each $n \leq N_\epsilon$, we choose $y_n$ from a finite subset of $B_\epsilon$ (only elements that satisfy $|y_n| \leq \frac{1}{n}$). With finitely many positions ($N_\epsilon$ total) and finite choices per position, the total number of elements in $S_\epsilon$ is finite.

Step 3: Verify $S_\epsilon$ Covers $S$ Within $\epsilon$ #

Take any arbitrary $x = (x_n) \in S$. We need to find a $y = (y_n) \in S_\epsilon$ such that $|x - y|_\infty \leq \epsilon$:

• For $n \leq N_\epsilon$: Since $x_n \in \left[-\frac{1}{n}, \frac{1}{n}\right] \subseteq [-1,1]$, there exists a $y_n \in B_\epsilon \cap \left[-\frac{1}{n}, \frac{1}{n}\right]$ such that $|x_n - y_n| \leq \frac{\epsilon}{2}$. This works because $B_\epsilon$ is a grid spaced by $\frac{\epsilon}{2}$, so every point in $[-1,1]$ is within $\frac{\epsilon}{2}$ of some grid point.
• For $n > N_\epsilon$: Set $y_n = 0$. Then $|x_n - y_n| = |x_n| \leq \frac{1}{n} < \frac{\epsilon}{2}$ by our choice of $N_\epsilon$.

Now calculate the infinite norm:
$$|x - y|\infty = \max\left( \max{n \leq N_\epsilon} |x_n - y_n|, \max_{n > N_\epsilon} |x_n - y_n| \right) \leq \max\left( \frac{\epsilon}{2}, \frac{\epsilon}{2} \right) = \epsilon$$

This means every element of $S$ is within $\epsilon$ of some element in the finite set $S_\epsilon$, so $S$ is totally bounded.

内容的提问来源于stack exchange，提问作者ʎpoqou