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本帖最后由 Czhang271828 于 2022-8-7 11:26 编辑 Hint: The finite product topology space of sequentially compact metric spaces is also sequentially compact.
(Formal definition in topology) $(X,\tau)$ is sequentially compact if and only if every $\{x_n\}_{n\geq 1}\subset X$ has a convergent subsequence with limit in $X$. We say $x_n\to x_0$ in $(X,\tau)$ whenever each neighbourhood of $x_0$ contains $\{x_n\}_{n\geq N}$ for some $N\in \mathbb N$ ($N$ depends on the choice of neighbourhoods).
When $(X,\tau)$ is a metric space, we can also use $\varepsilon$ - $\delta$ language to restate the convergence.
Proof of the hint. Let $\{X_i\}_{i=1}^d$ be a family of seq compact metric space. For each $\{x_n=(x_n^1,\ldots, x_n^d)\}_{n\geq 1}\subset \prod_{i=1}^d X_i$, we shall prove that $\{x_n\}$ has an accumulation point in the product space.
There exists a subsequence $\{x_{n_k}\}$ s.t. $\{x_{n_k}^1\}$ is convergent in $X_1$. Similarly, one can also find a subsub...subsequence whose projection on each $X_i$ is convergent in $X_i$. Therefore, let $x^i_n\to x^i_0$ for simplicity ($\forall i=1,\ldots, n$).
We claim that $x_0:=(x_0^1,\ldots,x_0^n)$ is a limit point of $\{x_n\}$. By definition of finite product topology, each open neighbourhood of $x_0$ contains some $\prod_{i=1}^d U_i$ ($U_i$ is an open neighbourhood of $x_0^i$ in $X_i$). There exists $\{N_i\}_{i=1}^d\subset \mathbb N$ s.t. $\{x^i_n\}_{n\geq N_i}\subset U_i$. As a result, $\{x_n\}_{n\geq \max_i N_i}\subset \prod_{i=1}^d U_i$.
One can also prove the compactness by Heine-Borel thm.
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Solution.
$\forall \{x_n\}\subset B(0;R)$. If $\{x_n\}\subset B(0;R)$ has subsequence converging to $0$, then $0$ is a accumulation point of $\{x_n\}$. Otherwise, $\inf_{n\geq 1} \|x_n\|:=\delta >0$.
$\forall x\in B(0;R)\setminus B(0;\delta)$, denote $x=(\omega ,r)$, where $\omega \in S$ and $r\in [\delta,R]$. We define the metric $d'(x_m,x_n):=\sqrt{d_{\mathrm{arc}}(\omega_m-\omega_n)^2+(r_m-r_n)^2}$, where $d_{\mathrm{arc}}$ is the arc length on $S$.
One can verify:
(i) $d'$ is a well defined metric.
(ii) $\exists C_1,C_2>0$ (only depending on $\delta$ and $R$) s.t. $C_1\|\omega_p-\omega_q\|\leq d_{\mathrm{arc}}(\omega_p,\omega_q)\leq C_2\|\omega_p-\omega_q\|$. Therefore, $d_{\mathrm{arc}}$ is comparable to the metric induced by norm $\|\cdot \|$ on $S$.
(iii) $d'$ is comparable to the metric induced by norm $\|\cdot\|$ on $B(0;R)\setminus B(0;\delta)$.
Since $\{r_n\}_{n\geq 1}$ has a accumulation point on $[\delta,R]$, there exists a subsequence $\{x_{n_k}\}_{k\geq 1}$ s.t. $\lim_{k\to\infty}r_{n_k}$ exists. Since $S_0$ is also seq compact, there exists a subsubsequence $\{x_{n_{k_s}}\}_{s\geq 1}$ s.t. $\lim_{s\to\infty}\omega_{n_{k_s}}$ exists.
By (i) (iii) (iii), $x_{n_{k_s}}$ converges in norm $\|\cdot \|$ due to the convergence of $\omega_{n_{k_s}}$ and $r_{n_{k_s}}$. As a consequence, $B$ is seq compact. |
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Czhang271828
发表于 2022-8-6 19:09
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