收敛序列是 $\ell^\infty$ 的闭子集

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lecture_blog page4
Our other example concerned the sequence space $\ell^\infty$. It contains the subset of convergent sequences, and we showed that this subset is also closed.

类似于（连续实值函数空间）$\mathcal C[a,b]$是（有界实值函数空间）$\mathcal B[a,b]$中的闭子集（具有 sup 范数）

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Convergent sequences are closed subspace of $l^{\infty}$
Example 46 Let $c$ denote the subset of $l^{\infty}$ consisting of all convergent sequences. Then $c$ is a closed subset of $l^{\infty}$.

Solution Say that $\left(x_n^{(k)}\right)$ is a sequence in $c$ which converges to $\left(X_n\right)$ in $l^{\infty}$ as $k \rightarrow \infty$. We aim to show that $\left(X_n\right)$ is in $c$; we will do this by showing $\left(X_n\right)$ is Cauchy and so convergent.
Let $\varepsilon>0$. As $\left(x_n^{(k)}\right) \rightarrow\left(X_n\right)$ in $l^{\infty}$ then
$$
\left\|\left(x_n^{(k)}\right)-\left(X_n\right)\right\|_{\infty}=\sup _n\left|x_n^{(k)}-X_n\right| \rightarrow 0 \text { as } k \rightarrow \infty
$$
So there exists $K$ such that
$$
\left|x_n^{(k)}-X_n\right|<\varepsilon / 3 \text { for } k \geqslant K \text { and all } n
$$
As $\left(x_n^{(K)}\right)$ is convergent then it is Cauchy. So there exists $N$ such that
$$
\left|x_n^{(K)}-x_m^{(K)}\right|<\varepsilon / 3 \quad \text { for } n, m \geqslant N .
$$
Thus for $m, n \geqslant N$ we have
$$
\left|X_m-X_n\right| \leqslant\left|X_m-x_m^{(K)}\right|+\left|x_m^{(K)}-x_n^{(K)}\right|+\left|x_n^{(K)}-X_n\right|<\varepsilon / 3+\varepsilon / 3+\varepsilon / 3=\varepsilon
$$

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这已经证明收敛序列在 $\ell^\infty$ 下的所有 Cauchy 列收敛于自身. 因此不论把收敛序列扔到哪个更大的空间中, 其本身都是完备的.

Metric Spaces: Completeness
A closed subset of a complete metric space is itself complete, when considered as a subspace using the same metric, and conversely. Note that this means, for example, that a closed interval in is a complete metric space. Theorem 5.3: Let be a complete metric space, and let be a subset of.

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