Example of a set that is closed and bounded but not compact

hbghlyj

A2-metricspaces.pdf page 55 bottom

To show that the converse is not true in general, consider the following example. Take $C_{b}(\mathbf{R})$ to be the normed space of continuous bounded functions on the real line equipped as usual with the $\|\cdot\|_{\infty}$-norm and the associated metric.

Define a function $\phi: \mathbf{R} \rightarrow \mathbf{R}$ by $$ \phi(t)=\left\{\begin{array}{cc} 2 t+1, & -1 / 2 \leqslant t \leqslant 0 \\ 1-2 t, & 0 \leqslant t \leqslant 1 / 2 \end{array}\right. $$ and $\phi(t)=0$ for $t \notin[-1 / 2,1 / 2]$. For each $n \in \mathbf{N}$ set$$f_{n}(t)=\phi(t-n)$$(we might call this sequence of functions a "moving bump"). All of the functions $f_{n}$ lie in $\bar{B}(0,1)$ (that is, have sup norm bounded by 1). However, if $n \neq m$ then $f_{n}(n)=1$, whilst $f_{m}(n)=0$, so $\left\|f_{n}-f_{m}\right\|_{\infty}=1$. Thus the sequence $\left(f_{n}\right)_{n=1}^{\infty}$ has no Cauchy subsequence, and hence certainly no convergent subsequence.

Another Example (from math.stackexchange.com):

If we consider $X=\left\{\frac1n:n\in\Bbb N^+\right\}$ in the discrete topology, then we can endow it with the metric $d:X\times X\to\Bbb R$ given by $$d(x,y)=\begin{cases}0 & x=y\\1 & \text{otherwise,}\end{cases}$$ which does indeed induce the discrete topology on $X$ (it's called the discrete metric for this reason). Then $X$ is certainly bounded, as any ball of radius greater than $1$ necessarily includes the whole set, and is certainly closed in itself (as all spaces are). However, it is not compact, since the open cover by singletons admits no finite subcover, as you've observed. More generally, any infinite discrete space admits a proper subspace that is closed and bounded, but not compact (delete any point).

We could come to the same conclusions if we considered $X$ as a space under the metric $$\rho(x,y)=|x-y|.$$ Indeed, $\rho$ induces the discrete topology on $X$, as well, and we similarly find that $X$ is bounded under $\rho$.

The kicker, here, is the boundedness. You need to specify a metric, or some other convention to determine boundedness, not just a topology. For example, $\Bbb Z$ considered as a subspace of $\Bbb R$ is indeed discrete, but while it is bounded in the discrete metric, it is not bounded in the standard metric on $\Bbb R$.

hbghlyj

For each $n ∈\mathbb N$ set $f_{n}(t)=\phi(t\color{red}+n)$

原文红色字应为 $-$ 号,我在1楼修改了.

hbghlyj

A2-metricspaces.pdf page 57 bottom:

Consider $X=[0,1]$ and the following sequence $\left(f_n\right)_{n=1}^{\infty}$ of functions in $C(X)$:

$f_n(x)=0$ if $x \notin\left(\frac{1}{n+1}, \frac{1}{n}\right),$ $f_n(t_n)=1$ where $t_{n}:=\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)$ is the midpoint of this interval, and is piecewise linear elsewhere. The $f_{n}$ are all continuous, but clearly $d(f_{m}, f_{n})=1$ whenever $m \neq n$, since $f_{m}(t_{m})=1$ whilst $f_{n}(t_{m})=0$. Thus this sequence has no convergent subsequence.

The issue here is that the functions $f_n$, whilst continuous, become “less and less continuous” as $n → ∞$; the gradient of the piecewise linear sequences tends to infinity.
Whilst the whole space $C(X)$ is not sequentially compact, interesting subsets of it may be. Roughly speaking, the two types of example we have just mentioned are the only obstruction to sequential compactness, an idea made precise by the Arzelà-Ascoli theorem.

Czhang271828

hbghlyj 发表于 2022-8-2 22:35
本帖最后由 hbghlyj 于 2022-8-2 20:39 编辑 A2-metricspaces.pdf page 57 bottom:
Consider $X=[0,1]$ an ...

One may equal the description "bounded & closed", "compact" and "sequentially compact" when studying the topology of $\mathbb R^n$. In fact, such relations become much more complicated when you study further.

It is highly recommended to verify the following statements in your commence study of functional analysis.

[1-2 are related to functional analysis.]
1. In any complete normed space (Banach space) $X$, the unit ball is compact whenever $X$ has finite dimension.
2. In metric spaces, compact is equivalent to sequentially compact, and is also equivalent to complete & totally bounded.

[3-4 is from the theory of Stone–Čech compactification, which has something to do with the dual of $\ell^\infty$. Learn it when you are familiar with ultrafilter.]
3. Sequentially completeness does not imply completeness.
4. Sequentially compactness does not imply compactness.

hbghlyj

Proposition 9.2.1. A compact metric space is sequentially compact.

Proof. Let $X$ be the space in question, and suppose that $\left(x_{n}\right)_{n=1}^{\infty}$ is a sequence of elements of $X$. We wish to find a convergent subsequence of this sequence. For each natural number $n$, set $A_{n}:=\left\{x_{n}, x_{n+1}, x_{n+2}, \ldots\right\}$. Obviously, $A_{2} \supseteq A_{3} \supseteq \cdots$, and so $\overline{A_1} \supseteq \overline{A_2} \supseteq \overline{A_3} \supseteq \cdots$. Applying Lemma 9.2.2, we see that $\bigcap_{n=1}^{\infty} \overline{A_n}$ is nonempty.

Let $a$ be a point in this intersection. We inductively construct a subsequence $\left(x_{n_{k}}\right)_{k=1}^{\infty}$ such that $d\left(x_{n_{k}}, a\right)<1 / k$ for all $k$; it is then clear that this subsequence converges (to $a$) and the proof will be complete. Suppose that $n_{1}, \ldots, n_{k}$ have already been constructed. Now $a$ lies in $\overline{A_{n_k+1}}$, that is to say the closure of the set $\left\{x_{n_{k}+1}, x_{n_{k}+2}, \ldots\right\}$. In particular, there is some element of this sequence at distance less than $1 /(k+1)$ from $a$, and we can take this to be our $x_{n_{k+1}}$.$\square$

hbghlyj

Set that is bounded but not totally bounded: Reading textbook
Bounded but Not Totally Bounded
Bounded Metric Space is not necessarily Totally Bounded/Proof 1

hbghlyj

Consider $[1,\sqrt2]∩ℚ=\{q∈ℚ|q≥1;q^2≤2\}$. In the metric space, $ℚ$, this set is closed for the same reason $[1,\sqrt2]$ is closed in $ℝ$. And it is bounded. But it is not compact for the same reason $[1,\sqrt2)$ is not compact in $ℝ$.

hbghlyj

Czhang271828 发表于 2022-8-3 09:15
...
2. In metric spaces, compact is equivalent to sequentially compact, and is also equivalent to complete & totally bounded.

$\newcommand{\set}[2]{\left\{ #1 : #2 \right\}}\newcommand{\seq}[2]{(#1_{#2})_{#2\in\N}}$ma30252
Theorem 4.17. If a metric space is complete and totally bounded, then it is compact.
Proof. Suppose that \((X,d)\) is complete and totally bounded. Assume, by way of contradiction, that it is not compact. Then there exists an open cover \(\set{U_\lambda}{\lambda \in \Lambda}\) that does not have a finite subcover. Define \(r_n = 1/2^n\) for \(n \in ℕ\text{.}\) Since \(X\) is totally bounded, there exists a finite set \(E_1 \subseteq X\) such that \begin{equation*} X = \bigcup_{x \in E_1} B_{r_1}(x). \end{equation*} Since \(E_1\) is finite, there exists at least one point \(x_1 \in E_1\) such that no finite subcollection of \(\set{U_\lambda}{\lambda \in \Lambda}\) covers \(B_{r_1}(x_1)\text{.}\) (If we had a finite subcollection covering \(B_{r_1}(x)\) for every \(x \in E_1\text{,}\) then the union of all of them would give a finite subcover of \(X\text{.}\)) By Lemma 4.9, the set \(B_{r_1}(x_1)\) is totally bounded, too. Hence there exists a finite set \(E_2 \subseteq B_{r_1}(x_1)\) such that \begin{equation*} B_{r_1}(x_1) \subseteq\bigcup_{x \in E_2} B_{r_2}(x). \end{equation*} Furthermore, there exists at least one point \(x_2 \in E_2\) such that no finite subcollection of \(\set{U_\lambda}{\lambda \in \Lambda}\) covers \(B_{r_2}(x_2)\text{.}\) We can continue this construction with \(r_3, r_4, \dotsc\text{,}\) and we obtain a sequence \((x_n)_{n \in ℕ}\) in \(X\) such that \(x_{n + 1} \in B_{r_n}(x_n)\) and \(B_{r_n}(x_n)\) is not covered by any finite subcollection of \(\set{U_\lambda}{\lambda \in \Lambda}\) for every \(n \in ℕ\text{.}\) We claim that \((x_n)_{n \in N}\) is a Cauchy sequence. In order to verify this, let \(m, n \in ℕ\) with \(n \gt m\text{.}\) Then \begin{align*} d(x_m, x_n) &\le d(x_m, x_{m + 1}) + \cdots + d(x_{n - 1}, x_n) \\ &\le \sum_{k = m}^\infty r_k = \sum_{k = m}^\infty \frac{1}{2^k} = \frac{1}{2^{m - 1}}. \end{align*} Thus \(d(x_m, x_n) \to 0\) as \(m, n \to \infty\text{,}\) which means that we have a Cauchy sequence. As \(X\) is complete, it follows that there exists a limit \(x_0 = \lim_{n \to \infty} x_n\text{.}\) Since \(\set{U_\lambda}{\lambda \in \Lambda}\) is a cover of \(X\text{,}\) there exists \(\lambda_0 \in \Lambda\) such that \(x_0 \in U_{\lambda_0}\text{.}\) Since \(U_{\lambda_0}\) is open, there exists \(s \gt 0\) such that \(B_s(x_0) \subseteq U_{\lambda_0}\text{.}\) Now choose \(N \in ℕ\) so large that \(d(x_N, x_0) \lt s/2\) and \(r_N \lt s/2\text{.}\) Then \(B_{r_N}(x_N) \subseteq B_s(x_0) \subseteq U_{\lambda_0}\text{.}\) But that means that \(B_{r_N}(x_N)\) is covered by a single element of \(\set{U_\lambda}{\lambda \in \Lambda}\text{,}\) contradicting the previous statement that no finite subcollection will cover it. This concludes the proof.
Proving a complete and totally bounded metric space is compact.