Section 4.3 Compactness
Definition 4.13.
Let \(S\) be a subset of a metric space \((X,d)\text{.}\) A collection \(\set{U_\lambda}{\lambda \in \Lambda}\) of open sets in \(X\) is called an open cover of \(S\) if \(S \subseteq \bigcup_{\lambda \in \Lambda} U_\lambda\text{.}\) If \(M \subseteq \Lambda\) and \(S \subseteq \bigcup_{\mu \in M} U_\mu\text{,}\) then \(\set{U_\mu}{\mu \in
M}\) is called a subcover of \(\set{U_\lambda}{\lambda
\in \Lambda}\text{.}\)
Warning 4.14.
Note that the term open cover refers to the collection of open sets \(\set{U_\lambda}{\lambda\in\Lambda}\) and not to their union \(\bigcup_{\lambda \in \Lambda} U_\lambda\text{,}\) which is a subset of \(X\text{.}\)
Remark 4.15.
The index set \(\Lambda\) has no other purpose than to label the sets \(U_\lambda\text{,}\) and we can replace it with any other set of the same size. In particular, for a finite open cover, we can always write \(\{U_1, \dotsc, U_N\}\) for some \(N \in \N \cup \{0\}\text{.}\)
Definition 4.16.
A subset \(K\) of a metric space \((X,d)\) is called compact if every open cover of \(K\) has a finite subcover. If \(K=X\) is compact, we say that the metric space \((X,d)\) is compact.
Example 4.17.
Consider \(S=(0,1)\) as a subset of \(X=\R\) with the Euclidean metric. Then for any \(n \in \N\) the subset \(U_n = (1/n,1) \subset X\) is open. Since \(S \subseteq \bigcup_{n \in \N}
U_n\text{,}\) \(\{U_n : n \in \N\}\) is an open cover of \(X\text{.}\) It has many infinite subcovers, for instance \(\{U_{2k} : k \in \N\}\text{.}\) But one can check that there are no finite subcovers, and hence that \(S\) is not compact subset.
Example 4.18.
Let \(X\) be an arbitrary set equipped with the discrete metric.
Suppose first that \(X\) is finite. Let \(\set{U_\lambda}{\lambda \in \Lambda}\) be an open cover of \(X\text{.}\) Then for every \(x \in X\text{,}\) there exists \(\lambda_x \in \Lambda\) with \(x \in U_{\lambda_x}\text{.}\) Then \(\set{U_{\lambda_x}}{x \in X}\) is a finite subcover. Thus \(X\) is compact.
Now suppose that \(X\) is infinite. Then we consider the open cover \(\set{U_x}{x \in X}\) of \(X\) with \(U_x = \{x\}\) for every \(x \in X\text{.}\) Clearly, any finite subcollection \(\{U_{x_1,} \dotsc, U_{x_N}\}\) satisfies
\begin{equation*}
\bigcup_{n=1}^N U_{x_n} = \{x_1, \dotsc, x_n\} \ne X.
\end{equation*}
So \(X\) is not compact in this case.
Lemma 4.19. Compactness and metric subspaces.
A subset \(K\) of a metric space \((X,d)\) is compact if and only if the associated metric subspace \((K,d')\) is compact.
Proof.
Theorem 4.20.
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 \subseteq \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 4.11, 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)
\amp\le d(x_m, x_{m + 1}) + \cdots + d(x_{n - 1}, x_n) \\
\amp\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.
Theorem 4.21.
If a subset of a metric space is compact, then it is sequentially compact.
Proof.
Let \(K\) be a compact subset of a metric space \((X,d)\text{.}\) Assume for contradiction that there exists a sequence \((x_n)_{n \in \N}\) in \(K\) that has no subsequence converging to a limit in \(K\text{.}\) In other words, every \(y \in K\) is not the limit of any subsequence of \((x_n)_{n \in \N}\text{.}\) According to Exercise 1.3.6, this means that for every \(y \in K\) there exists \(r_y \gt 0\) with
\begin{equation*}
B_{r_y}(y) \cap \set{x_n}{n \in \N} \subseteq\{y\}.
\end{equation*}
Now the collection \(\set{B_{r_y}(y)}{y \in K}\) is an open cover of \(K\text{.}\) Since \(K\) is compact, there exist \(y_1, \dotsc, y_I \in X\) with
\begin{equation*}
K \subseteq \bigcup_{i = 1}^I B_{r_{y_i}}(y_i).
\end{equation*}
Hence
\begin{equation*}
\set{x_n}{n \in \N} \subseteq\bigcup_{i = 1}^I B_{r_{y_i}}(y_i) \cap \set{x_n}{n \in \N} \subseteq\{y_1, \dotsc, y_I\}.
\end{equation*}
Therefore, we actually have a sequence in the finite subspace \(\{y_1,
\dotsc, y_I\}\) of \(K\text{.}\) But clearly every finite metric space is sequentially compact, and we conclude that \((x_n)_{n \in \N}\) does have a subsequence converging to a limit in \(K\text{,}\) in contradiction to the assumption.
We now summarise some of the results on compactness.
Theorem 4.22.
Let \((X,d)\) be a metric space and let \(K \subseteq X\text{.}\) Then the following are equivalent.
- \(K\) is compact.
- \(K\) is totally bounded and the metric subspace \((K,d')\) is complete.
- \(K\) is sequentially compact.
Proof.
Theorem 4.21 states that (i) implies (iii). It follows from Lemma 4.3, Theorem 4.4 and Theorem 4.10 that (iii) implies (ii). Finally, Theorem 4.20 states that (ii) implies (i), where here we also use Lemma 4.19.
Remark 4.23.
It may seem silly to have different names for concepts that are equivalent. However, while they are equivalent for metric spaces, the two notions of compactness can also be defined in different contexts, and then they are no longer equivalent.
Since compactness and sequential compactness are equivalent, we can also restate some of the results from the previous sections.
Corollary 4.24.
Every compact subset of a metric space is bounded.
Proof.
Combine Theorem 4.22 with Lemma 4.9.
Corollary 4.25.
Suppose that \((X,d)\) is a compact metric space and \(Y \subset X\text{.}\) If \(Y \subset X\) is closed, then it is compact.
Proof.
Combine Theorem 4.22 with Theorem 4.6.
Exercises Exercises
1. (PS9) Direct proof of non-compactness.
Let \(S\) be the unit ball \(B_1(0)\) as a subset of the metric space \(\R^2\) with the usual Euclidean metric. Directly prove that \(S\) is not compact. That is, find an open cover \(\set{U_\lambda}{\lambda \in \Lambda}\) of \(S\) which has no finite subcovers.
Hint.
If you’re feeling stuck, examples from the lectures, lecture notes and problems classes may be useful inspiration.
Solution.
Consider the open cover \(\set{B_r(0)}{r \in (0,1)}\text{.}\) Clearly this is a collection of open sets, and it is a cover since any \(x \in B_1(0)\) also lies in \(B_r(0)\) for all \(r \in (\n x,1)\text{.}\) Let \(\{B_{r_1}(0),\ldots,B_{r_N}(0)\}\) be a finite subcollection, and set \(R = \max\{r_1,\ldots,r_N\} \lt 1\text{.}\) Then
\begin{equation*}
\bigcup_{n=1}^N B_{r_n}(0) = B_R(0) \not \supseteq B_1(0),
\end{equation*}
and so this cannot be a subcover.
Comment 1.
There are of course many different open covers that one can use for this problem. For instance, many students this year used something like \(\{B_{1-1/n}(0) : n \in \N \without \{1\}\}\text{,}\) which is a (countably infinite) subcover of the cover in the official solutions
Comment 2.
On exams I am inclined to accept without proof statements about sets which are both correct and relatively straightforward. For instance, in the context of this problem (i.e. in \(\R^2\) with the Euclidean metric) I would accept \(\bigcup_{r \in (0,1)} B_r(0) = B_1(0)\) and \(B_{1/2}(0) \not \supseteq B_1(0)\) without proof. On the other hand, I would not accept a statement like ‘clearly there are no finite subcovers’ without some further justification.
2. Compactness and metric subspaces.
Let \(Y\) be a subset of a metric space \((X,d)\text{.}\) Show that \(Y\) is compact if and only the metric subspace \((Y,d')\) is compact.
Hint.
Use Exercise 1.2.2.
Solution.
Suppose that \(Y \subseteq X\) is compact, and let \(\set{U_\lambda}{\lambda \in \Lambda}\) be a collection of open subsets of \(X\) with \(Y \subseteq \bigcup_{\lambda \in \Lambda} U_\lambda\text{.}\) By Exercise 1.2.2, the sets \(V_\lambda = U_\lambda \cap Y\) are open in \((Y,d_Y)\text{,}\) and intersecting the above inclusion with \(Y\) yields \(Y = \bigcup_{\lambda \in \Lambda} V_\lambda\text{.}\) Since \((Y,d_Y)\) is compact, we can therefore find a finite subset \(M \subseteq \Lambda\) such that
\begin{equation*}
Y = \bigcup_{\mu \in M} V_\mu \subseteq \bigcup_{\mu \in M} U_\mu
\end{equation*}
as desired.
Conversely, suppose that for every collection \(\set{U_\lambda}{\lambda \in \Lambda}\) of open subsets of \(X\) with \(Y \subseteq \bigcup_{\lambda \in \Lambda} U_\lambda\text{,}\) there exists a finite subset \(M \subseteq \Lambda\) such that \(Y \subseteq \bigcup_{\mu \in M} U_\mu\text{.}\) To see that \((Y,d_Y)\) is compact, let \(\set{V_\lambda}{\lambda \in \Lambda}\) be a collection of open sets in \(Y\) with \(Y = \bigcup_{\lambda \in \Lambda} V_\lambda\text{.}\) By Exercise 1.2.2, we can find open sets \(U_\lambda\) in \(X\) such that \(V_\lambda = U_\lambda \cap Y\text{.}\) In particular, \(U_\lambda \subseteq V_\lambda\text{,}\) and so we have \(Y
\subseteq \bigcup_{\lambda \in \Lambda} U_\lambda\text{.}\) By assumption, we can then find a finite subset \(M \subseteq \Lambda\) such that \(Y \subseteq \bigcup_{\mu \in M} U_\mu\text{.}\) Intersecting both sides with \(Y\) we conclude that \(Y = \bigcup_{\mu \in M} V_\mu\) as desired.
3. (PS9) Continuity on compact spaces.
Let \((X, d_X)\) and \((Y, d_Y)\) be a metric spaces and \(f \maps X \to Y\) a continuous map. Show that \(f\) is uniformly continuous if \(X\) is compact.
Hint 1.
This is relatively big hint. The official solution uses Definition 4.16 directly. The general idea is to construct an open cover in the following way. Fix \(\varepsilon \gt 0\text{.}\) Since \(f\) is continuous, for all \(x \in X\) there exists a \(\delta(x) \gt 0\) such that \(d_X(x,x') \lt \delta(x)\) implies that \(d_Y(f(x),f(x')) \lt \varepsilon\text{.}\) Then the set of all balls of the form \(B_{\delta(x)}(x)\) forms an open cover of \(X\text{,}\) and so using compactness we can find a finite subcover.
This is only the general idea, though, and you will likely find that you need to do things like fiddle around with \(\varepsilon\) versus \(\varepsilon/2\text{,}\) and also with the radii of the balls in the open cover.
Hint 2.
Alternatively, thanks to Theorem 4.22, we can argue instead using sequential continuity. For this, you will want to write down what it would mean for \(f\) not to be uniformly continuous, and then use that to construct sequences with a certain property. Sequential compactness will let you find convergent subsequences, which will then lead to a contradiction with the (pointwise) continuity of \(f\text{.}\)
Solution 1. Using covers and subcovers
Let \(\varepsilon \gt 0\text{.}\) Since \(f\) is continuous, for every point \(x \in X\) we have a number \(\delta(x) \gt 0\) such that, for all \(x' \in X\text{,}\) the inequality \(d_X(x, x') \lt \delta(x)\) implies \(d_Y(f(x), f(x')) \lt
\frac{\varepsilon}{2}\text{.}\)
The collection of balls \(\set{B_{\delta(x)/2}(x)}{x \in X}\) is an open cover of \(X\text{.}\) If \(X\) is compact, then there exist \(x_1, \dotsc, x_N\) such that
\begin{equation*}
X = \bigcup_{n=1}^N B_{\delta(x_n)/2}(x_n).
\end{equation*}
Set \(\delta = \frac 12 \min\{\delta(x_1), \dotsc, \delta(x_N)\}\text{.}\)
Now suppose that we have two points \(a, b \in X\) with \(d_X(a,b) \lt \delta\text{.}\) Then there exists \(n \in \{1, \dotsc, N\}\) such that \(a \in B_{\delta(x_n)/2}(x_n)\text{.}\) Thus \(d_X(a, x_n) \lt \delta(x_n)\) and also \(d_X(b, x_n) \le d_X(a,b) + d_X(a, x_n) \lt \delta(x_n)\text{.}\) Hence by the choice of \(\delta(x_n)\text{,}\)
\begin{equation*}
d_Y(f(x_n), f(a)) \lt \frac{\varepsilon}{2} \quad \text{and} \quad d_Y(f(x_n), f(b)) \lt \frac{\varepsilon}{2}.
\end{equation*}
The triangle inequality then implies that
\begin{equation*}
d_Y(f(a), f(b)) \le d_Y(f(x_n), f(a)) + d_Y(f(x_n), f(b)) \lt \varepsilon.
\end{equation*}
Solution 2. Using sequential compactness
Suppose for the sake of contradiction that \(f\) is not uniformly continuous. Then there exists \(\varepsilon \gt 0\) such that, for all \(n \in \N\text{,}\) there exists \(x_n,y_n \in X\) with \(d_X(x_n,y_n) \lt 1/n\) but \(d_Y(f(x_n),f(y_n)) \ge \varepsilon\text{.}\)
Since \(X\) is compact, Theorem 4.22 implies it is sequentially compact, and so the sequence \(\seq xn\) has a subsequence \(\subseq xnk\) converging to a limit \(x_0 \in X\text{.}\) Unfortunately, there is no guarantee that the associated subsequence \(\subseq ynk\) of \(\seq yn\) converges. But we can apply sequential compactness to \(\subseq ynk\) to find a further subsequence \(\subsubseq ynk\ell\) which does converge, say to some \(y_0 \in X\text{.}\) We note (by Theorem 1.34) that \(\subsubseq xnk\ell\) still converges to \(x_0\text{.}\)
By the triangle inequality
\begin{gather*}
d_X(x_0,y_0) \le d_X(x_0,x_{n_{k_\ell}}) + d_X(x_{n_{k_\ell}},y_{n_{k_\ell}})
+ d_X(y_{n_{k_\ell}},y_0) \to 0
\end{gather*}
as \(\ell \to \infty\text{,}\) where the first and last terms on the left hand side converge to zero since \(x_{n_{k_\ell}} \to x_0\) and \(y_{n_{k_\ell}} \to y_0\) while the middle term is bounded by \(1/n_{k_\ell}\) which also converges to zero as \(\ell \to \infty\text{.}\) Thus \(d_X(x_0,y_0) = 0\) and hence \(x_0 =
y_0\text{.}\)
By the sequential continuity of \(f\) and the sequential continuity of \(d_Y \maps
Y \times Y \to \R\) from Exercise 1.3.1, we have
\begin{gather*}
\lim_{\ell \to \infty} d_Y\big(f(x_{n_{k_\ell}}),f(y_{n_{k_\ell}})\big) = d_Y(f(x_0),f(y_0)) = 0
\end{gather*}
as \(x_0 = y_0\text{.}\) On the other hand by our choice of \(\seq xn\) and \(\seq yn\) we have \(d_Y\big(f(x_{n_{k_\ell}}),f(y_{n_{k_\ell}})\big) \ge \varepsilon\) for all \(\ell\text{,}\) and so the above limit must also be \(\ge \varepsilon\text{,}\) which is a contradiction.
Comment.
Note that in the official solution just having \(a\) and \(b\) lie in the same ball \(B_{\delta(x_n)/2}(x_n)\) is not enough to immediately estimate \(d_Y(f(a),f(b))\text{.}\) This is because \(\delta(x_n)\) is defined using the continuity of \(f\) at \(x_n\) and not \(a\) or \(b\text{.}\) However, we can get what we need with a bit of extra work using the triangle inequality.
4. (PS9) Images of compact sets.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces and suppose \(f\maps
X \to Y\) is continuous.
(a)
Suppose \((X,d_X)\) is sequentially compact. Prove using the definition of sequential compactness that the image \(f(X)\) is sequentially compact.
Hint.
Given a sequence in \(f(X)\text{,}\) relate it to a sequence in \(X\text{,}\) and then use sequential compactness.
Solution.
Suppose \((X,d_X)\) is sequentially compact. We want to show that \(f(X)\) is sequentially compact (as a subset of the metric space \((Y,d_Y)\)). So let \(\seq yn\) be a sequence in \(f(X)\text{.}\) Then for each \(n \in \N\) there exists \(x_n \in X\) with \(f(x_n) = y_n\text{,}\) giving rise to a sequence \(\seq xn\) in \(X\text{.}\) As \((X,d_X)\) is sequentially compact, we can find a subsequence \(\subseq xnk\) which converges to some limit \(x_0 \in X\text{.}\) Since \(f\) is continuous, Theorem 1.46 then implies that
\begin{equation*}
y_{n_k} = f(x_{n_k}) \to f(x_0) \in f(X)
\end{equation*}
as \(k \to \infty\text{.}\) Thus \(\subseq ynk\) is a subsequence of \(\seq
yn\) which converges (as a sequence in \(Y\)) to a limit in \(f(X)\text{.}\) We conclude that \(f(X)\) is sequentially compact.
Comment.
Some students gave arguments involving \(\varepsilon\)s, \(\delta\)s and \(N\)s, essentially repeating part of the proof of Theorem 1.46. Citing Theorem 1.46 directly saves some ink and mental energy; could also simply say “by the sequential continuity of \(f\)”.
(b)
Suppose \(U \subseteq Y\) is open in \(Y\text{.}\) Show that the preimage \(f^{-1}(U)\) is open in \(X\text{.}\)
Hint.
Look at the definition of continuity (Definition 1.42) and rewrite the relevant inequalities in terms of points being in certain balls.
Solution.
Suppose \(U \subseteq Y\) is open in \(Y\text{.}\) Let \(x_0 \in f^{-1}(U)\text{.}\) Then \(f(x_0) \in U\text{,}\) and, since \(U\) is open, there exists \(\varepsilon \gt 0\) such that \(B_\varepsilon(f(x_0)) \subseteq U\text{.}\) Since \(f\) is continuous at \(x_0\) we know that there exists \(\delta \gt 0\) such that for every \(x \in X\text{,}\) \(d_X(x,x_0)\lt\delta\) implies \(d_Y(f(x_0),f(x))\lt\varepsilon\text{.}\) Hence, for every \(x \in B_\delta(x_0)\) we have \(d_Y(f(x_0),f(x))\lt\varepsilon\text{,}\) that is \(f(x) \in B_\varepsilon(f(x_0))
\subseteq U\text{.}\) It follows that \(B_\delta(x_0) \subseteq f^{-1}(U)\text{.}\) Since \(x_0 \in f^{-1}(U)\) was arbitrary, we conclude \(f^{-1}(U)\) is open.
(c)
Suppose \((X,d_X)\) is compact. Prove using the definition of compactness that \(f(X)\) is compact.
Hint.
Part b is useful here.
Solution.
Suppose \((X,d_X)\) is compact, and let \(\set{U_{\lambda}}{\lambda\in\Lambda}\) be an open cover of the metric subspace \(f(X)\) (thought of as a subset of \((Y,d_Y)\)). Then by Part b, the collection \(\{f^{-1}(U_{\lambda}):\lambda \in \Lambda\}\) is an open cover of \(X\text{.}\) As \(X\) is compact we find a finite subcollection \(\{f^{-1}(U_{1}), \ldots
,f^{-1}(U_{N})\}\) with \(X \subseteq \bigcup_{k = 1}^Nf^{-1}(U_{k})\text{.}\) We claim \(\{ U_{1} ,\ldots, U_{N} \}\) is a finite cover of \(f(X)\text{.}\) To see this, let \(y \in f(X)\text{.}\) Then there is \(x \in X\) such that \(f(x) =
y\text{.}\) Since \(X \subseteq \bigcup_{k = 1}^Nf^{-1}(U_{k})\) we have \(x \in
f^{-1}(U_{k})\) for some \(k \in \{1,\ldots,N\}\) and hence \(y = f(x) \in
U_{k}\text{.}\) We conclude \(f(X) \subseteq \bigcup_{k = 1}^N U_{k} \text{,}\) and so \(\{ U_{1}
,\ldots, U_{N} \}\) is a finite subcollection of \(\{U_{\lambda}:\lambda \in \Lambda\}\) which covers \(f(X)\text{.}\) Hence \(f(X)\) is compact.
5. (PS8) A closed and bounded set which is not compact.
Consider the sequence \(\seq fn\) in \(C^0([0,\pi])\) defined by
\begin{equation*}
f_n(t) = \sin(2^n t) \text{,}
\end{equation*}
together with the corresponding subset
\begin{equation*}
F = \set{f_n}{n \in \N} \subset C^0([0,\pi]) \text{.}
\end{equation*}
(a)
Show that \(F\) is a bounded subset of \(C^0([0,\pi])\text{.}\)
Solution.
Since \(\abs{\sin x} \le 1\) for all \(x \in \R\text{,}\) we clearly have \(\n{f}_{C^0([0,\pi])} \le 1\) for any \(f \in F\text{.}\) Thus \(\sup_{f \in F}
\n f_{C^0([0,\pi])} \le 1\) is finite, and so \(F\) is bounded; see Definition 1.15 and Exercise 1.1.11. Alternatively, the triangle inequality gives
\begin{gather*}
\diam F = \sup_{f,g \in F} \n{f-g}_{C^0([0,\pi])} \le 1 + 1 = 2 \lt \infty\text{.}
\end{gather*}
Comment.
The set \(F\) is introduced as a subset of the metric space \(C^0([0,\pi])\text{.}\) Thus the term bounded should be interpreted as in Definition 1.15.
Other quick comments:
- It is not enough for every function in \(f \in F\) to have \(\n f_{C^0([0,\pi])} \lt \infty\text{.}\)
- On an exam it would also not be sufficient to say something like ‘this follows from some properties of \(\sin\)’.
- Remember that \(\n f_{C^0([0,\pi])} = \sup_{t \in [0,\pi]} \abs{f(t)}\) involves absolute values. It is not enough to estimate \(\sup_{t \in [0,\pi]} f(t)\text{.}\)
- The notation \(\sup F\) for subsets \(F\) of \(C^0([0,\pi])\) has not been defined, nor is there a standard definition I am aware of.
(b)
Show that
\begin{gather}
\n{f_n-f_m}_{C^0([0,\pi])}
\ge 1 \text{ if } n \ne m\text{.}\tag{✶}
\end{gather}
Hint.
To get some intuition, draw graphs for small values of \(n\) and \(m\text{.}\) Since you only need a lower bound, it suffices to find a point \(t \in [0,\pi]\) where \(\abs{f_n(t)-f_m(t)} \ge 1\text{.}\)
Solution.
Let \(n,m \in \N\) with \(n \lt m\text{,}\) and set \(t = 2^{-n-1}\pi \in
[0,\pi]\text{.}\) Then
\begin{equation*}
f_n(t) = \sin \tfrac \pi2 = 1
\end{equation*}
while
\begin{equation*}
f_m(t) = \sin 2^{m-n-1}\pi = 0
\end{equation*}
since \(2^{m-n-1} \in \Z\text{.}\) Thus
\begin{equation*}
\n{f_n-f_m}_{C^0([0,\pi])}
\ge \abs{f_n(t)-f_m(t)} = 1
\end{equation*}
as desired.
Comment.
Note that it is not the case that \(\n{f_n-f_m}_{C^0([0,\pi])} =
1\) for \(n \ne m\text{.}\) Indeed, we have
\begin{equation*}
\n{f_1-f_2}_{C^0([0,\pi])} \ge \abs{f_1(3\pi/4)-f_2(3\pi/4)} = 1 + \frac 1{\sqrt
2}\text{.}
\end{equation*}
(c)
Use (✶) to show that \(\seq fn\) has no convergent subsequences.
Hint.
Theorem 1.37 is useful.
Solution.
Suppose for the sake of contradiction that \((f_{n_k})_{k \in \N}\) is a convergent subsequence. Then in particular it is Cauchy, and so there exists \(K \in \N\) such that \(\n{f_{n_k}-f_{n_\ell}}_{C^0([0,\pi])}
\lt 1\) for all \(k,\ell \ge K\text{.}\) But by (✶) this can only happen if \(n_k = n_\ell\text{,}\) i.e. if \(k=\ell\text{,}\) and so this is a contradiction.
Comment.
Note that the fact that \(\seq fn\) is not Cauchy is not enough on its own. For instance the alternating sequence \((0,1,0,1,\ldots)\) in \(\R\) is not Cauchy, but it has many convergent subsequences.
(d)
Use (✶) to show that \(F\) is not totally bounded.
Hint 1.
Suppose that \(r \in (0,1)\) and let \(f \in F\text{.}\) What can you say about the open ball \(B_r(f)\text{?}\)
Hint 2.
Solution 1. \(F\) as a subset
We will show that \(F\) is not a totally bounded subset of \(C^0([0,\pi])\text{.}\) Fix \(r \in (0,1]\text{.}\) Then (✶) shows that, in \(C^0([0,\pi])\text{,}\) the ball \(B_r(f)\) satisfies \(B_r(f) \cap F = \{f\}\) for all \(f \in F\text{.}\) Thus, since \(F\) is infinite, it cannot possibly be contained in a finite union of balls of this form (by the same argument as in Example 4.18).
Solution 2. \(F\) as a metric subspace
We will show that \(F\) is not totally bounded as a metric subspace of \(C^0([0,\pi])\text{.}\) Fix \(r \in (0,1]\text{.}\) Then (✶) shows that, in the metric subspace \(F\) of \(C^0([0,\pi])\text{,}\) \(B_r(f) = \{f\}\) for all \(f \in F\text{.}\) Thus, since \(F\) is infinite, it cannot possibly be a finite union of balls of this form (by the same argument as in Example 4.18).
(e)
Use (✶) to find an open cover of \(F\) with no finite subcover.
Hint.
Same hints as for the previous part.
Solution 1. \(F\) as a subset
Here we are considering \(F\) as a subset of the metric space \(C^0([0,\pi])\text{.}\) Clearly \(\set{B_1(f)}{f \in F}\) is an open cover of \(F\text{.}\) By the previous part, for any \(g_1,\ldots,g_N \in F\) the subcollection \(\{B_1(g_1),\ldots,B_1(g_N)\}\) must have
\begin{equation*}
F \cap \Big(B_1(g_1)\cup \cdots \cup B_1(g_N)\Big)
= \{g_1,\ldots,g_N\}.
\end{equation*}
In particular, as \(F\) is infinite, this subcollection does not cover \(F\text{.}\)
Solution 2. \(F\) as a subspace
Here we are considering \(F\) as a metric subspace of \(C^0([0,\pi])\text{.}\) By our above argument (and Lemma 1.21), the set \(\{f\}\) is open in \(F\) for all \(f \in F\text{.}\) Thus \(\set{\{f\}}{f \in F}\) is an open cover of \(F\text{.}\) As \(F\) is infinite, this cover has no finite subcover.
(f)
Use Theorem 1.35 to show that \(F\) is closed.
Hint 1.
Once again, (✶) is useful.
Hint 2.
It is tempting to think that the only sequences in \(F\) are subsequences of \(\seq fn\text{,}\) but this is not quite true. For instance, the constant sequence \((f_1,f_1,\ldots)\) is a sequence in \(F\text{,}\) but it is certainly not a subsequence of \(\seq fn\text{.}\)
Solution.
Suppose that \(\seq gn\) is a sequence in \(F\) which converges, as a sequence in \(C^0([0,\pi])\text{,}\) to some \(g \in C^0([0,\pi])\text{.}\) Then \(\seq gn\) is also Cauchy, and so there exists \(N \in \N\) such that \(\n{g_n-g_m}_{C^0([0,\pi])} \lt 1\) for all \(n,m \ge
N\text{.}\) By (✶), this can only happen if \(g_n=g_m\text{,}\) i.e. if \(g_n=g_N\) for all \(n \ge N\text{.}\) In particular, this means that
\begin{equation*}
g = \lim_{n \to \infty} g_n = \lim_{n \to \infty} g_N = g_N \in F\text{.}
\end{equation*}
Applying Theorem 1.35, we conclude that \(F\) is closed.
