Exercises Attempts

Exercise after §43

1. Let $X$ be a metric space.

   (a) Suppose that for some $\varepsilon>0$, every $\varepsilon$-ball in $X$ has compact closure. Show that $X$ is complete.

   (b) Suppose that for each $x \in X$ there is an $\varepsilon > 0$ such that the ball $B(x,\varepsilon)$ has compact closure. Show by means of an example that $X$ need not be complete.

(a) Let $(x_n)$ be a Cauchy sequence in $X$, then for every $\varepsilon>0$, there exists a positive integer $N$ such that $d(x_n,x_m)<\varepsilon$ whenever $n,m \geqslant N$. In particular, we have that $d(x_n,x_N) < \varepsilon$ for any $n \geqslant N$.

Take those $\varepsilon$ such that the closure of every $B(x,\varepsilon)$, where $x \in X$, is compact. By Theorem 28.2, this implies that every sequence in $\overline{B(x,\varepsilon)}$ has a convergent subsequence. In particular, there exists a subsequence of $(x_n)$ that is contained in $\overline{B(x_N,\varepsilon)}$, which is convergent. By Lemma 43.1, this means that $X$ is complete.

(b) Note that in this case, the choice of $\varepsilon>0$ depends on the choice of $x \in X$. Take the subspace $\mathbb{Q}$ of $\mathbb{R}$ under the standard topology. For every $x \in \mathbb{Q}$, there exists an $\varepsilon>0$, say, $|x|/2$, such that the ball $B(x,\varepsilon)$ has compact closure since the closure is the intersection of $\mathbb{Q}$ with a closed interval in $\mathbb{R}$. However, we know that $\mathbb{Q}$ is not complete.

2. Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces; let $Y$ be complete. Let $A \subseteq X$. Show that if $f : A \to Y$ is uniformly continuous, then $f$ can be uniquely extended to a continuous function $g : \overline{A} \to Y$, and $g$ is uniformly continuous.

Suppose that $f : A \to Y$ is uniformly continuous, then for any $\varepsilon>0$, there exists a $\delta>0$ such that for each pair of points $x_0,x_1$ of $X$, we have

$$d_X(x_0,x_1)<\delta \implies d_Y(f(x_0),f(x_1))<\varepsilon.$$

Let $a \in \overline{A}$, then $a$ is the limit (not necessarily in $A$) of a convergent sequence $(a_n)$ in $A$ by the sequence lemma. This implies that $(a_n)$ is Cauchy. Since $f$ is uniformly continuous (which applies to every pair of points of the domain), it follows that the sequence $(f(a_n))$ in $Y$ is also Cauchy. Since $Y$ is complete, $(f(a_n))$ is convergent.

In view of this, let us define $g : \overline{A} \to Y$ by $g(a)=\lim f(a_n)$, where $(a_n)$ is any sequence of points in $A$ that converges to $a$.

To show that $g$ is well-defined, we prove that if $a_n \to a$ and $b_n \to a$, then $\lim f(a_n)=\lim f(b_n)$. To see this, note that the hypothesis implies that $a_n-b_n \to 0$. Since $f$ is uniformly continuous, it follows that $f(a_n)-f(b_n) \to 0$¹ and as both $\lim f(a_n)$ and $\lim f(b_n)$ exist, it follows that $\lim f(a_n)=\lim f(b_n)$. Now, $g$ is also an extension of $f$, since if $a \in A$, then we can just take the constant sequence $(a,a,\ldots)$ so that $g(a)=\lim f(a)=f(a)$.

Next, we show that $g$ is uniformly continuous. Let $\varepsilon>0$ be arbitrary. Since $f$ is uniformly continuous, for any $\varepsilon>0$, there exists $\delta>0$ such that for every $x, y \in A$, we have $d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon$.

Choose $x,y \in \overline{A}$ such that $d_X(x,y)<\delta/3$, and let $(x_n)$ and $(y_n)$ be sequences in $A$ that converge to $x$ and $y$ respectively. There then exists positive integers $N_1,N_2$ such that

$$d_X(x_n,x) < \delta/3 \text{ whenever } n \geqslant N_1,$$$$d_X(y_n,y) < \delta/3 \text{ whenever } n \geqslant N_2.$$

Take $N=\max\{N_1,N_2\}$. It follows by triangle inequality that $d(x_n,y_n) \leqslant d_X(x_n,x)+d_X(x,y)+d_X(y_n,y)<\delta$, so by the uniform continuity of $f$, we have that $d_Y(f(x_n),f(y_n))<\varepsilon$.

Taking limits over $n$ we have that $d_Y(g(x),g(y))<\varepsilon$. Therefore, $g$ is uniformly continuous by definition.

Finally, we show that $g$ is unique. Indeed, suppose that $h : \overline{A} \to Y$ is also a continuous function extended from $f : A \to Y$. Again, if $a \in \overline{A}$, then there exists a sequence $(a_n)$ in $A$ such that $a_n \to a$. Since $h$ is continuous, by the sequential criterion of continuity, we have that $h(a_n) \to h(a)$. However, since $a_n \in A$ for each $n$, it follows that $h(a_n)=f(a_n)$ for each $n$, whilst $f(a_n) \to g(a)$. Therefore, we have that $h(a)=g(a)$ for every $a \in \overline{A}$.

3. Two metric $d$ and $d'$ on a set $X$ are said to be metrically equivalent if the identity map $i : (X,d) \to (X,d')$ and its inverse are both uniformly continuous.

   (a) Show that $d$ is metrically equivalent to the standard bounded metric $\overline{d}$ derived from $d$.

   (b) Show that if $d$ and $d'$ are metrically equivalent, then $X$ is complete under $d$ if and only if it is complete under $d'$.

(a) Let $\varepsilon>0$ be arbitrary and $i : (X,d) \to (X,\bar{d})$ be the identity map. Set $\delta = \varepsilon$. Note that by definition of $\bar{d}$, we have that $\bar{d}(x,y) \leqslant d(x,y)$ for every $x,y \in X$. It follows that for every pair of points $x_1,x_2 \in (X,d)$, we have that

$$d(x_1,x_2)<\delta \implies \bar{d}(i(x_1),i(x_2))=\bar{d}(x_1,x_2) \leqslant d(x_1,x_2) < \varepsilon.$$

Note that for any $x,y \in X$, if $d(x,y) \geqslant 1$, then $\bar{d}(x,y)=d(x,y)$; if $d(x,y)>1$, then $\bar{d}(x,y)=1$. Thus, for the inverse $i^{-1}$, set the corresponding $\delta$ to be $\min\{\varepsilon,1\}$. It follows that for every pair of points $x_1, x_2 \in (X,\bar{d})$, we have that

$$\bar{d}(x_1,x_2)<\delta \implies d(i(x_1),i(x_2))=d(x_1,x_2)=\begin{cases} \bar{d}(x_1,x_2) & \text{if } d(x_1,x_2) \leqslant 1 \\ d(x_1,x_2) & \text{if } d(x_1,x_2)>1 \end{cases}<\varepsilon.$$

Since both $i$ and $i^{-1}$ are uniformly continuous, it follows that $d$ and $\bar{d}$ are metrically equivalent.

(b) Suppose that $d$ and $d'$ are metrically equivalent. Let $(x_n)$ be a Cauchy sequence that converges to a point $x$ under $d$. It follows that for every $\delta>0$, there exists positive integers $N_1,N_2$ such that

$$d(x_n,x_m)<\delta \text{ whenever } m,n \geqslant N_1,$$$$d(x_n,x)<\delta \text{ whenever } n \geqslant N_2.$$

Since $d$ and $d'$ are metrically equivalent, the same relations above can be carried over to $(X,d')$ via the uniformly continuous identity map (since uniform continuity applies for every pair of points in $X$), so that if $(x_n)$ is a Cauchy sequence under $d'$, for every $\varepsilon>0$, by setting $\delta=\varepsilon$, we have that

$$d'(x_n,x_m)<\varepsilon \text{ whenever } m,n \geqslant N_1,$$$$d'(x_n,x)<\varepsilon \text{ whenever } n \geqslant N_2.$$

This implies that $(x_n)$ also converges under $d'$, and thus $X$ is complete under $d'$.

The converse also holds by applying the same argument above, just with the identity map replaced by its inverse.

4. Show that the metric space $(X,d)$ is complete if and only if for every nested sequence $A_1 \supseteq A_2 \supseteq \cdots$ of nonempty closed sets of $X$ such that $\mathrm{diam} \; A_n \to 0$, the intersection of the sets $A_n$ is nonempty.

note

This is a generalised version of the nested intervals theorem for real numbers.

It is also a special case of Theorem 26.9 (see Chapter 3) for complete metric spaces.

5. If $(X,d)$ is a metric space, recall that a map $f : X \to X$ is called a contraction if there is a number $\alpha < 1$ such that

$$d(f(x),f(y)) \leqslant \alpha d(x,y)$$

for all $x, y \in X$. Show that if $f$ is a contraction of a complete metric space, then there is a unique point $x \in X$ such that $f(x)=x$. Compare Exercise 7 of §28.

Compared to Exercise 7 of §28, this is a stronger result than that stated in part (a) of the exercise in which compactness is assumed, because compactness implies completeness but the converse needs not be true.

6. A space $X$ is said to be topologically complete (a.k.a. completely metrizable) if there exists a metric for the topology of $X$ relative to which $X$ is complete.

   (a) Show that a closed subspace of a topologically complete space is topologically complete.

   (b) Show that a countable product of topologically complete spaces is topologically complete (in the product topology).

   (c) Show that an open subspace of a topologically complete space is topologically complete. [Hint: If $U \subseteq X$ and $X$ is complete under the metric $d$, define $\phi : U \to \mathbb{R}$ by the equation $\phi(x)=1/d(x,X \backslash U)$. Embed $U$ in $X \times \mathbb{R}$ by setting $f(x)=(x,\phi(x))$.]

   (d) Show that if $A$ is a $G_\delta$ set in a topologically complete space, then $A$ is topologically complete. [Hint: Let $A$ be the intersection of the open sets $U_n$, for $n \in \mathbb{Z}_+$. Consider the diagonal embedding $f(a) = (a,a,\ldots)$ of $A$ into $\prod U_n$.] Conclude that the irrationals are topologically complete.

7. Show that the set of all sequences $(x_1,x_2,\ldots)$ such that $\sum x_i^2$ converges is complete in the $l^2$-metric. (See Exercise 8 of §20.)

9. Let $(X,d)$ be a metric space. Show that there is an isometric embedding (see Exercise 21.2) $h$ of $X$ into a complete metric space $(Y,D)$, as follows: Let $\tilde{X}$ denote the set of all Cauchy sequences

$$\mathbf{x}=(x_1,x_2,\ldots)$$

of points of $X$. Define $\mathbf{x} \sim \mathbf{y}$ if

$d(x_n,y_n) \to 0 .$

Let $[\mathbf{x}]$ denote the equivalence class of $\mathbf{x}$; and let $Y$ denote the set of equivalence classes. Define a metric $D$ on $Y$ by the equation

$$D([\mathbf{x}],[\mathbf{y}])=\lim_{n \to \infty}d(x_n,y_n).$$

(a) Show that $\sim$ is an equivalence relation, and show that $D$ is a well-defined metric.

(b) Define $h : X \to Y$ by letting $h(x)$ be the equivalence class of the constant sequence $(x,x,\ldots)$:

$$h(x)=[(x,x,\ldots)].$$

Show that $h$ is an isometric embedding.

(c) Show that $h(X)$ is dense in $Y$; indeed, given $\mathbf{x}=(x_1,x_2,\ldots) \in \tilde{X}$, show that the sequence $h(x_n)$ of points of $Y$ converges to the point $[\mathbf{x}]$ of $Y$.

(d) Show that if $A$ is a dense subset of a metric space $(Z,\rho)$, and if every Cauchy sequence in $A$ converges in $Z$, then $Z$ is complete.

(e) Show that $(Y,D)$ is complete.

note

Fun fact: if we set $X = \mathbb{Q}$, i.e. the set of rational numbers, then the cardinality of $\tilde{X}$ is the same as that of the set of real numbers, which is expected as the real numbers can be constructed via the completion of the rational numbers using Cauchy sequences.

The process done in this exercise is also called metric completion.

10. Theorem (Uniqueness of the completion). Let $h : X \to Y$ and $h' : X \to Y'$ be isometric embeddings of the metric space $(X,d)$ in the complete metric spaces $(Y,D)$ and $(Y',D')$ respectively, then there is an isometry $(\overline{h(X)},D)$ with $(\overline{h'(X)},D')$ that equals $h'h^{-1}$ on the subspace $h(X)$.

Exercise after §44

1. Given $n$, show there is a continuous surjective map $g : I \to I^n$. [Hint: Consider $f \times f : I \times I \to I^2 \times I^2$].

2. Show there is a continuous surjective map $f : \mathbb{R} \to \mathbb{R}^n$.

Exercise after §45

1. If $X_n$ is metrizable with metric $d_n$, then

$$D(\mathbf{x},\mathbf{y})=\sup\{\overline{d_i}(x_i,y_i)/i\}$$

is a metric for the product space $X = \prod X_n$. Show that $X$ is totally bounded under $D$ if each $X_n$ is totally bounded under $d_n$. Conclude without using the Tychonoff theorem that a countable product of compact metrizable spaces is compact.

2. Let $(Y,d)$ be a metric space; let $\mathcal{F}$ be a subset of $\mathcal{C}(X,Y)$.

   (a) Show that if $\mathcal{F}$ is finite, then $\mathcal{F}$ is equicontinuous.

   (b) Show that if $f_n$ is a sequence of elements of $\mathcal{C}(X,Y)$ that converges uniformly, then the collection $\{f_n\}$ is equicontinuous.

   (c) Suppose that $\mathcal{F}$ is a collection of differentiable functions $f : \mathbb{R} \to \mathbb{R}$ such that each $x \in \mathbb{R}$ lies in a neighbourhood $U$ on which the derivatives of the functions in $\mathcal{F}$ are uniformly bounded. [This means that there is an $M$ such that $|f'(x)| \leqslant M$ for all $f$ in $\mathcal{F}$ and all $x \in U$.] Show that $\mathcal{F}$ is equicontinuous.

3. Prove the following:

Theorem (Arzela's theorem). Let $X$ be compact; let $f_n \in \mathcal{C}(X,\mathbb{R}^k)$. If the collection $\{f_n\}$ is pointwise bounded and equicontinuous, then the sequence $f_n$ has a uniformly convergent subsequence.

4. (a) Let $f_n : I \to \mathbb{R}$ be the function $f_n(x)=x^n$. The collection $\mathcal{F}=\{f_n\}$ is pointwise bounded but the sequence $(f_n)$ has no uniformly convergent subsequence; at what point or points does $\mathbb{F}$ fail to be equicontinuous?

   (b) Repeat (a) for the functions $f_n$ of Exercise 9 of §21.

5. Let $X$ be a space. A subset $\mathcal{F}$ of $\mathcal{C}(X,\mathbb{R})$ is said to vanish uniformly at infinity if given $\varepsilon>0$, there is a compact subspace $C$ of $X$ such that $|f(x)|<\varepsilon$ for $x \in X \backslash C$ and $f \in \mathcal{F}$. If $\mathcal{F}$ consists of a single function $f$, we simply say that $f$ vanishes at infinity. Let $\mathcal{C}_0(X,\mathbb{R})$ denote the set of continuous functions $f : X \to \mathbb{R}$ that vanish at infinity.

   Theorem. Let $X$ be locally compact Hausdorff; give $\mathcal{C}_0(X,\mathbb{R})$ the uniform topology. A subset $\mathcal{F}$ of $\mathcal{C}_0(X,\mathbb{R})$ has compact closure if and only if it is pointwise bounded, equicontinuous, and vanishes uniformly at infinity.

   [Hint: Let $Y$ denote the one-point compactification of $X$. Show that $\mathcal{C}_0(X,\mathbb{R})$ is isometric with a closed subspace of $\mathcal{C}(Y,\mathbb{R})$ if both are given the sup metric.]

Exercise after §46

1. Show that the sets $B_C(f,\varepsilon)$ form a basis for the topology on $Y^X$.

2. Prove Theorem 46.7.

3. Show that the set $\mathcal{B}(\mathbb{R},\mathbb{R})$ of bounded functions $f : \mathbb{R} \to \mathbb{R}$ is closed in $\mathbb{R}^\mathbb{R}$ in the uniform topology, but not in the topology of compact convergence.

4. Consider the sequence of continuous functions $f_n : \mathbb{R} \to \mathbb{R}$ defined by

$$f_n(x)=x/n.$$

In which of the three topologies of Theorem 46.7 does this sequence converge?

Answer the same question for the sequence given in Exercise 9 of §21.

5. Consider the sequence of functions $f_n : (-1,1) \to \mathbb{R}$, defined by

   $$f_n(x)=\sum_{k=1}^{n}kx^k.$$

   (a) Show that $(f_n)$ converges in the topology of compact convergence; conclude that the limit function is continuous. (This is a standard fact about power series.)

   (b) Show that $(f_n)$ does not converge in the uniform topology.

6. Show that in the compact-open topology, $\mathcal{C}(X,Y)$ is Hausdorff if $Y$ is Hausdorff, and regular if $Y$ is regular. [Hint: If $\overline{U} \subseteq V$, then $\overline{S(C,U)} \subseteq S(C,V)$.]

7 (reworded). Show that if $Y$ is locally compact Hausdorff, then composition of maps $\circ$ given by

$$\circ : \mathcal{C}(X,Y) \times \mathcal{C}(Y,Z) \to \mathcal{C}(X,Z)$$$$(f,g) \mapsto g \circ f$$

is continuous, provided the compact-open topology is used throughout. [Hint: If $g \circ f \in S(C,U)$, find $V$ such that $f(C) \subseteq V$ and $g(\bar{V}) \subseteq U$.]

10. A space is locally compact if it can be covered by open sets each of which is contained in a compact subspace of $X$. It is said to be $\sigma$-compact if it can be covered by countably many such open sets.

    (a) Show that if $X$ is locally compact and second-countable, it is $\sigma$-compact.

    (b) Let $(Y,d)$ be a metric space. Show that if $X$ is $\sigma$-compact, there is a metric for the topology of compact convergence on $Y^X$ such that if $(Y,d)$ is complete, $Y^X$ is complete in this metric. [Hint: Let $A_1,A_2,\ldots$ be a countable collection of compact subspaces of $X$ whose interiors cover $X$. Let $Y_i$ denote the set of all functions from $A_i$ to $Y$, in the uniform topology. Define a homeomorphism of $Y^X$ with a closed subspace of the product space $Y_1 \times Y_2 \times \cdots$.]

(a) Suppose that $X$ is locally compact and second-countable. Since $X$ is locally compact, for each $x \in X$, there exists a compact subspace $C$ of $X$ such that a neighbourhood $U$ of $x$ is contained in $C$. Since $X$ is second-countable, $X$ has a countable basis. This implies that $U$ contains at least one of the basis elements, at least one of which contains the point $x$. Let $B$ be one of such basis elements, then $B$ is also a neighbourhood of $x$ that is contained in a compact subspace $C$ of $X$. There exists such basis elements for each arbitrary choice of $x \in X$, and countably many of them cover the whole space $X$. It follows that $X$ can be covered by countably many open sets, each of which is contained in some compact subspace of $X$, and is thus $\sigma$-compact.

Exercise after §47

1. Which of the following subsets of $\mathcal{C}(\mathbb{R},\mathbb{R})$ are pointwise bounded? Which are equicontinuous?

   (a) The collection $\{f_n\}$, where $f_n(x)=x+ \sin nx$.

   (b) The collection $\{g_n\}$, where $g_n(x)=n+\sin nx$.

   (c) The collection $\{h_n\}$, where $h_n(x)=|x|^{1/n}$.

   (d) The collection $\{k_n\}$, where $k_n(x)=n \sin (x/n)$.

3. Show that the general version of Ascoli's theorem implies the classical version (Theorem 45.4) when $X$ is Hausdorff.

4. Prove the following:

   Theorem (Arzela's theorem, general version). Let $X$ be a Hausdorff space that is $\sigma$-compact; let $f_n$ be a sequence of functions $f_n : X \to \mathbb{R}^k$. If the collection $\{f_n\}$ is pointwise bounded and equicontinuous, then the sequence $f_n$ has a subsequence that converges, in the topology of compact convergence, to a continuous function.

   [Hint: Show $\mathcal{C}(X,\mathbb{R}^k)$ is first-countable.]

5. Let $(Y,d)$ be a metric space; let $f_n : X \to Y$ be a sequence of continuous functions; let $f : X \to Y$ be a function (not necessarily continuous). Suppose $f_n$ converges to $f$ in the topology of pointwise convergence. Show that if $\{f_n\}$ is equicontinuous, then $f$ is continuous and $f_n$ converges to $f$ in the topology of compact convergence.

Footnotes

1. Why? The fact that $a_n - b_n \to 0$ implies that for every $\varepsilon>0$, there exists a positive integer $N$ such that whenever $n \geqslant N$, we have $d_X(a_n,b_n)<\varepsilon$. Meanwhile, the uniform continuity of $f$ implies that for every $\varepsilon>0$, there exists $\delta >0$ such that for every pair of points in $X$, whenever $d_X(x,y)<\delta$, we have $d_Y(f(x),f(y))<\varepsilon$. Substitute $x,y$ here to be $a_n,b_n$, then for any $\varepsilon>0$, whenever $n \geqslant N$, we have that $d_Y(f(a_n),f(b_n))<\varepsilon$, so $f(a_n)-f(b_n) \to 0$. ↩