Notes

Complete Metric Spaces

Definition. Let $(X,d)$ be a metric space. A sequence $(x_n)$ of points of $X$ is said to be a Cauchy sequence in $(X,d)$ if it has the property that for any $\varepsilon>0$, there exists an integer $N$ such that

$$d(x_n,x_m)<\varepsilon \quad \text{whenever } n,m \geqslant N.$$

Definition. The metric space $(X,d)$ is said to be complete if every Cauchy sequence in $X$ converges (in the space $X$ itself).

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• A closed subset $A$ of a complete metric space $(X,d)$ is necessarily complete in the restricted metric (inherited from the original space), as a Cauchy sequence in $A$ is also Cauchy in $X$, so it converges in $X$, and the limit must lie in $A$ because $A$ is closed (see Theorem 17.6 in Chapter 2).
• If $X$ is complete under the metric $d$, then $X$ is also complete under the standard bounded metric $\bar{d}(x,y)=\min\{d(x,y),1\}$ with respect to $d$, and conversely, because a sequence is Cauchy (resp. converges) under $d$ if and only if it is Cauchy (resp. converges) under $d$ (since $\bar{d}(x,y) \leqslant d(x,y)$ for any $x,y \in X$ and the equality holds when $d(x,y) \leqslant 1$, see also Exercise 43.3).

[Lemma 43.1] A metric space $X$ is complete if every Cauchy sequence in $X$ has a convergent subsequence.

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In fact, it also follows from this theorem and Theorem 28.2 (see Chapter 3) that every compact metric space is complete. However, the converse does not hold: a complete metric space need not be compact (e.g. $\mathbb{R}$ in its standard topology).

Let $(x_n)$ be a Cauchy sequence in $(X,d)$. It follows that given $\varepsilon > 0$, there exists a positive integer $N'$ such that $d(x_n, x_m)< \varepsilon/2$ whenever $n,m \geqslant N'$.

Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$ with the limit $x$, then for any $\varepsilon>0$, there exists a positive integer $K$ such that $d(x_{n_k},x)<\varepsilon/2$ whenever $k \geqslant K$.

Take $N= \max\{N',n_K\}$. Using the fact that $n_k \geqslant k$ for any subindex $k$, it follows that whenever $n,n_k \geqslant N$, we have

$$d(x_n,x) \leqslant d(x_n,x_{n_k})+d(x_{n_k},x) < \varepsilon/2+\varepsilon/2=\varepsilon.$$

[Theorem 43.2] The Euclidean space $\mathbb{R}^k$ is complete in the Euclidean metric $d$ or the square metric $\rho$.

To show that the metric space $(\mathbb{R}^k,d)$ is complete, let $(x_n)$ be a Cauchy sequence in $(\mathbb{R}^k,d)$, then the set $\{x_n\}$ is a bounded subset of $(\mathbb{R}^k,d)$, because if we choose $N$ such that

$$d(x_n,x_m) \leqslant 1 \text{ for all } n,m \geqslant N,$$

then the number

$$M =\max\{d(x_1,\mathbf{0}),\ldots,d(x_{N-1},\mathbf{0}),d(x_N,\mathbf{0})+1\}$$

is an upper bound for $d(x_n,\mathbf{0})$, by triangle inequality ($d(x_n,\mathbf{0}) \leqslant d(x_n,x_m)+d(x_m,\mathbf{0})$). This implies that the points of the sequence $(x_n)$ all lie in the $k$-ball $\{ x \in \mathbb{R}^k : d(x,\mathbf{0}) \leqslant M+1\}$. Since this sphere is compact (see Theorem 27.3 in Chapter 3), the sequence $(x_n)$ has a convergent subsequence, by Theorem 28.2 (in Chapter 3). Thus, $(\mathbb{R}^k,d)$ is complete.

Note that a sequence is Cauchy (resp. converges) relative to $\rho$ if and only if it is Cauchy (resp. converges) relative to $d$, hence $(\mathbb{R}^k,\rho)$ is also complete.

[Lemma 43.3] Let $X$ be the product space $X = \prod X_\alpha$ and $\mathbf{x}_n$ be a sequence of points of $X$, then $\mathbf{x}_n \to \mathbf{x}$ if and only if $\pi_\alpha(\mathbf{x}_n) \to \pi_\alpha(\mathbf{x})$ for each $\alpha$.

Since the projection mapping $\pi_\alpha : X \to X_\alpha$ is continuous, it preserves convergent sequences (see Theorem 21.3 in Chapter 2); the 'only if' ($\implies$ direction) part of the lemma follows.

To prove the converse ($\impliedby$ direction), suppose that $\pi_\alpha(\mathbf{x}_n) \to \pi_\alpha(\mathbf{x})$ for each $\alpha \in J$. Let $U = \prod U_\alpha$ be a basis element for $X$ that contains $\mathbf{x}$. For each $\alpha$ such that $U_\alpha$ does not equal the entire space $X_\alpha$ (see Theorem 19.1 in Chapter 2), choose a positive integer $N_\alpha$ so that $\pi_\alpha(\mathbf{x}_n) \in U_\alpha$ whenever $n \geqslant N_\alpha$. Let $N$ be the largest of such numbers $N_\alpha$, then for all $n \geqslant N$, we have that $\mathbf{x}_n \in U$.

[Theorem 43.4] There is a metric for the product space $\mathbb{R}^\omega$ relative to which $\mathbb{R}^\omega$ is complete.

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Another way of phrasing this theorem is:

$\mathbb{R}^\omega$ is completely metrizable.

See Exercise 43.6.

Let $\bar{d}(a,b)=\min\{|a-b|,1\}$ be the standard bounded metric on $\mathbb{R}$. Let $D$ be the metric on $\mathbb{R}^\omega$ defined by $D(\mathbf{x},\mathbf{y})=\sup\{\bar{d}(x_i,y_i)/i\}$, then $D$ induces the product topology on $\mathbb{R}^\omega$ (see Theorem 20.5 in Chapter 2).

We verify that $\mathbb{R}^\omega$ is complete under $D$. Let $\mathbf{x}_n$ be a Cauchy sequence in $(\mathbb{R}^\omega, D)$. Since by definition

$$\bar{d}(\pi_i(\mathbf{x}),\pi_i(\mathbf{y})) \leqslant i D(\mathbf{x},\mathbf{y}),$$

we see that for fixed $i$ the sequence $\pi_i(\mathbf{x}_n)$ is a Cauchy sequence in $\mathbb{R}$ (consider $N$ such that $D(\mathbf{x}_n,\mathbf{x}_m)<\varepsilon/i$ whenever $n,m \geqslant N$, given $\varepsilon>0$), so it converges, say to $a_i$ (see Theorem 43.2 above). The sequence $\mathbf{x}_n$ then converges to the point $\mathbf{a}=(a_1,a_2,\ldots)$ in $\mathbb{R}^\omega$.

Definition. Let $(Y,d)$ be a metric space; let $\bar{d}(a,b)=\min\{d(a,b),1\}$ be the standard bounded metric on $Y$ derived from $d$. If $\mathbf{x}=(x_\alpha)_{\alpha \in J}$ and $\mathbf{y}=(y_\alpha)_{\alpha \in J}$ are points of the Cartesian product $Y^J$, let

$$\bar{\rho}(\mathbf{x},\mathbf{y})=\sup\{\bar{d}(x_\alpha,y_\alpha) : \alpha \in J\}.$$

One can check that $\bar{\rho}$ is a metric; it is called the uniform metric on $Y^J$ corresponding to the metric $d$ on $Y$.

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This is the generalisation of the definition of uniform metric given in Chapter 2.

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Since the elements of $Y^J$ are in fact functions from $J$ to $Y$, one could also use functional notation for them, which will be used from hereon.

The functional notation for the uniform metric is as such: If $f,g : J \to Y$, then

$$\bar{\rho}(f,g)=\sup\{\bar{d}(f(\alpha),g(\alpha)) : \alpha \in J\}.$$

[Theorem 43.5] If the space $Y$ is complete in the metric $d$, then the space $Y^J$ is complete in the uniform metric $\bar{\rho}$ corresponding to $d$.

[Theorem 43.6] Let $X$ be a topological space and let $(Y,d)$ be a metric space. The set $\mathcal{C}(X,Y)$ of continuous functions is closed in $Y^X$ under the uniform metric, and so is the set $\mathcal{B}(X,Y)$ of bounded functions. Therefore, if $Y$ is complete, these spaces are complete in the uniform metric.

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A function $f$ is said to be bounded if its image $f(X)$ is a bounded subset of the metric space $(Y,d)$.

To show the first part of this theorem, we first show that if a sequence of elements $f_n$ of $Y^X$ converges to the element $f$ of $Y^X$ relative to the metric $\bar{\rho}$ on $Y^X$, then it converges to $f$ uniformly in the sense defined in §21, relative to the metric $\bar{d}$ on $Y$. Let $\varepsilon>0$ be arbitrary, then due to the assumed convergence, we obtain an integer $N$ such that

$$\bar{\rho}(f,f_n)<\varepsilon \text{ for all } n \geqslant N.$$

It follows that for all $x \in X$ and all $n \geqslant N$, we have

$$\bar{d}(f_n(x),f(x)) \leqslant \bar{\rho}(f_n,f)<\varepsilon.$$

Thus, $(f_n)$ converges uniformly to $f$.

Now we show that $\mathcal{C}(X,Y)$ is closed in $Y^X$ relative to the metric $\bar{\rho}$. Let $f$ be an element of $Y^X$ that is a limit point of $\mathcal{C}(X,Y)$, then there is a sequence $(f_n)$ of elements of $\mathcal{C}(X,Y)$ converging to $f$ in the metric $\bar{\rho}$ (see Lemma 21.2 in Chapter 2). By the uniform limit theorem, $f$ is continuous, so that $f \in \mathcal{C}(X,Y)$. This implies that the closure of $\mathcal{C}(X,Y)$ is itself, and thus $\mathcal{C}(X,Y)$ is closed in $Y^X$.

Finally, we show that $\mathcal{B}(X,Y)$ is closed in $Y^X$. If $f$ is a limit point of $\mathcal{B}(X,Y)$, then there is a sequence of elements $f_n$ of $\mathcal{B}(X,Y)$ converging to $f$. Choose a sufficiently large $N$ such that $\bar{\rho}(f_N,f)<1/2$, then for $x \in X$, we have $\bar{d}(f_N(x),f(x))<1/2$, which implies that $d(f_N(x),f(x))<1/2$. It follows that if $M$ is the diameter of the set $f_N(x)$, then $f(X)$ has diameter at most $M+1$. Thus, $f \in \mathcal{B}(X,Y)$.

If $Y$ is complete, we conclude that $\mathcal{C}(X,Y)$ and $\mathcal{B}(X,Y)$ are complete in the metric $\bar{\rho}$.

Definition. If $(Y,d)$ is a metric space, one can define another metric on the set $\mathcal{B}(X,Y)$ of bounded functions from $X$ to $Y$ by the equation

$$\rho(f,g)=\sup\{d(f(x),g(x)) : x \in X\}.$$

Since $f$ and $g$ are bounded, the set $f(X) \cup g(X)$ is bounded (so that the said supremum exists) if both $f(X)$ and $g*(X)$ are bounded, it follows that $\rho$ is well-defined. The metric $\rho$ here is called the sup metric.

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• There is a simple relation between the sup metric and the uniform metric, which is that if $f,g \in \mathcal{B}(X,Y)$, then $\bar{\rho}(f,g)=\min\{\rho(f,g),1\}$.

  This is because if $\rho(f,g)>1$, then there exists $x_0 \in X$ such that $d(f(x_0),g(x_0))>1$, so by definition, $\bar{d}(f(x_0),g(x_0)) = 1 \implies \bar{\rho}(f,g)=1$. On the other hand, if $\rho(f,g) \leqslant 1$, then $\bar{d}(f(x),g(x)) = d(f(x),g(x)) \leqslant 1$ for all $x$, so that $\bar{\rho}(f,g)=\rho(f,g)$.

  Thus, on $\mathcal{B}(X,Y)$, the metric $\bar{\rho}$ is in fact the standard bounded metric derived from the uniform metric $\rho$.

• If $X$ is a compact space, then every continuous function $f : X \to Y$ is bounded, so the sup metric is well-defined on $\mathcal{C}(X,Y)$.

• If $Y$ is complete under $d$, then $\mathcal{C}(X,Y)$ is complete under the corresponding uniform metric $\bar{\rho}$ (see Theorem 43.5), so it is also complete under the sup metric $\rho$ (see the first 'Info' block of this section).

[Theorem 43.7] Let $(X,d)$ be a metric space. There is an isometric embedding of $X$ into a complete metric space.

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An alternative proof can be found in Exercise 43.9.

Definition. Let $X$ be a metric space. If $h : X \to Y$ is an isometric embedding of $X$ into a complete metric space $Y$, then the subspace $\overline{h(X)}$ of $Y$. is a complete metric space. It is called the completion of $X$.

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The completion of $X$ is uniquely determined up to an isometry. See Exercise 43.10.

A Space-Filling Curve

[Theorem 44.1] (Peano space-filling curve) Let $I=[0,1]$. There exists a continuous map $f : I \to I^2$ whose image fills up the entire square $I^2$.

Step 1. We construct the map $f$ as the (uniform) limit of a sequence of a continuous functions $f_n$, through a series of triangular paths.

Step 2. We define a sequence of functions $f_n : I \to I^2$ through recursively defined triangular paths.

Step 3. We claim that the sequence of functions $(f_n)$ defined in Step 2 is a Cauchy sequence under the sup metric on $\mathcal{C}(I,I^2)$, which we denote by $\rho$.

Step 4. Since $\mathcal{C}(I,I^2)$ is complete, the sequence $f_n$ converges to a continuous function $f : I \to I^2$. We prove that $f$ is surjective.

Compactness in Metric Spaces

Definition. A metric space $(X,d)$ is said to be totally bounded if for every $\varepsilon>0$, there is a finite covering of $X$ by $\varepsilon$-balls.

[Theorem 45.1] A metric space $(X,d)$ is compact if and only if it is complete and totally bounded.

Definition. Let $(Y,d)$ be a metric space. Let $\mathcal{F}$ be a subset of the function space $\mathcal{C}(X,Y)$. If $x_0 \in X$, the set $\mathcal{F}$ of functions is said to be equicontinuous at $x_0$ if given $\varepsilon > 0$, there is a neighbourhood $U$ of $x_0$ such that for all $x \in U$ and all $f \in \mathcal{F}$, $d(f(x),f(x_0))<\varepsilon$.

If the set $\mathcal{F}$ is equicontinuous at each point in $X$, it is said simply to be equicontinuous.

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Equicontinuity is stronger than continuity, but is a separate concept to uniform continuity.

While continuity of the function $f$ at a point $x_0$ implies the existence of a neighbourhood $U$ of $x_0$ such that $d(f(x),f(x_0))<\varepsilon$ for all $x \in U$ with a given $\varepsilon>0$, which may change depending on the choice of $x_0$ and $\varepsilon$, equicontinuity of $\mathcal{F}$ ensures the existence of a single neighbourhood that works for all functions $f$ in the collection $\mathcal{F}$.

In other words, equicontinuity gives rise to the existence of such neighbourhood that does not depend on the choice of functions $f$ amongst a family of functions, which is not to be confused with uniform continuity, for which the existence of such neighbourhood does not depend on the choice of points in the domain $X$.

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Note that equicontinuity depends on the specific metric $d$ rather than just on the topology of $Y$.

[Lemma 45.2] Let $X$ be a space and $(Y,d)$ be a metric space. If the subset $\mathcal{F}$ of $\mathcal{C}(X,Y)$ is totally bounded under the uniform metric correspond to $d$, then $\mathcal{F}$ is equicontinuous under $d$.

[Lemma 45.3] Let $X$ be a space and $(Y,d)$ a metric space; assume $X$ and $Y$ are compact. If the subset $\mathcal{F}$ of $\mathcal{C}(X,Y)$ is equicontinuous under $d$, then $\mathcal{F}$ is totally bounded under the uniform and sup metrics corresponding to $d$.

Definition. If $(Y,d)$ is a metric space, a subset $\mathcal{F}$ of $\mathcal{C}(X,Y)$ is said to be pointwise bounded under $d$ if for each $x \in X$ (fixed), the subset

$$\mathcal{F}_x = \{f(x) : f \in \mathcal{F}\}$$

of $Y$ is bounded under $d$.

[Theorem 45.4] (Ascoli's theorem, classical version). Let $X$ be a compact space; let $(\mathbb{R}^n,d)$ denote Euclidean space in either the square metric or the Euclidean metric; give $\mathcal{C}(X,\mathbb{R}^n)$ the corresponding uniform topology. A subspace $\mathcal{F}$ of $\mathcal{C}(X,\mathbb{R}^n)$ has compact closure if and only if $\mathcal{F}$ is equicontinuous and pointwise bounded under $d$.

[Corollary 45.5] Let $X$ be compact and $d$ denote either the square metric or the Euclidean metric on $\mathbb{R}^n$. Give $\mathcal{C}(X,\mathbb{R}^n)$ the corresponding uniform topology. A subspace $\mathcal{F}$ of $\mathcal{C}(X,\mathbb{R}^n)$ is compact if and only if it is closed, bounded under the sup metric $\rho$, and equicontinuous under $d$.

Suppose that $\mathcal{F}$ is compact, then it must be closed and bounded (see Theorem 27.3 in Chapter 3); since the closure of $\mathcal{F}$ here is also compact (see Theorem 26.3 in Chapter 3 and recall that every metric space is Hausdorff), Theorem 45.4 implies that it is also equicontinuous.

Conversely, if $\mathcal{F}$ is closed, it is equal to its closure $\mathcal{G}$; if it is bounded under $\rho$, it is also pointwise bounded under $d$; it is also assumed that $\mathcal{F}$ is equicontinuous, so Theorem 45.4 implies that it is compact.

Pointwise and Compact Convergence

Definition. Given a point $x$ of the set $X$ and an open set $U$ of the space $Y$, let

$$S(x,U) = \{f : f \in Y^X \text{ and } f(x) \in U\}.$$

The sets $S(x,U)$ are a subbasis for topology on $Y^X$, which is called the topology of pointwise convergence (or the point-open topology).

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The general basis element for the topology of pointwise convergence is a finite intersection of subbasis elements $S(x,U)$, so a typical basis element about the function $f$ consists of all functions $g$ that are 'close' to $f$ at finitely many points.

In fact, the topology of pointwise convergence on $Y^X$ is just the product topology. The set $S(\alpha,U)$ of all functions $\mathbf{x} : J \to Y$ such that $\mathbf{x}(\alpha) \in U$ is just the subset $\pi_\alpha^{-1}(U)$ of $Y^J$, which is the standard subbasis element for the product topology.

[Theorem 46.1] A sequence $f_n$ of functions converges to the function $f$ in the topology of pointwise convergence if and only if for each $x$ in $X$, the sequence $f_n(x)$ of points of $Y$ converges to the point $f(x)$.

This result is just a rephrasing of Lemma 43.3 in function space notation.

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A sequence of continuous functions that converges only in the topology of pointwise convergence need not have a continuous limit. See here for (counter)examples.

Is there a topology intermediate between uniform topology and topology of pointwise convergence that preserves continuity in limits? Yes, by imposing a restriction that the space $X$ is compactly generated, we can define the topology of compact convergence.

Definition. Let $(Y,d)$ be a metric space and $X$ be a topological space. Given an element $f$ of $Y^X$, a compact subspace $C$ of $X$ and a number $\varepsilon>0$, let $B_C(f,\varepsilon)$ denote the set of all the elements $g$ of $Y^X$ for which

$$\sup\{d(f(x),g(x)) : x \in C\}<\varepsilon.$$

The sets $B_C(f,\varepsilon)$ form a basis for a topoogy on $Y^X$, which is called the topology of compact convergence (or 'the topology of uniform convergence on compact sets').

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• To show that $B_C(f,\varepsilon)$ satisfies the condition for a basis, the key step is to note that if $g \in B_C(f,\varepsilon)$, then if we take

  $$\delta := \varepsilon - \sup\{d(f(x),g(x)) : x \in C\},$$

  we then have $B_C(g,\delta) \subseteq B_C(f,\varepsilon)$.

• The topology of compact convergence is 'stronger' than the topology of pointwise convergence in the sense that the general basis element containing $f$ are functions that are 'close' to f$f$ not just at finitely many points, but at all points of some compact set.

[Theorem 46.2] A sequence $f_n : X \to Y$ of functions converges to the function $f$ in the topology of compact convergence if and only if for each compact subspace $C$ of $X$, the sequence $f_n|_{C}$ converges uniformly to $f|_{C}$.

Let $f_n : X \to Y$ be a sequence of functions in the function space $Y^X$ under the topology of compact convergence.

Suppose that for each compact subspace of $C$ of $X$, the sequence $f_n|_{C}$ converges uniformly to $f|_{C}$, then given $\varepsilon>0$, there exists an integer $N$, such that

$$d(f_n|_{C}(x),f|_{C}(x))<\varepsilon \text{ whenever } n \geqslant N, \text{ for all } x \in C.$$

This implies that whenever $n \geqslant N$, the set $\{d(f_n(x),f(x)) : x \in C\}$ is bounded above by $\varepsilon$, so we have that

$$\sup\{d(f_n(x),f(x)) : x \in C\}<\varepsilon.$$

Note that each neighbourhood of $f$ in $Y^X$ under the topology of compact convergence must contain some basis element of the form $B_C(f,\varepsilon)$. Note as well that if we range over all such neighbourhoods, we will run through all $\varepsilon>0$; conversely, if we range over all $\varepsilon>0$, we will run through all the basis elements in the form of $B_C(f,\varepsilon)$, which generates all neighbourhoods of $f$.

Therefore, the above implies that for each such neighbourhood $U$, there exists an integer $N$ such that $f_n \in U$ for all $n \geqslant N$. Therefore, the sequence $f_n : X \to Y$ of functions converges to the function $f$ in the topology of compact convergence.

The steps above are reversible, so the converse also holds.

Definition. A space $X$ is said to be compactly generated (or is called a k-space), if it satisfies the following condition:

A set $A$ is open (equivalently, closed) in $X$ if $A \cap C$ is open (equiv. closed) in $C$ for each compact subspace $C$ of $X$.

[Lemma 46.3] If $X$ is locally compact, or if $X$ satisfies the first countability axiom, then $X$ is compactly generated.

Locally compact

Suppose that $X$ is locally compact, then for every point $x$ in $X$, there exists some compact subspace $C$ of $X$ that contains a neighbourhood of $x$.

Let $A \cap C$ be open in $C$ for every compact subspace $C$ of $X$. We want to show that $A$ is open in $X$, so that $X$ is compactly generated. Given $x \in A$, choose a neighbourhood $U$ of $x$ that lies in a compact subspace $C$ of $X$. Note that by hypothesis, $A \cap C$ is open in $C$, and thus $A \cap U$ is open in $U$, implying that $A \cap U$ is open in $X$ as well (see Lemma 16.2 in Chapter 2). It follows that $A \cap U$ is a neighbourhood of $x$ contained in $A$, so $A$ is open in $X$ (since the choice of $x \in A$ here is arbitrary).

First-countable

Suppose that $X$ satisfies the first countability axiom. If $B \cap C$ is closed in $C$ for every compact subspace $C$ of $X$, we want to show that $B$ is closed in $X$. Let $x$ be a point of $\overline{B}$; we show that $x \in B$ (so that $B = \overline{B}$ and thus $B$ is closed in $X$). Since $X$ has a countable basis at $x$, there exists a sequence $(x_n)$ of points of $B$ converging to $x$ (see Theorem 30.1 in Chapter 4). The subspace

$$C = \{x\} \cup \{x_n : n \in \mathbb{Z}_+\}$$

is compact (why?), so that $B \cap C$ is by assumption closed in $C$. Since $B \cap C$ contains $x_n$ for every $n$, it contains $x$ as well. Therefore, $x \in B$, as desired.

[Lemma 46.4] If $X$ is compactly generated, then a function $f : X \to Y$ is continuous if for each compact subspace $C$ of $X$, the restricted function $f|_{C}$ is continuous.

We show that if $V$ is an open subset of $Y$, then $f^{-1}(V)$ is also open in $X$. For any subspace $C$ of $X$, we have

$$f^{-1}(V) \cap C = (f|_{C})^{-1}(V).$$

If $C$ is compact, then by the continuity of $f|_{C}$, the set $(f|_{C})^{-1}(V)$ is open in $C$. Since $X$ is compactly generated, it follows that $f^{-1}(V)$ is open in $X$.

[Theorem 46.5] Let $X$ be a compactly generated space and $(Y,d)$ be a metric space, then $\mathcal{C}(X,Y)$ is clsoed in $Y^X$ in the topology of compact convergence.

Let $f \in Y^X$ be a limit point of $\mathcal{C}(X,Y)$; we wish to show that $f \in \mathcal{C}(X,Y)$, i.e. $f$ is continuous.

It suffices to show that $f|_{C}$ is continuous for each compact subspace $C$ of $X$ (see Theorem 46.2 above). For each $n$, consider the neighbourhood $B_C(f,1/n)$ of $f$; it intersects $\mathcal{C}(X,Y)$ (see Theorem 20.1 in Chapter 2), so we can choose a function $f_n \in \mathcal{C}(X,Y)$ lying in this neighbourhood. The sequence of functions $f_n|_{C} : C \to Y$ converges uniformly to the function $f|_{C}$ (as the definition of $B_C(f,1/n)$ implies that the choice of $N$ does not depend on the points $x \in X$), so by the uniform limit theorem, $f|_{C}$ is continuous.

[Corollary 46.6] Let $X$ be a compactly generated space and $(Y,d)$ be a metric space. If a sequence of continuous functions $f_n : X \to Y$ converges to $f$ in the topology of compact convergence, then $f$ is continuous.

Theorem 46.5 implies that the closure of $\mathcal{C}(X,Y)$ is itself, so all limit points will also be in $\mathcal{C}(X,Y)$, and this implies the desired result.

[Theorem 46.7] (Relationship between uniform, compact convergence, and pointwise convergence) Let $X$ be a space and $(Y,d)$ be a metric space. For the function space $Y^X$, one has the following inclusion of topologies:

$$(\text{uniform}) \supseteq (\text{compact convergence}) \supseteq (\text{pointwise convergence}).$$

If $X$ is compact, then (uniform) = (compact convergence), and if $X$ is discrete, then (compact convergence) = (pointwise convergence).

A proof of this is in Exercise 46.2.

Definition. Let $X$ and $Y$ be topological spaces. If $C$ is a compact subspace of $X$ and $U$ is an open subset of $Y$, define

$$S(C,U)=\{ f : f \in \mathcal{C}(X,Y) \text{ and } f(C) \subseteq U\}.$$

The sets $S(C,U)$ form a subbasis for a topology on $\mathcal{C}(X,Y)$ that is called the compact-open topology.

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From the definition, we see that the compact-open topology is finer than the pointwise convergence topology (since $S(C,U) \subseteq S(x,U)$ for any open subset of $Y$ and compact subspace of $X$ and note that $\mathcal{C}(X,Y) \subseteq Y^X$).

The compact-open topology can actually be defined on the entire function space $Y^X$, but it is not the focus of this section.

[Theorem 46.8] Let $X$ be a space and let $(Y,d)$ be a metric space. On the set $\mathcal{C}(X,Y)$, the compact-open topology and the topology of compact convergence coincide.

[Corollary 46.9] (from Theorem 46.8) Let $Y$ be a metric space. The compact convergence topology on $\mathcal{C}(X,Y)$ does not depend on the metric of $Y$. Therefore, if $X$ is compact, the uniform topology on $\mathcal{C}(X,Y)$ does not depend on the metric of $Y$.

The first part of this corollary comes from Theorem 46.8 and the definition of compact-open topology, which does not depend on the choice of said metric. The second part of this corollary comes from the first part and Theorem 46.7.

[Theorem 46.10] Let $X$ be a locally compact Hausdorff space and $\mathcal{C}(X,Y)$ have the compact-open topology, then the evaluation map, given by

$$e : X \times \mathcal{C}(X,Y) \to Y$$

which is defined by the equation

$$e(x,f)=f(x)$$

is continuous.

Let $(x,f)$ be a point of $X \times \mathcal{C}(X,Y)$ and $V$ be an open set in $Y$ that contains the image point $e(x,f)=f(x)$. We want to find an open set $U$ containing $(x,f)$ such that $e(U) \subseteq V$.

First, due to the continuity of $f$ and the fact that $X$ is locally compact Hausdorff, we can choose an open set $U$ about $x$ having compact closure $\overline{U}$, such that $f(\overline{U}) \subseteq V$ (see Theorem 29.2 in Chapter 3). Next, consider the open set $U \times S(\overline{U},V)$ in $X \times \mathcal{C}(X,Y)$. It is an open set containing $(x,f)$. If $(x',f')$ belongs to this set, then $e(x',f')=f'(x')$ belongs to $V$, as desired.

Definition. Given a function $f : X \times Z \to Y$, there is a corresponding function $F : Z \to \mathcal{C}(X,Y)$, defined by the equation

$$(F(z))(x)=f(x,z).$$

Conversely, given $F : Z \to \mathcal{C}(X,Y)$, this equation defines a corresponding function $f : X \times Z \to Y$. We say that $F$ is the map of $Z$ into $\mathcal{C}(X,Y)$ that is induced by $f$.

[Theorem 46.11] Let $X$ and $Y$ be spaces; give $\mathcal{C}(X,Y)$ the compact-open topology. If $f : X \times Z \to Y$ is continuous, then so is the induced function $F : Z \to \mathcal{C}(X,Y)$. The converse holds if $X$ is locally compact Hausdorff.

Ascoli's Theorem (see Arzelà–Ascoli theorem)

[Theorem 47.1] (Ascoli's theorem, general version). Let $X$ be a space and $(Y,d)$ be a metric space. Give $\mathcal{C}(X,Y)$ the topology of compact convergence; let $\mathcal{F}$ be a subset of $\mathcal{C}(X,Y)$.

(a) If $\mathcal{F}$ is equicontinuous under $d$ and the set

$$\mathcal{F}_a = \{ f(a) : f \in \mathcal{F} \}$$

has compact closure for each $a \in X$, then $\mathcal{F}$ is contained in a compact subspace of $\mathcal{C}(X,Y)$.

(b) The converse holds if $X$ is locally compact Hausdorff.

Proof of (a). Throughout, we give $Y^X$ the product topology. Recall that this is the same as the topology of pointwise convergence, so $Y^X$ is a Hausdorff space. The space $\mathcal{C}(X,Y)$, which has the topology of compact convergence, is not a subspace of $Y^X$. Let $\mathcal{G}$ be the closure of $\mathcal{F}$ in $Y^X$.

Step 1. We show that $\mathcal{G}$ is a compact closure of $Y^X$.

Step 2. We show that each function belonging to $\mathcal{G}$ is continuous, and that $\mathcal{G}$ itself is equicontinuous under $d$.

Step 3. We show that the product topology on $Y^X$ and the compact convergence topology on $\mathcal{C}(X,Y)$ coincide on the subset $\mathcal{G}$.

Step 4. The set $\mathcal{G}$ contains $\mathcal{F}$ and is contained in $\mathcal{C}(X,Y)$. We know from Step 1 that it is compact as a subspace of $Y^X$ in the product topology. By Step 3, it is also compact as a subspace of $\mathcal{C}(X,Y)$ in the compact convergence topology.

Proof of (b). Let $\mathcal{H}$ be a compact subspace of $\mathcal{C}(X,Y)$ that contains $\mathcal{F}$. We show that $\mathcal{H}$ is equicontinuous and that for each $a \in X$, the set $\mathcal{H}_a$ is compact, so that $\mathcal{F}$ is equicontinuous (since $\mathcal{F} \subseteq \mathcal{H})$ and that $\mathcal{F}_a$ lies in the compact subspace $\mathcal{H}_a$ of $Y$, implying that $\overline{\mathcal{F}_a}$ is compact.