Function Space Topologies and the Exponential Law

Let $X$ be a set and let $(Y, \mathcal{T}_Y)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). Write

$Y^{X} \;:=\; \prod_{x \in X} Y ,$

the product of the constant family whose factor at every index $x \in X$ is $Y$ (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Unwinding that definition, an element of $Y^{X}$ is a function with domain $X$ taking its value at each $x$ in $Y$; so $Y^{X}$ is the set of all functions $X \to Y$, and the projection at $x$ is evaluation,

$\pi_x : Y^{X} \to Y, \qquad \pi_x(f) = f(x) .$

The topology of pointwise convergence on $Y^{X}$ is the product topology: the initial topology of the family $(\pi_x)_{x \in X}$ (The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology), that is the topology generated by the subbasis

$\mathcal{G} \;:=\; \{\, \pi_x^{-1}[V] \;:\; x \in X,\ V \in \mathcal{T}_Y \,\}, \qquad \pi_x^{-1}[V] = \{\, f \in Y^{X} : f(x) \in V \,\} .$

By A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis the finite intersections of members of $\mathcal{G}$ form a basis for it (Basis and subbasis for a topology, and the topology generated by a family of sets), so the basic open sets are exactly the sets

$\{\, f \in Y^{X} \;:\; f(x_j) \in V_j \text{ for every } j < n \,\} \qquad (n \in \mathbb{N},\ x_0, \dots, x_{n-1} \in X,\ V_0, \dots, V_{n-1} \in \mathcal{T}_Y),$

the value $n = 0$ giving the empty intersection $Y^{X}$ itself. A basic open set therefore constrains a member of $Y^{X}$ at finitely many points only, and that is the whole content of the topology.

The restriction to the continuous maps. Suppose in addition that $X$ carries a topology, and write

$C(X,Y) \;:=\; \{\, f \in Y^{X} : f \text{ is continuous} \,\}$

(Continuity of a map of topological spaces at a point and globally). The topology of pointwise convergence on $C(X,Y)$ is the subspace topology inherited from $Y^{X}$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace); its subbasic open sets are the traces $\pi_x^{-1}[V] \cap C(X,Y)$, since tracing carries a subbasis to a subbasis (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Nothing on this page gives $C(X,Y)$ a default topology. The set $C(X,Y)$ carries several different topologies below, and every statement names the one it means at the point of use.

• Why the name. A sequence converges in this topology exactly when it converges at every point of $X$; that is the next item, and it is what justifies calling the product topology on a set of functions the topology of pointwise convergence. Nothing about sequences is built into the definition, which is purely the product topology of The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space.

• The box topology is a different topology on the same set, and a finer one (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). It is not used here: the topology of pointwise convergence is the product topology, and the product topology is what has the characteristic property of A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice.

• $Y^{X}$ here carries no algebra. The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ writes $F^{X}$ for the same set of functions when the target is a field, and equips it with pointwise addition and scalar multiplication to make a vector space. That is a different structure on the same underlying set: nothing on this page uses those operations, and nothing on this page requires the target to be a field or even to have a single algebraic operation. Where both are in play the algebraic structure is named explicitly.

• Degenerate cases are not excluded. For $X = \varnothing$ the set $Y^{X}$ has exactly one element, the empty function, and carries its unique topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). For $Y$ empty and $X$ nonempty, $Y^{X}$ is empty. Both are consistent with the definition and neither is used below; the items on this page that need $X$ nonempty say so.

Let $X$ be a set, let $(Y, \mathcal{T}_Y)$ be a topological space, and give $Y^{X}$ the topology of pointwise convergence (The topology of pointwise convergence on $Y^{X}$, which is the product topology, and its restriction to $C(X,Y)$). Let $(f_k)$ be a sequence in $Y^{X}$ and let $f \in Y^{X}$. Then

$f_k \to f \text{ in } Y^{X} \qquad \Longleftrightarrow \qquad f_k(x) \to f(x) \text{ in } Y \text{ for every } x \in X ,$

convergence being that of Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure on both sides.

No uniqueness of limits is asserted on either side. In a general topological space a sequence may converge to several points, and the equivalence above is between two conditions on the pair $((f_k), f)$, not between two values (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure). No choice principle is used: the only selection made below is of a least natural number and of a maximum among finitely many.

• This is what the name of the topology records. The topology of pointwise convergence is defined as the product topology (The topology of pointwise convergence on $Y^{X}$, which is the product topology, and its restriction to $C(X,Y)$), with no reference to sequences; the lemma above is what makes the name accurate, and it is the reason the product topology, rather than the box topology, is the one used on a set of functions.

• The corresponding statement for the box topology is false. A basic box constrains a member of $Y^{X}$ at every index at once, so a sequence converging in the box topology must converge in a much stronger sense; the failure of the characteristic property of the box topology is recorded on the page that builds it (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

• Nothing here makes the pointwise topology well behaved for limits of continuous functions. A pointwise limit of continuous functions need not be continuous, so $C(X,Y)$ is in general not closed in $Y^{X}$ for this topology; that failure is what the uniform topology of this page repairs, and it is witnessed on the companion page.

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric), with balls as in Open ball, closed ball and sphere in a metric space and compact subsets as in Open cover, subcover, compact metric space, and compact subset of a metric space.

$(X,d)$ is locally compact if for every $x \in X$ there are a compact subset $K \subseteq X$ and a real $r > 0$ with

$B(x,r) \;\subseteq\; K .$

This is the condition "every point has a compact neighbourhood", written out. Give $X$ its metric topology $\mathcal{T}_d$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), so that $(X, \mathcal{T}_d)$ is a topological space (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). A set $K \subseteq X$ is a neighbourhood of $x$ in the sense of Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open exactly when some open $U$ satisfies $x \in U \subseteq K$, and by The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement that holds exactly when some ball $B(x,r)$ satisfies $B(x,r) \subseteq K$. So the displayed condition says precisely that $x$ has a compact neighbourhood, and the two readings are the same condition and not two notions.

Two conventions are fixed here, because both are live in the literature.

• Neighbourhoods need not be open. This library's convention is Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open's, and it is what makes "compact neighbourhood" a useful phrase at all: a compact set is rarely open. Note that The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement uses the word neighbourhood for an open set containing the point; that narrower usage is confined to that item, and the present definition never relies on it.
• Compactness of a subset is intrinsic. $K$ compact means the metric subspace $(K, d_K)$ is a compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space, Isometry, isometric embedding, and the subspace metric on a subset); the equivalent description by families of open subsets of the ambient $X$ is A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it and is cited wherever it is used.

Every compact metric space is locally compact, since $K := X$ and any $r > 0$ serve at every point. The empty metric space is locally compact, the condition being vacuous.

• Dictionary: this was the metric special case of a notion this library did not yet define in general, and now does. Locally compact is ordinarily defined for an arbitrary topological space, by the same words: every point has a compact neighbourhood, compactness there being the open-cover condition for arbitrary topological spaces. General topological compactness is now available at this point in the reading order, alongside the metric notion (Open cover, subcover, compact metric space, and compact subset of a metric space) this page uses, so the definition above is stated for a metric space and for nothing else, and never claimed to be the general one.

  The agreement is immediate and not a theorem about metrics. A metric space is a topological space whose topology is $\mathcal{T}_d$ (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and the metric topology is the topology: the open sets of $(X,d)$ used by Open cover, subcover, compact metric space, and compact subset of a metric space and by A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it are literally the members of $\mathcal{T}_d$. So a subset of $X$ is compact in the metric sense exactly when it is compact in the topological sense, and "has a compact neighbourhood" means the same thing on both sides. This agreement is now recorded explicitly, exactly as Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not records the agreement of the metric and topological notions of neighbourhood, closure, convergence and continuity: see For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide, which discharges the standing obligation this Remark used to record. Stating both here and there is what stops the library from acquiring two unrelated notions under one name.

• What local compactness is used for on this page. Two things, and only two: continuity of the evaluation map, and the converse half of the exponential law. The compact-open topology itself, the comparison of the three topologies, the uniform limit theorem, completeness and Dini's theorem all do without it.

• The real line is locally compact and the rationals are not. For $\mathbb{R}$ with its usual metric a closed bounded interval around a point is compact (Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), which is the hypothesis discharged. For $\mathbb{Q}$ with the metric of $\mathbb{R}$ it fails at every point, and that failure is exactly what the false statement about the evaluation map on this page exploits.

Let $(X,d)$ be a locally compact metric space (Locally compact metric space: every point has a compact neighbourhood) and let $x \in X$. Then there is a real $r_0 > 0$ such that

1. $\bar B(x,r)$ is a compact subset of $X$ (Open ball, closed ball and sphere in a metric space, Open cover, subcover, compact metric space, and compact subset of a metric space) for every real $r$ with $0 < r < r_0$; and
2. the family $\{\, \bar B(x,r) : 0 < r < r_0 \,\}$ is a neighbourhood base at $x$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open) consisting of compact sets: every neighbourhood of $x$ contains one of these closed balls.

Note that the closed ball itself is compact, not merely its closure; a closed ball is already closed (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed). No choice principle is used.

• Where local compactness is spent. Once, at step 1.1, to produce a single compact set with nonempty interior around $x$. Everything after that is the hereditary behaviour of compactness: a closed subset of a compact space is compact (A closed subset of a compact metric space is compact), and closed balls are closed (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

• The bound $r < r_0$ is not cosmetic. The closed ball $\bar B(x,r_0)$ itself need not be contained in $K$, since $B(x,r_0) \subseteq K$ says nothing about points at distance exactly $r_0$, so the argument is run strictly below $r_0$. That costs nothing, because arbitrarily small radii are what the neighbourhood base of claim 2 needs.

• Compactness of every closed ball is a strictly stronger property. The lemma asserts compactness of the small closed balls at each point, with the threshold depending on the point. In $\mathbb{R}^{n}$ every closed ball is compact (Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), but that is a feature of $\mathbb{R}^{n}$ and not a consequence of local compactness.

Let $(X,d)$ be a metric space carrying its metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), let $(Z, \mathcal{T}_Z)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and give $X \times Z$ the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Let $K \subseteq X$ be a compact subset (Open cover, subcover, compact metric space, and compact subset of a metric space), let $z_0 \in Z$, and let $N \subseteq X \times Z$ be open with

$K \times \{z_0\} \;\subseteq\; N .$

Then there is an open $W \subseteq Z$ with $z_0 \in W$ and

$K \times W \;\subseteq\; N .$

The set $K \times W$ is the tube of the name. The case $K = \varnothing$ is included and is settled by $W = Z$. No choice principle is used at all: the cover produced below is indexed by pairs of open sets, so the ambient form of compactness returns the second entries together with the indices and nothing has to be selected afterwards.

• Why the pairs and not the open sets. A single open $U \subseteq X$ may be the first entry of many admissible pairs, and recovering a suitable $W$ from $U$ alone would be a selection over an infinite family. Indexing the cover by the pairs rather than by the sets is what makes the second entries come back with the indices, and it is the same device the ambient form of compactness uses (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

• Compactness of the metric factor is what the lemma is about. No hypothesis whatever is placed on $Z$, and none is needed: the finite intersection of the $W_j$ is taken in $Z$ and uses only axiom (T3). What compactness of $K$ buys is that finitely many boxes already cover the slice $K \times \{z_0\}$, so that finitely many second entries have to be intersected.

• The hypothesis cannot be moved to the other factor. With $K$ replaced by a non-compact set the conclusion fails: the region under the graph of a positive function tending to $0$ contains a whole slice and no tube around it. Nothing on this page needs that witness, and it is not constructed here.

• The general tube lemma, for a compact factor in an arbitrary topological product, is now available in this library, on an earlier page (Tube lemma: if $K$ is compact and an open $N \subseteq X \times Z$ contains $K \times \{z_0\}$, then $N$ contains $K \times W$ for some open $W \ni z_0$). The proof above is the metric special case of that general lemma, narrowed to a metric factor $X$ and written independently of it: nothing above cites the general statement, and nothing needs to, since compactness of a metric-space subset is the same notion under either reading (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) carrying its metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), let $(Y, \mathcal{T}_Y)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and let

$C(X,Y) \;=\; \{\, f : X \to Y \;:\; f \text{ is continuous} \,\}$

(Continuity of a map of topological spaces at a point and globally). For a compact subset $K \subseteq X$ (Open cover, subcover, compact metric space, and compact subset of a metric space) and an open $V \in \mathcal{T}_Y$ put

$S(K,V) \;:=\; \{\, f \in C(X,Y) \;:\; f[K] \subseteq V \,\} \;=\; \{\, f \in C(X,Y) : f(x) \in V \text{ for every } x \in K \,\} .$

The compact-open topology on $C(X,Y)$ is the topology generated by

$\mathcal{S}_{\mathrm{co}} \;:=\; \{\, S(K,V) \;:\; K \subseteq X \text{ compact},\ V \in \mathcal{T}_Y \,\}$

as a subbasis (Basis and subbasis for a topology, and the topology generated by a family of sets). By A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis the finite intersections

$S(K_0,V_0) \cap \dots \cap S(K_{n-1},V_{n-1}) \qquad (n \in \mathbb{N})$

form a basis for it, the value $n = 0$ giving the empty intersection $C(X,Y)$. Nothing has to be checked for this to be a topology: a topology generated by an arbitrary family exists and is the coarsest one containing it (Basis and subbasis for a topology, and the topology generated by a family of sets).

Two degenerate members, recorded because they are used. $S(\varnothing, V) = C(X,Y)$ for every open $V$, the empty set being compact and $f[\varnothing] = \varnothing$; and $S(K, Y) = C(X,Y)$ for every compact $K$. Both are the whole space, so neither constrains anything, and arguments below dispose of them separately rather than dividing by a distance that does not exist.

The domain is metric, and the target is not. Compactness of $K$ is Open cover, subcover, compact metric space, and compact subset of a metric space, which is defined for subsets of a metric space and, at this point in the reading order, for nothing else; that is why $X$ carries a metric here. The target $Y$ needs only its open sets, so it is an arbitrary topological space throughout this definition and wherever the compact-open topology alone is in play. Where a distance in the target is used — the uniform metric, compact convergence, the comparison theorem — $Y$ is required to be metric and the requirement is stated.

Compactness of $K$ is intrinsic (Open cover, subcover, compact metric space, and compact subset of a metric space): it means that the metric subspace $(K, d_K)$ is a compact metric space. The equivalent description by families of open subsets of $X$ covering $K$ is A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, and it is cited at every step that uses it.

Notation. The letter $S$ carries two unrelated meanings in this library: $S(x,r)$ is the sphere of centre $x$ and radius $r$ in a metric space (Open ball, closed ball and sphere in a metric space), and $S(K,V)$ is the set defined above. The two are never ambiguous, because the first argument of a sphere is a point and its second a positive real, while the first argument of $S(K,V)$ is a compact set and its second an open set; no item on this page writes a sphere.

• Why compact sets and open sets. $S(K,V)$ says "$f$ maps all of $K$ into $V$". Taking $K$ to be a single point recovers the subbasis of the topology of pointwise convergence (The topology of pointwise convergence on $Y^{X}$, which is the product topology, and its restriction to $C(X,Y)$), so the compact-open topology is at least as fine as that one; taking $K$ large makes the condition a uniform one over $K$, which is what the comparison with the topology of compact convergence on this page makes precise.

• The definition is on $C(X,Y)$ and not on $Y^{X}$. The sets $S(K,V)$ could be written down for arbitrary functions, but the theory of this page uses that $f[K]$ is compact when $K$ is and $f$ is continuous (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset), which is false for a discontinuous $f$. The compact-open topology in this library is therefore a topology on continuous maps only.

• Metrizability is not asserted. The compact-open topology need not be metrizable, and this page records that as a false statement with an explicit witness. What is proved here is that for a metric target it coincides with the topology of compact convergence, which for a suitable $X$ is metrizable by a metric this library does not construct.

Let $X$ be a nonempty set, let $(Y,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and write

$\bar d(u,v) \;:=\; \min\{\, d(u,v),\ 1 \,\} \qquad (u, v \in Y),$

which is a metric on $Y$ with $\bar d \le 1$ everywhere ($\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$, so every metric space carries a bounded metric with the same topology, claims 1 and 2). For $f, g \in Y^{X}$ (The topology of pointwise convergence on $Y^{X}$, which is the product topology, and its restriction to $C(X,Y)$) put

$R(f,g) \;:=\; \{\, \bar d\big(f(x), g(x)\big) : x \in X \,\} \subseteq \mathbb{R}, \qquad \bar\rho(f,g) \;:=\; \sup R(f,g) .$

This is well defined: $R(f,g)$ is nonempty because $X$ is, and $1$ is an upper bound of it, so the least upper bound exists (Complete ordered field (least-upper-bound property)) and is unique (Suprema and infima are unique).

Then $\bar\rho$ is a metric on $Y^{X}$ (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric), the uniform metric, and $\bar\rho(f,g) \le 1$ for all $f, g$.

Both hypotheses are used and neither is decoration. Nonemptiness of $X$ is what makes $R(f,g)$ nonempty; for $X=\varnothing$ the set $Y^X$ has a single element, but $\sup\varnothing$ is undefined under the real-valued supremum convention used here (Conventions: $\sup \emptyset$, unbounded sets, and the extended reals). The extended real line is introduced later and is not the codomain of this metric. Truncating $d$ at $1$ is what makes $R(f,g)$ bounded above with no boundedness hypothesis on $f$ and $g$; that is the whole reason the truncation is there.