LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A sequence converges in the topology of pointwise convergence exactly when it converges at every point

Statement

Let $X$ be a set, let $(Y, T_{Y})$ 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

$f_{k} \to f in Y^{X} ⟺ f_{k} (x) \to f (x) in Y for every x \in X,$

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.

Facts & Assumptions

Given: A set $X$ , a topological space $(Y, T_{Y})$ , the space $Y^{X}$ with the topology of pointwise convergence, a sequence $(f_{k})$ in $Y^{X}$ and a point $f \in Y^{X}$ ; $ι$ is the canonical natural of $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L1]

For $x \in X$ and $V \in T_{Y}$ the set $π_{x}^{- 1} [V] = {g \in Y^{X} : g (x) \in V}$ is open in $Y^{X}$ , and the sets ${g \in Y^{X} : g (x_{j}) \in V_{j} for every j < n}$ , for $n \in N$ , points $x_{0}, \dots, x_{n - 1} \in X$ and open $V_{0}, \dots, V_{n - 1} \subseteq Y$ , form a basis for the topology of pointwise convergence (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ , 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, 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).

[L2]

A set $N$ is a neighbourhood of a point $p$ exactly when there is an open $U$ with $p \in U \subseteq N$ ; in particular an open set containing $p$ is a neighbourhood of $p$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[L3]

$g_{k} \to g$ in a topological space means: for every neighbourhood $N$ of $g$ there is $K \in N$ with $g_{k} \in N$ for every $k \geq K$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

[L4]

If $B$ is a basis for a topology and $N$ is a neighbourhood of $g$ , then there is $B \in B$ with $g \in B \subseteq N$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, Basis and subbasis for a topology, and the topology generated by a family of sets).

[L5]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L6]

For $n \geq 1$ and natural numbers $k_{0}, \dots, k_{n - 1}$ there is an index $j^{*} < n$ with $k_{j} \leq k_{j^{*}}$ for every $j < n$ : the nonempty finite set of reals ${ι (k_{0}), \dots, ι (k_{n - 1})}$ has a maximum, attained at some index, and $ι$ is strictly increasing on $N$ , hence reflects the order (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Canonical naturals are positive and strictly increasing, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

Proof

technique · direct

1.1

Suppose $f_{k} \to f$ in $Y^{X}$ ; fix $x \in X$ and a neighbourhood $N$ of $f (x)$ in $Y$ , and fix an open $V \subseteq Y$ with $f (x) \in V \subseteq N$ .

assume-hypL2choose

1.2

Suppose instead that $f_{k} (x) \to f (x)$ in $Y$ for every $x \in X$ , and let $N$ be a neighbourhood of $f$ in $Y^{X}$ .

assume-hypL2

2.1

Under the assumption of step 1.1: $π_{x}^{- 1} [V]$ is open in $Y^{X}$ and contains $f$ , hence is a neighbourhood of $f$ , so there is $K \in N$ with $f_{k} \in π_{x}^{- 1} [V]$ for every $k \geq K$ , that is $f_{k} (x) \in V \subseteq N$ for every $k \geq K$ .

step 1.1L1L2L3

2.2

Under the assumption of step 1.2: there are $n \in N$ , points $x_{0}, \dots, x_{n - 1} \in X$ and open $V_{0}, \dots, V_{n - 1} \subseteq Y$ such that $f \in B \subseteq N$ , where $B : = {g \in Y^{X} : g (x_{j}) \in V_{j} for every j < n}$ .

step 1.2L1L4choose

3.1

Since $N$ was an arbitrary neighbourhood of $f (x)$ and $x$ an arbitrary point of $X$ , step 2.1 says exactly that $f_{k} (x) \to f (x)$ in $Y$ for every $x \in X$ ; this is the forward implication.

step 2.1L3

3.2

If $n = 0$ in step 2.2 then $B$ is the empty intersection $Y^{X}$ , so $f_{k} \in B \subseteq N$ for every $k \in N$ .

step 2.2L1

3.3

If $n \geq 1$ in step 2.2 then for each $j < n$ the set $A_{j} : = {m \in N : f_{k} (x_{j}) \in V_{j} for every k \geq m}$ is nonempty, because $f \in B$ gives $f (x_{j}) \in V_{j}$ with $V_{j}$ open, hence $V_{j}$ is a neighbourhood of $f (x_{j})$ , and $f_{k} (x_{j}) \to f (x_{j})$ ; put $N_{j} : = \min A_{j}$ .

step 1.2step 2.2L2L3L5

4.1

If $n \geq 1$ : there is $j^{*} < n$ with $N_{j} \leq N_{j^{*}}$ for every $j < n$ , and then every $k \geq N_{j^{*}}$ satisfies $k \geq N_{j}$ for every $j < n$ , so $f_{k} (x_{j}) \in V_{j}$ for every $j < n$ , that is $f_{k} \in B \subseteq N$ .

step 2.2step 3.3L6

5.1

By steps 3.2 and 4.1 there is in either case a $K \in N$ with $f_{k} \in N$ for every $k \geq K$ , namely $K = 0$ when $n = 0$ and $K = N_{j^{*}}$ when $n \geq 1$ ; as $N$ was an arbitrary neighbourhood of $f$ , this says $f_{k} \to f$ in $Y^{X}$ , which is the converse implication.

step 3.2step 4.1L3

6.1

Steps 3.1 and 5.1 are the two implications, so the two conditions are equivalent.

step 3.1step 5.1∎

Remarks

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.

Depends on

Used by

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Topology of pointwise convergence (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §46 (standard reference, not scraped)