ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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The functions $f_{n} (k) = 1$ for $k \geq n$ and $0$ otherwise converge pointwise but not uniformly on $N$

Example

Give ${0, 1}$ the uniformity of its zero-one metric. For $n, k \in N$ , let $f_{n} (k) = 1$ when $k \geq n$ and $f_{n} (k) = 0$ otherwise. Then $f_{n}$ converges pointwise, but not uniformly, to the zero function.

Facts & Assumptions

Given: The function set ${0, 1}^{N}$ .

[L1]

For $d (u, v) = 0$ when $u = v$ and $d (u, v) = 1$ otherwise, separation and symmetry are immediate, while the triangle inequality follows because $u \neq w$ forces $u \neq v$ or $v \neq w$ . Thus $d$ is a metric; its radius- $1 / 2$ balls are singletons, so its topology is discrete, and its radius- $1 / 2$ entourage is equality (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, A metric on a nonempty set generates an entourage uniformity whose induced topology and uniformly continuous maps are the usual metric notions, and this uniformity is separated, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L2]

Pointwise and uniform convergence are the two function-space uniformities of The pointwise and uniform-convergence uniformities on a function set $Y^{X}$ .

[L3]

Sequence convergence means eventual membership in every neighbourhood (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

Verification

technique · direct

1.1

For fixed $k$ , all $n > k$ have $f_{n} (k) = 0$ , so the coordinate sequence converges to $0$ .

L1L3

1.2

For every $n$ , $f_{n} (n) = 1$ , so $f_{n}$ is not in the uniform entourage induced by equality on ${0, 1}$ .

L1L2

2.1

Thus $f_{n}$ converges pointwise to zero by [L2].

step 1.1L2

3.1

Hence convergence is not uniform.

step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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The functions fn(k)=1fn​(k)=1 for k≥nk≥n and 00 otherwise converge pointwise but not uniformly on NN