ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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The sequential fan is Fréchet–Urysohn and not first countable

Example

Let $S_\omega=(\mathbb N\times\mathbb N)\cup\{\infty\}$ , with all $(n,m)$ isolated. A neighbourhood of $\infty$ contains $\infty$ and, for every $n$ , all but finitely many $(n,m)$ on the $n$ -th spoke. This is the sequential fan. It is Fréchet–Urysohn but not first countable.

Facts & Assumptions

Given: The sequential fan and a subset $A\subseteq S_\omega$ .

[A1]

A space is Fréchet–Urysohn when closure points are limits of sequences from the set, and first countability means a countable local base (Fréchet–Urysohn spaces and sequential spaces, First countable space: a countable neighbourhood base at every point).

[L1]

Every nonempty finite subset of $\mathbb N$ has a maximum, every nonempty subset of $\mathbb N$ has a least member, and recursion defines sequences from uniquely specified successive terms (Every nonempty finite set of reals has a maximum and a minimum, The well-ordering principle, The recursion theorem).

Verification

technique · constructive

1.1

Suppose $\infty\in\overline A$ . If every spoke met $A$ only finitely, define $f(n)= \begin{cases} 0,&\{m:(n,m)\in A\}=\varnothing,\\ 1+\max\{m:(n,m)\in A\},&\text{otherwise}. \end{cases}$ This is a canonically defined function by [L1], and the neighbourhood containing on spoke $n$ exactly the points $(n,m)$ with $m\ge f(n)$ misses $A$ , a contradiction. Hence one spoke meets $A$ infinitely.

A1L1construct

1.2

Suppose $(B_k)$ were a countable neighbourhood base at $\infty$ . For each $k,n$ , let $f_k(n)$ be the least threshold such that $(n,m)\in B_k$ for every $m\ge f_k(n)$ ; it exists and is unique by [L1]. Form the neighbourhood whose threshold on spoke $k$ is $g(k)=f_k(k)+1$ .

A1L1construct

2.1

On the infinite spoke supplied by step 1.1, recursion and least elements from [L1] list the second coordinates increasingly. The resulting sequence in $A$ is eventually beyond every threshold on that spoke, hence converges to $\infty$ . Isolated closure points already lie in $A$ , so $S_\omega$ is Fréchet–Urysohn.

step 1.1A1L1

2.2

The point $(k,f_k(k))$ lies in $B_k$ but not in this neighbourhood, so no $B_k$ is contained in it. This contradicts the base property.

step 1.2A1

3.1

Therefore the sequential fan is Fréchet–Urysohn and not first countable.

step 2.1step 2.2discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

D. Ma, A note on products of sequential fans (Dan Ma's Topology Blog) (standard reference, not scraped)
Fréchet–Urysohn space (Wikipedia) (standard reference, not scraped)