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Assuming countable choice, every first countable space is Fréchet–Urysohn; in ZF every Fréchet–Urysohn space is sequential
Statement
Assume countable choice. Every first countable space is Fréchet–Urysohn. In ZF, every Fréchet–Urysohn space is sequential.
Facts & Assumptions
Given: A topological space .
Under countable choice, first countability gives for every (Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there, The Axiom of Countable Choice ()).
A space is Fréchet--Urysohn when for every subset , and it is sequential when every sequentially closed subset is closed (Fréchet–Urysohn spaces and sequential spaces).
Proof
Under countable choice, [L1] is exactly the defining equality for a first countable space to be Fréchet–Urysohn.
Now suppose is Fréchet–Urysohn and is sequentially closed. Then , because the constant sequence gives and sequential closedness gives the reverse inclusion.
Fréchet–Urysohnness gives , so is closed. Therefore is sequential.
Depends on
- Fréchet–Urysohn spaces and sequential spaces
- Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- The sequential closure is contained in the closure, continuity implies sequential continuity, and sequential limits need not be unique
Used by
Nothing in the library uses this result yet.
Dependency tree · next 3 levels
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Sources
- Sequential space (Wikipedia) (standard reference, not scraped)
- Fréchet–Urysohn space (Wikipedia) (standard reference, not scraped)
- First-countable space (Wikipedia) (standard reference, not scraped)