Metamath Proof Explorer


Theorem neiss2

Description: A set with a neighborhood is a subset of the base set of a topology. (This theorem depends on a function's value being empty outside of its domain, but it will make later theorems simpler to state.) (Contributed by NM, 12-Feb-2007)

Ref Expression
Hypothesis neifval.1 𝑋 = 𝐽
Assertion neiss2 ( ( 𝐽 ∈ Top ∧ 𝑁 ∈ ( ( nei ‘ 𝐽 ) ‘ 𝑆 ) ) → 𝑆𝑋 )

Proof

Step Hyp Ref Expression
1 neifval.1 𝑋 = 𝐽
2 elfvdm ( 𝑁 ∈ ( ( nei ‘ 𝐽 ) ‘ 𝑆 ) → 𝑆 ∈ dom ( nei ‘ 𝐽 ) )
3 2 adantl ( ( 𝐽 ∈ Top ∧ 𝑁 ∈ ( ( nei ‘ 𝐽 ) ‘ 𝑆 ) ) → 𝑆 ∈ dom ( nei ‘ 𝐽 ) )
4 1 neif ( 𝐽 ∈ Top → ( nei ‘ 𝐽 ) Fn 𝒫 𝑋 )
5 4 fndmd ( 𝐽 ∈ Top → dom ( nei ‘ 𝐽 ) = 𝒫 𝑋 )
6 5 eleq2d ( 𝐽 ∈ Top → ( 𝑆 ∈ dom ( nei ‘ 𝐽 ) ↔ 𝑆 ∈ 𝒫 𝑋 ) )
7 6 adantr ( ( 𝐽 ∈ Top ∧ 𝑁 ∈ ( ( nei ‘ 𝐽 ) ‘ 𝑆 ) ) → ( 𝑆 ∈ dom ( nei ‘ 𝐽 ) ↔ 𝑆 ∈ 𝒫 𝑋 ) )
8 3 7 mpbid ( ( 𝐽 ∈ Top ∧ 𝑁 ∈ ( ( nei ‘ 𝐽 ) ‘ 𝑆 ) ) → 𝑆 ∈ 𝒫 𝑋 )
9 8 elpwid ( ( 𝐽 ∈ Top ∧ 𝑁 ∈ ( ( nei ‘ 𝐽 ) ‘ 𝑆 ) ) → 𝑆𝑋 )