Metamath Proof Explorer


Theorem recnperf

Description: Both RR and CC are perfect subsets of CC . (Contributed by Mario Carneiro, 28-Dec-2016)

Ref Expression
Hypothesis recnperf.k 𝐾 = ( TopOpen ‘ ℂfld )
Assertion recnperf ( 𝑆 ∈ { ℝ , ℂ } → ( 𝐾t 𝑆 ) ∈ Perf )

Proof

Step Hyp Ref Expression
1 recnperf.k 𝐾 = ( TopOpen ‘ ℂfld )
2 elpri ( 𝑆 ∈ { ℝ , ℂ } → ( 𝑆 = ℝ ∨ 𝑆 = ℂ ) )
3 oveq2 ( 𝑆 = ℝ → ( 𝐾t 𝑆 ) = ( 𝐾t ℝ ) )
4 1 reperf ( 𝐾t ℝ ) ∈ Perf
5 3 4 eqeltrdi ( 𝑆 = ℝ → ( 𝐾t 𝑆 ) ∈ Perf )
6 oveq2 ( 𝑆 = ℂ → ( 𝐾t 𝑆 ) = ( 𝐾t ℂ ) )
7 1 cnfldtopon 𝐾 ∈ ( TopOn ‘ ℂ )
8 7 toponunii ℂ = 𝐾
9 8 restid ( 𝐾 ∈ ( TopOn ‘ ℂ ) → ( 𝐾t ℂ ) = 𝐾 )
10 7 9 ax-mp ( 𝐾t ℂ ) = 𝐾
11 1 cnperf 𝐾 ∈ Perf
12 10 11 eqeltri ( 𝐾t ℂ ) ∈ Perf
13 6 12 eqeltrdi ( 𝑆 = ℂ → ( 𝐾t 𝑆 ) ∈ Perf )
14 5 13 jaoi ( ( 𝑆 = ℝ ∨ 𝑆 = ℂ ) → ( 𝐾t 𝑆 ) ∈ Perf )
15 2 14 syl ( 𝑆 ∈ { ℝ , ℂ } → ( 𝐾t 𝑆 ) ∈ Perf )