Metamath Proof Explorer


Theorem perfcls

Description: A subset of a perfect space is perfect iff its closure is perfect (and the closure is an actual perfect set, since it is both closed and perfect in the subspace topology). (Contributed by Mario Carneiro, 26-Dec-2016)

Ref Expression
Hypothesis lpcls.1 X = J
Assertion perfcls J Fre S X J 𝑡 S Perf J 𝑡 cls J S Perf

Proof

Step Hyp Ref Expression
1 lpcls.1 X = J
2 1 lpcls J Fre S X limPt J cls J S = limPt J S
3 2 sseq2d J Fre S X cls J S limPt J cls J S cls J S limPt J S
4 t1top J Fre J Top
5 1 clslp J Top S X cls J S = S limPt J S
6 4 5 sylan J Fre S X cls J S = S limPt J S
7 6 sseq1d J Fre S X cls J S limPt J S S limPt J S limPt J S
8 ssequn1 S limPt J S S limPt J S = limPt J S
9 ssun2 limPt J S S limPt J S
10 eqss S limPt J S = limPt J S S limPt J S limPt J S limPt J S S limPt J S
11 9 10 mpbiran2 S limPt J S = limPt J S S limPt J S limPt J S
12 8 11 bitri S limPt J S S limPt J S limPt J S
13 7 12 bitr4di J Fre S X cls J S limPt J S S limPt J S
14 3 13 bitr2d J Fre S X S limPt J S cls J S limPt J cls J S
15 eqid J 𝑡 S = J 𝑡 S
16 1 15 restperf J Top S X J 𝑡 S Perf S limPt J S
17 4 16 sylan J Fre S X J 𝑡 S Perf S limPt J S
18 1 clsss3 J Top S X cls J S X
19 eqid J 𝑡 cls J S = J 𝑡 cls J S
20 1 19 restperf J Top cls J S X J 𝑡 cls J S Perf cls J S limPt J cls J S
21 18 20 syldan J Top S X J 𝑡 cls J S Perf cls J S limPt J cls J S
22 4 21 sylan J Fre S X J 𝑡 cls J S Perf cls J S limPt J cls J S
23 14 17 22 3bitr4d J Fre S X J 𝑡 S Perf J 𝑡 cls J S Perf