Metamath Proof Explorer


Theorem metcld

Description: A subset of a metric space is closed iff every convergent sequence on it converges to a point in the subset. Theorem 1.4-6(b) of Kreyszig p. 30. (Contributed by NM, 11-Nov-2007) (Revised by Mario Carneiro, 1-May-2014)

Ref Expression
Hypothesis metcld.2 J = MetOpen D
Assertion metcld D ∞Met X S X S Clsd J x f f : S f t J x x S

Proof

Step Hyp Ref Expression
1 metcld.2 J = MetOpen D
2 1 mopntop D ∞Met X J Top
3 1 mopnuni D ∞Met X X = J
4 3 sseq2d D ∞Met X S X S J
5 4 biimpa D ∞Met X S X S J
6 eqid J = J
7 6 iscld4 J Top S J S Clsd J cls J S S
8 2 5 7 syl2an2r D ∞Met X S X S Clsd J cls J S S
9 19.23v f f : S f t J x x S f f : S f t J x x S
10 simpl D ∞Met X S X D ∞Met X
11 simpr D ∞Met X S X S X
12 1 10 11 metelcls D ∞Met X S X x cls J S f f : S f t J x
13 12 imbi1d D ∞Met X S X x cls J S x S f f : S f t J x x S
14 9 13 bitr4id D ∞Met X S X f f : S f t J x x S x cls J S x S
15 14 albidv D ∞Met X S X x f f : S f t J x x S x x cls J S x S
16 dfss2 cls J S S x x cls J S x S
17 15 16 bitr4di D ∞Met X S X x f f : S f t J x x S cls J S S
18 8 17 bitr4d D ∞Met X S X S Clsd J x f f : S f t J x x S