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Metamath Proof Explorer

Theorem totbndss

Description: A subset of a totally bounded metric space is totally bounded. (Contributed by Jeff Madsen, 2-Sep-2009) (Revised by Mario Carneiro, 12-Sep-2015)

Ref Expression
Assertion totbndss ( ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) ∧ 𝑆 ⊆ 𝑋 ) → ( 𝑀 ↾ ( 𝑆 × 𝑆 ) ) ∈ ( TotBnd ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 istotbnd ⊢ ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) ↔ ( 𝑀 ∈ ( Met ‘ 𝑋 ) ∧ ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( ∪ 𝑣 = 𝑋 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) ) )
2 1 simprbi ⊢ ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) → ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( ∪ 𝑣 = 𝑋 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) )
3 sseq2 ⊢ ( ∪ 𝑣 = 𝑋 → ( 𝑆 ⊆ ∪ 𝑣 ↔ 𝑆 ⊆ 𝑋 ) )
4 3 biimprcd ⊢ ( 𝑆 ⊆ 𝑋 → ( ∪ 𝑣 = 𝑋 → 𝑆 ⊆ ∪ 𝑣 ) )
5 4 anim1d ⊢ ( 𝑆 ⊆ 𝑋 → ( ( ∪ 𝑣 = 𝑋 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) → ( 𝑆 ⊆ ∪ 𝑣 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) ) )
6 5 reximdv ⊢ ( 𝑆 ⊆ 𝑋 → ( ∃ 𝑣 ∈ Fin ( ∪ 𝑣 = 𝑋 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) → ∃ 𝑣 ∈ Fin ( 𝑆 ⊆ ∪ 𝑣 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) ) )
7 6 ralimdv ⊢ ( 𝑆 ⊆ 𝑋 → ( ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( ∪ 𝑣 = 𝑋 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) → ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( 𝑆 ⊆ ∪ 𝑣 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) ) )
8 2 7 mpan9 ⊢ ( ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) ∧ 𝑆 ⊆ 𝑋 ) → ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( 𝑆 ⊆ ∪ 𝑣 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) )
9 totbndmet ⊢ ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) → 𝑀 ∈ ( Met ‘ 𝑋 ) )
10 eqid ⊢ ( 𝑀 ↾ ( 𝑆 × 𝑆 ) ) = ( 𝑀 ↾ ( 𝑆 × 𝑆 ) )
11 10 sstotbnd ⊢ ( ( 𝑀 ∈ ( Met ‘ 𝑋 ) ∧ 𝑆 ⊆ 𝑋 ) → ( ( 𝑀 ↾ ( 𝑆 × 𝑆 ) ) ∈ ( TotBnd ‘ 𝑆 ) ↔ ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( 𝑆 ⊆ ∪ 𝑣 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) ) )
12 9 11 sylan ⊢ ( ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) ∧ 𝑆 ⊆ 𝑋 ) → ( ( 𝑀 ↾ ( 𝑆 × 𝑆 ) ) ∈ ( TotBnd ‘ 𝑆 ) ↔ ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( 𝑆 ⊆ ∪ 𝑣 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) ) )
13 8 12 mpbird ⊢ ( ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) ∧ 𝑆 ⊆ 𝑋 ) → ( 𝑀 ↾ ( 𝑆 × 𝑆 ) ) ∈ ( TotBnd ‘ 𝑆 ) )