Theorem istotbnd2

Description: The predicate "is a totally bounded metric space." (Contributed by Jeff Madsen, 2-Sep-2009)

	Ref	Expression
Assertion	istotbnd2	⊢ ( 𝑀 ∈ ( Met ‘ 𝑋 ) → ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) ↔ ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( ∪ 𝑣 = 𝑋 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) ) )

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	baib	⊢ ( 𝑀 ∈ ( Met ‘ 𝑋 ) → ( 𝑀 ∈ ( TotBnd ‘ 𝑋 ) ↔ ∀ 𝑑 ∈ ℝ+ ∃ 𝑣 ∈ Fin ( ∪ 𝑣 = 𝑋 ∧ ∀ 𝑏 ∈ 𝑣 ∃ 𝑥 ∈ 𝑋 𝑏 = ( 𝑥 ( ball ‘ 𝑀 ) 𝑑 ) ) ) )