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Metamath Proof Explorer
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 ) ) ) ) |