English
If edist(x,y) ≠ ∞ and edist(x,y) + ε1 ≤ ε2, then B(x,ε1) ⊆ B(y,ε2).
Русский
Если edist(x,y) ≠ ∞ и edist(x,y) + ε1 ≤ ε2, тогда B(x,ε1) ⊆ B(y,ε2).
LaTeX
$$$\\operatorname{edist}(x,y) \\neq \\infty \\wedge (\\operatorname{edist}(x,y) + \\varepsilon_1 \\le \\varepsilon_2) \\Rightarrow \\mathrm{ball}(x,\\varepsilon_1) \\subseteq \\mathrm{ball}(y,\\varepsilon_2)$$$
Lean4
theorem ball_subset (h : edist x y + ε₁ ≤ ε₂) (h' : edist x y ≠ ∞) : ball x ε₁ ⊆ ball y ε₂ := fun z zx =>
calc
edist z y ≤ edist z x + edist x y := edist_triangle _ _ _
_ = edist x y + edist z x := (add_comm _ _)
_ < edist x y + ε₁ := (ENNReal.add_lt_add_left h' zx)
_ ≤ ε₂ := h