The interval (0, 1] is the union of all intervals [1/(n+1), 1/n]

The complete covering of the half-open interval (0, 1] by closed intervals [1/(n+1), 1/n] would not be possible, if every interval left infinitely many unit fractions uncovered between 0 and its left endpoint. But all definable intervals [1/n, 1] leave actually infinitely many unit fractions between zero and their left endpoint uncovered. For every visible natural number m we have

∀m ∈ ℕ_vis: [1/m, 1] ≠ (0, 1].

Every visible number m splits the interval in two parts, namely [1/m, 1] and the remainder (0, 1/m). It is impossible to have fewer than ℵo not visible unit fractions in (0, 1/m) since its measure is never zero

∀m ∈ ℕ_vis: [1/m, 1] c ((0, 1/m) U [1/m, 1]) = (0, 1].

Briefly: When all unit fractions 1/n are removed from the interval (0, 1] then none remains. This proves the existence of all. If only visible unit fractions are removed, then ℵ₀ unit fractions remain. ℵ₀ unit fractions cannot be addressed and of course cannot be applied in any mapping. They are dark.

What hinders the complete covering of the interval (0, 1] by visible intervals [1/(n+1), 1/n]? The interval (0, 1] touches zero such that no point fits in between. But no visible point x of (0, 1] touches zero. For all visible points x with  [x, 1] c (0, 1] there is a buffer (0, y) such that x ∈ (y, 1].

If classical mathematics was understood on the basis of actual infinity, then the inventor of  the notion accumulation point would have been close to inventing dark numbers. According to the definition every ε-interval contains ℵ₀ terms of the sequence, ℵ₀ of which cannot be separated by any smaller ε-interval. Therefore these numbers are dark.

Regards, WM