3.2.28. Show that x is a limit point [of] E iff 𝑑𝐸 𝑥 = 0. Conclude that 𝑥 ∈ 𝐴 iff 𝑑𝐴 𝑥 = 0.
In particular, if A is closed, then 𝑑𝐴 𝑥 = 0 iff 𝑥 ∈ 𝐴.
( ) Suppose x is a limit point of E.
If 𝑥 ∈ 𝐸, then 𝑑𝐸 𝑥 = 0. If 𝑥 ∉ 𝐸, then for  > 0, 𝐵(𝑥, )⋂𝐸 ≠ . So 𝑑𝐸 𝑥 < . As  is
made arbitrarily small, 𝑑𝐸 𝑥 < , so 𝑑𝐸 𝑥 = 0. In English; no matter how small  is, the
distance from x to E is less, and the only number smaller than something that is arbitrarily
small (and positive) is 0.
() Suppose 𝑑𝐸 𝑥 = 0.
Then  > 0, 𝐵(𝑥, )⋂𝐸 ≠  since 𝑑𝐸 𝑥 < . So x is a limit point of E. In English; the
distance from x to E is 0, so E is part of every neighborhood of x.
Furthermore, since the closure of A (𝐴) contains all its limit points, and since every element
of 𝐴 is a limit point, by the preceding, 𝑥 ∈ 𝐴 iff 𝑑𝐴 𝑥 = 0. Particularly, if A is closed, then
𝐴 = 𝐴, so 𝑑𝐴 𝑥 = 0 iff 𝑥 ∈ 𝐴.