Is the closure of a discrete subspace of $[0,1]$ necessarily countable?

Question

Question is in the title.

According to this, every discrete subspace of $[0,1]$ must be countable. But what about its closure?

See also this question. – Noah Schweber Jun 24 '17 at 23:05 — Noah Schweber, Jun 24 '17 at 23:05

hmakholm left over Monica · Accepted Answer · 2017-06-24T21:27:22.750

11

Consider the middle-thirds Cantor set, and take the midpoint of every interval that is removed during its construction.

These midpoints evidently form a discrete set.

However, every point in the Cantor set is the limit of a sequence of midpoints, and the Cantor set is uncountable.

edited Jun 24 '17 at 21:27

answered Jun 24 '17 at 21:21

In fact, every closed nowhere dense subset of $[0,1]$ is the closure of a discrete set. – Henno Brandsma Jun 25 '17 at 21:27

1 Answers1