I'm the book Linear functional analysis by Rynne and Youngson, Second edition gives a summary of topology of metric spaces concepts
Definition 1.41.
A set $X$ is countable if it contains either a finite number of elements or infinitely many elements and can be written in the form $X = \{x_n : n \in \mathbb{N}\}$; in the latter case X is said to be countably infinite.
A metric space $(M, d)$ is separable if it contains a countable, dense subset.
The empty set $\phi$ is regarded as separable.
Then I started wondering if the empty set being separable was consistent:
Since $\phi$ is separable, and the only set contained in it is $\phi$ itself, it must be dense (I can also proof that arguing that the empty set is always closed in any topology, so it is clearly equal to its closure , hence dense in itself) and countable. Since any space S contains the empty set, I have found that S contains a dense countable set, so by definition of separable space, S is separable. S was arbitrary, then any space is separable.
This can't be true,cleary non-separable spaces do exist, but then what is wrong with my proof? It seems to me that declaring the empty set as separable is inconsistent. Perhaps I am missunderstanding the statement or I am supposed to assume something
Best Answer
A space $X$ is seperable if it contains a countable subset $A$ such that $\overline{A}=X$; in other words A is dense in $X$.
Clearly if $X$ is non empty then the empty set is never dense in $X$.