7

I have a general question on how to prove a space is not separable.

I read some posts on this site and it seems like it suffices to find an uncountable family of pairwise disjoint open sets to prove a space is not separable. (here: The space of bounded continuous functions are not separable)

Why an uncountable family of pairwise disjoint open sets is enough?

Community
  • 1
Kenneth.K
  • 1,485
  • 1
  • 16
  • 36
  • 1
    Imagine that a dense subset exists, then it intersects every subset on the family you described, so it can't be countable. – Smurf Feb 09 '17 at 19:44

2 Answers2

13

A separable space $X$ is a space with a countable dense subset $S$. Label the elements of the $S$ with $x_i$. Since $S$ is dense, $\bar{S}=X$. Thus any open set $U$ must contain some $x_i$. However, if $X$ contains uncountably many disjoint opens, this is not possible, as there must be some $U_\alpha$ not containing any of the $x_i$.

A Nonny Mouse
  • 886
3

Let $(U_i)_{i\in I}$be a family of pairwise disjoints open sets, where $I$ is uncountable. If the space is separable , fix $D=\{x_n| n\in \Bbb N\}$ a dense subset.

For each $i\in I$, choose (there is) $n_i\in \Bbb N$, such that $x_{n_i}\in U_i$.

No the map $I\to \Bbb N$, $i\mapsto n_i$ is injective.

Hamou
  • 6,915