A toplogical space $X$ is said to be second countable
if there exists a countable basis for the topology.$X$ is separable if there exist a countable dense subset.
Show that a second countable space always is separable.Give an example of a space that is separable but not
second countable
The first part was easy and i managed to solve it on my own
Assume that $X$ is second countable and let $B_1, B_2, B_3…$
be a countable basis. Construct the set $A$ by taking one
element $x_n$ out of every $B_n$. Then $A$ becomes dense because
every open set $U$ contain a $B_n$ which in turn contain the
element $x_n$ in $A$.
Again, like in the question i posted yesterday, i had trouble to come up with an example but our
professor gave an example during the last lecture but
i dont fully understand it:
"For an example of a separable but not second countable
space let $X$ be the real line provided with the cofinite
topology, i.e. where a set is open if its either empty
or if its complement is finite.
Then every infinite set is dense and hence $X$ is separable.
But for an arbitrary countable family of open sets
$U_1, U_2…$ write $U_i = X$\ $A_i$, where $A_i$ is finite,
there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,
such that the open set $X$\ $\{x\}$ is then not able
to contain any $U_i$. Consequently $X$ is not second countable."
We asked him to explain it in more detail but his english is poor
and nobody understood anything…:)
Def: A subset $A$ of a topological space $X$ is dense in $X$ if for any point $x$ in $X$, any neighborhood of $x$ contains at least one point from $A$.
$1.$ "Then every infinite set is dense"
Take $A$ to be one of these infinite sets. If $x \in A \subset X$ then it trivially fulfills the definition. But if we take a $x \not\in A$ why is it not possible to find a neighborhood in these finite sets that does not contain an element in $A$?
edit: its obvious if it is only open neighborhoods, open sets in the topology, but that is not specified in the definition.
$2.$ I am not fully understanding this either
"But for an arbitrary countable family of open sets
$U_1, U_2…$ write $U_i = X$\ $A_i$, where $A_i$ is finite,
there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,
such that the open set $X$\ $\{x\}$ is then not able
to contain any $U_i$. Consequently $X$ is not second countable."
Is it possible to express this in a simpler way?
Best Answer
Take your favorite uncountable set $X$ and fix $x\in X$. Now consider the following topology: $$\{U\subseteq X\mid \varnothing\neq U\leftrightarrow x\in U\}$$
Namely, every non-empty open set contains $x$, and vice versa. This is not a second-countable topology, since a second-countable topology implies being Lindelöf, but $\{\{x,y\}\mid y\in X\}$ is an open cover without a countable subcover.
But $\{x\}$ is dense, since it meets every non-empty open set. So this topology is separable.