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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?

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Let $\tau$ be the cofinite topology on $\Bbb R$. The members of $\tau$ (i.e., the open sets in this space) are $\varnothing$, the empty set, and all subsets of $\Bbb R$ of the form $\Bbb R\setminus F$, where $F$ is any finite subset of $\Bbb R$.

Now let $A$ be any infinite subset of $\Bbb R$, and let $x\in\Bbb R\setminus A$. Suppose that $x\in U\in\tau$, i.e., that $U$ is an open nbhd of $x$. Then by definition there is some finite set $F\subseteq\Bbb R$ such that $U=\Bbb R\setminus F$. That is, there are only finitely many real numbers — the members of $F$ — that are not in $U$. The set $A$ is infinite, so there must be some $a\in A\setminus F$. This real number $a$ is not in $F$, so it is in $U$. Thus, $U\cap A\ne\varnothing$. $U$ was an arbitrary open nbhd of $x$, so we’ve shown that every open nbhd of $x$ contains some element of $A$, and therefore $x\in\operatorname{cl}A$. As you’ve already pointed out, $A\subseteq\operatorname{cl}A$, so we’ve now shown that every real number is in $\operatorname{cl}A$: $\operatorname{cl}A=\Bbb R$, and hence $A$ is dense in $\Bbb R$.

This is true for every infinite subset $A$ of $\Bbb R$. In particular, it’s true for the countably infinite set $\Bbb N$: $\Bbb N$ is dense in $\Bbb R$ in this topology, and since $\Bbb N$ is countable, $\langle\Bbb R,\tau\rangle$ is a separable space.

Now we need to show that the topology $\tau$ has no countable base. Suppose that $\mathscr{B}=\{B_n:n\in\Bbb N\}$ is a countable family of open sets in this space; we’ll show that $\mathscr{B}$ is not a base for $\tau$ by finding a $U\in\tau$ that is not the union of members of $\mathscr{B}$.

We can assume that the members of $\mathscr{B}$ are all non-empty: there’s never any need to include the empty set in a base for a topology. Thus, by the definition of the topology $\tau$, for each $n\in\Bbb N$ there is a finite $F_n\subseteq\Bbb R$ such that $B_n=\Bbb R\setminus F_n$. Let $A=\bigcup_{n\in\Bbb N}F_n$; $A$ is the union of countably many finite sets, so $A$ is countable. $\Bbb R$ is uncountable, so there is some $x\in\Bbb R\setminus A$. Let $U=\Bbb R\setminus\{x\}$; the set $\{x\}$ is finite, so by definition $U\in\tau$. Let $n\in\Bbb N$ be arbitrary; then $x\notin A\supseteq F_n$, so $x\notin F_n$, and therefore $x\notin B_n$. But this means that $x\in U\setminus B_n$ for each $n\in\Bbb N$. That is, $x$ is not in any member of the family $\mathscr{B}$, so no matter what subcollection of $\mathscr{B}$ we take, its union cannot contain $x$ and therefore cannot be equal to $U$. This shows that $\mathscr{B}$ is not in fact a base for $\tau$. And since we’ve done this far an arbitrary countable subset $\mathscr{B}$ of $\tau$, we’ve shown that $\tau$ has no countable base and hence that $\langle\Bbb R,\tau\rangle$ is not second countable.

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  • $\begingroup$ I call your cofinite topology, and raise you with separability by a dense point. $\endgroup$ – Asaf Karagila Feb 13 '16 at 9:51
  • $\begingroup$ @Asaf: I wonder what the simplest Hausdorff example is. If we want a nice example we have Mrówka’s $\Psi$, and if we want a really nice example we have $\beta\omega$, but those are both a bit complicated for a beginner. $\endgroup$ – Brian M. Scott Feb 13 '16 at 9:56
  • $\begingroup$ Yeah, it's one of those things that the less $T_i$ the space, the easier the example becomes. $\endgroup$ – Asaf Karagila Feb 13 '16 at 9:58
  • $\begingroup$ I'd vote for the Sorgenfrey line. Separable is easy, not second countable not too hard. And it's hereditarily normal. $\endgroup$ – Henno Brandsma Feb 13 '16 at 11:41
  • $\begingroup$ @Henno: And it’s an example commonly found in elementary courses. Yes, that’s a good choice. $\endgroup$ – Brian M. Scott Feb 13 '16 at 17:18
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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.

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  • $\begingroup$ Thanks to you too! It is always helpful to look a multiple examples. $\endgroup$ – JKnecht Feb 14 '16 at 5:15

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