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Is it possible to define a topology $\mathcal{T}$ on $\mathbb{R}$ such that $\{0\}$ and $\mathbb{Z}^*=\mathbb{Z}\setminus\{0\}$ are dense but no finite $F\subseteq \mathbb{Z}^*$ is dense?

This is what I know. I know that $\{0\}$ nor $\mathbb{Z}^*$ can be open. Therefore, $\{0\}$ can't be close (since that would imply that $\{0\}$ is nowhere dense) which means that $(\mathbb{R},\mathcal{T})$ can't be $T_1$. I also know that, if the said topology were to exist, it would contain a $G_\delta$ set $E$ with empty interior such that $E\cap\mathbb{Z}=\{0\}$. Could it be that $E=\{0\}$, i.e. that $\{0\}$ is $G_\delta$?

Proof of the things I know. First, $\{0\}$ nor $\mathbb{Z}^*$ can't be open since they are disjoint dense sets. Now, since $F$ isn't dense $\forall F\subseteq_{fin}\mathbb{Z}^*$, we know that $\forall n\in\mathbb{Z}^*\exists \emptyset\neq O_n\in\mathcal{T}:n\not\in O_n$ and, since $\{0\}$ is dense, $\forall n\in\mathbb{Z}^*:0\in O_n$. Then, we define $E:=\bigcap\limits_{n\in\mathbb{Z}} O_n$ which is a $G_\delta$ set such that $E\cap\mathbb{Z}=\{0\}$ and with empty interior since $$\text{Int}(E)\neq\emptyset\Rightarrow \text{Int}(E)\cap\mathbb{Z}^*\neq\emptyset\Rightarrow E\cap\mathbb{Z}^*\neq\emptyset$$

Tyrone
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K. Makabre
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    Doesn't it work to just define the topology whose open sets are simply every set of the form ${0} \cup (a,\infty)$ for $a>0$? – Greg Martin Feb 05 '23 at 22:59

2 Answers2

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Yes. First consider the standard $\mathbb{R}$. Then $\mathbb{Q}$ is dense and countable but no finite subset is dense. Now consider $A$ to be some countable dense subset of irrationals, e.g. $\mathbb{Q}+\sqrt{2}$ and finally consider the quotient space $X=\mathbb{R}/A$ by collapsing $A$ to a point.

Image of dense is dense and so the image of $\mathbb{Q}$ is dense in $X$. It is also countable. But none of its finite subsets is dense, because points in $[\mathbb{Q}]$ are closed.

Analogously the image of $A$, which is a point in $X$, is dense.

Finally $X$ is uncountable (that's why I choose $A$ to be countable) and so we can transfer its topology to $\mathbb{R}$ through any bijection that maps $[A]$ to $0$ and $[\mathbb{Q}]$ to $\mathbb{Z}^*$.

freakish
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Yes, such a topology does exist. Consider $\tau_{cof}$ the cofinite topology on $\mathbb{R}$, defined by taking the collection $\{A\subset\mathbb{R} : A\space\text{is finite}\}\cup\{\emptyset,\mathbb{R}\}$ as closed sets. Then $\tau_{cof}$ satisfies the second and the third property :

  • $\mathbb{Z}^*=\mathbb{Z}-\{0\}$ is dense in $(\mathbb{R},\tau_{cof})$. In fact, $\mathbb{Z}^*$ is infinite and each closed set containing $\mathbb{Z}^*$ is finite, thus the smallest closed set containing $\mathbb{Z}^*$ is $\mathbb{R}$.
  • Each finite subset $F\subset\mathbb{Z}^*$ is not dense, since $F$ is finite, then is closed in $\tau_{cof}$.

Clearly $\{0\}$ is not dense in in $(\mathbb{R},\tau_{cof})$, since is finite, then closed. But we can refine $\tau_{cof}$ in order to obtain a new topology on $\mathbb{R}$.

Now, consider $\tau'$ defined by taking the collection $\{A\subset\mathbb{R} :0\space{\notin}\space A,\space A\space\text{is finite}\}\cup\{\emptyset,\mathbb{R}\}$ as collection of closed sets. These requests define a topology on $\mathbb{R}$ because :

  • $\emptyset,\mathbb{R}$ are closed
  • An arbitrary intersection of finite sets not containing $0$ is still a finite set not containing $0$.
  • A finite union of finite sets not containing $0$ is still a finite set not containing $0$.

Finally, $\tau'$ satisfies all the three properties, since the smallest closed set containing $\{0\}$ is $\mathbb{R}$, and the other two properties are verified since we have refined $\tau_{cof}$.

simo210
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