# Determine if $\tau_2$ and $\varepsilon_2$ are coarser, finer, the same or not comparable to $\tau$.

Let $$\tau_1$$ be the cofinite topology over $$\mathbb{R}$$.

Let's consider the product topology $$\tau=\tau_1 \times \tau_1$$over $$\mathbb{R}^2$$

Let $$\tau_2$$ be the cofinite topology over $$\mathbb{R}^2$$.

and $$\varepsilon_2$$ the euclidean topology over $$\mathbb{R}^2$$

Determine if $$\tau_2$$ and $$\varepsilon_2$$ are coarser, finer, the same or not comparable to $$\tau$$.

Up to now I have shown that che $$\tau_1 \times \tau_1 \nsubseteq \tau_2$$ by taking $$\mathbb{R}\setminus\{0\} \times \mathbb{R}\setminus\{0\} \in \tau_1 \times \tau_1$$ which contains infinite elements and therefore is not the complement of a finite subset. I am trying to show $$\tau_2 \subseteq \tau_1 \times \tau_1$$, but no luck: graphically it seems it is that way, but I am taking an open set $$\tau_2$$ and an element in it and trying to show it belongs to $$\tau_1 \times \tau_1$$, but I am stuck

You are correct in saying that $$(\Bbb R\setminus\{0\})\times(\Bbb R\setminus\{0\})\notin\tau_2$$, but the reason that you gave is not correct. The reason that this set is not in $$\tau_2$$ is that it is a non-empty set whose complement is not finite: its complement is $$(\{0\}\times\Bbb R)\cup(\Bbb R\times\{0\})$$, which isn’t even countable, let alone finite.
To show that $$\tau_2\subseteq\tau_1\times\tau_1$$, let $$U\in\tau_2$$; then $$\Bbb R^2\setminus U$$ is finite. Let $$F=\Bbb R^2\setminus U$$. For each $$p=\langle x,y\rangle\in F$$ let $$V_p=(\Bbb R\setminus\{x\})\times\Bbb R$$ and $$W_p=\Bbb R\times(\Bbb R\setminus\{y\})$$.
• Show that $$V_p\cup W_p=\Bbb R^2\setminus\{p\}$$.
• Show that $$U=\bigcap_\limits{p\in F}(V_p\cup W_p)\in\tau_1\times\tau_1$$.
• I am having trouble showing $\Bbb R\setminus[(\Bbb R\setminus\{\langle 0,0\rangle\})\times(\Bbb R\setminus\{\langle 0,0\rangle\})]=(\{0\}\times\Bbb R)\cup(\Bbb R\times\{0\})$ in the first part. Grafically is clear, but analytically is messy, is there any De Morgan-like identity for this? Jul 9, 2020 at 19:13
• @J.C.VegaO: Just chase elements: $\langle x,y\rangle\notin U$ iff $x\notin\Bbb R\setminus\{0\}$ or $y\notin\Bbb R\setminus\{0\}$ iff $x=0$ or $y=0$. Jul 9, 2020 at 19:15
• I don't think $\tau_2\subseteq\tau_1\times\tau_1$ is possible:an open set in the first one is everything minus, for instance, a point, then I can't say it is included in the second one which is everything minus the two vertical and horizontal lines that intersect on it. Jul 9, 2020 at 19:29