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


1 Answer 1


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 $U=\bigcap_\limits{p\in F}(V_p\cup W_p)\in\tau_1\times\tau_1$.
  • $\begingroup$ 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? $\endgroup$ Jul 9, 2020 at 19:13
  • $\begingroup$ @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 $y=0$. $\endgroup$ Jul 9, 2020 at 19:15
  • $\begingroup$ 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. $\endgroup$ Jul 9, 2020 at 19:29
  • 1
    $\begingroup$ @J.C.VegaO: It is not just possible: it is true, and I sketched out a proof in my answer. $\endgroup$ Jul 9, 2020 at 19:35

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