Topology on $\mathbb R^2$ that is not a product topology

Find an example of a topology on $$\mathbb R^2$$ that is not a product topology.

I feel like an open set in $$\mathbb R^2$$ with any topology can be written as a union of open balls so we can arrange ourselves to write it as a product $$U \times V$$.

But maybe another idea that I thought of is the co-finite topology on $$\Bbb R^2$$. Is it correct ? I think yes because an open set in the co-finite topology on $$\Bbb R^2$$ can be written for example as $$\Bbb R^2 \backslash \{(x,y)\}$$, i.e. the plane without a point and I feel like this cannot be written as a union of boxes (squares).

• Any set is a union of singletons , hence a union of rectangles. Oct 19, 2021 at 8:04
• Oct 19, 2021 at 8:06
• @KaviRamaMurthy In fact, you're right. Maybe the co-countable one ? Oct 19, 2021 at 8:06
• The slotted plane and the cross topology are well known topologies on $\Bbb R^2$ that are not product topologies. Also the topology induced by the river metric E.g. Oct 19, 2021 at 9:32

We cannot work without a definition of the statement we are trying to prove. I say that a (non-trivial) product topology on $$X$$ is a topology $$\tau$$ such that there is an indexed family $$\{X_i\}_{i\in I}$$ of non-one-point spaces such that $$\lvert I\rvert\ge2$$ and $$\prod_{j\in I}X_j$$ is homeomorphic to $$(X,\tau)$$.

Now, your claim that the cofinite topology on $$\Bbb R^2$$ is not a product topology is spot-on: the cofinite topology on an infinite set is never a product topology.

In point of fact, let $$\prod_{j\in I}X_j=X$$ as per the definition of $$X$$ having a non-trivial product topology. Since $$\lvert X\rvert\ne \varnothing$$ consider some $$h\in X$$. Each map $$\iota_j:X_j\to X$$, $$\iota_i(y)=\begin{cases}y&\text{if }i=j\\ h_i&\text{if }i\ne j\end{cases}$$ is a topological embedding (id est, a homeomorphism onto its image). Since subspaces of spaces with the cofinite topology have the cofinite topology, each $$X_j$$ must have the cofinite topology. Again, pick some $$h\in X$$. For $$k\in I$$ call $$U^{(k)}$$ the set $$\{x\in X\,:\, x_k\ne h_k\}$$. Notice that $$U^{(k)}$$ is open because $$U^{(k)}=(X_k\setminus\{h_k\})\times\prod\limits_{j\in I\setminus\{k\}}X_j$$. Notice that $$U^{(k)}\ne\varnothing$$ because $$\lvert X_k\rvert\ge2$$, and that $$X\setminus U^{(k)}=\{h_k\}\times\prod\limits_{j\in I\setminus\{k\}}X_j$$. I claim that, if $$X$$ is infinite, then there is some $$k\in I$$ such that $$\prod_{j\in I\setminus\{k\}}X_j$$ is infinite. Two cases:

• if one of the $$X_j$$-s, namely $$X_i$$, is infinite, then consider any $$k\ne i$$ and you'll have $$\lvert X_i\rvert\le\left\lvert\prod_{j\in I\setminus\{k\}}X_j\right\rvert$$.
• if all the $$X_j$$-s are finite, then $$I$$ must be infinite, and for any $$k$$ you have $$2^{\aleph_0}\le 2^{\lvert I\rvert}\le\left\lvert\prod_{j\in I\setminus\{k\}}X_j\right\rvert$$.

For such $$k$$, the set $$X\setminus U^{(k)}$$ is then an infinite closed set.

The topology on $$\Bbb R^2$$ defined by the jungle river metric or the French railroad metric are not product metrics as some thought will show.

The co-finite or co-countable topology will also work (but are not "nice" like metrics are).

The so-called slotted plane (a classic example of a Hausdorff non-regular topology) or the "cross"-topology (a set is open iff it contains a cross (parallel to both axes, of finite length) around each of its points) are refinements of the Euclidean topology that also (I'm pretty sure) cannot be written as a product of topologies on the components. Both have been used as examples in papers and books.