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).

  • $\begingroup$ Any set is a union of singletons , hence a union of rectangles. $\endgroup$ Oct 19, 2021 at 8:04
  • $\begingroup$ Related. $\endgroup$ Oct 19, 2021 at 8:06
  • $\begingroup$ @KaviRamaMurthy In fact, you're right. Maybe the co-countable one ? $\endgroup$
    – Kilkik
    Oct 19, 2021 at 8:06
  • $\begingroup$ 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. $\endgroup$ Oct 19, 2021 at 9:32

2 Answers 2


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.