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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).

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  • $\begingroup$ 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

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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.

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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.

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