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It is well-known that the operation of addition of two ultrafilters on the set $\mathbb{N}$ of natural numbers which extends the natural addition on $\mathbb{N}$ to $\beta\mathbb{N}$, the Cech-Stone compactification of $\mathbb{N}$, is not continuous (it is only right-continuous).

I am thus looking for examples of continuous binary operations on $\beta\mathbb{N}$ i.e. such functions $\beta\mathbb{N}\times\beta\mathbb{N}\to\beta\mathbb{N}$ which are (jointly) continuous (not only separately continuous). I am only interested in operations that engage both variables (so e.g. projections and their compositions with mappings $\beta\mathbb{N}\to\beta\mathbb{N}$ drop out), but they need not have any additional properties like associativity etc.

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  • $\begingroup$ Compact Hausdorff spaces possess exponentials (set of all continuous maps with compact-open topology). Moreover (here I am not 100% confident, that's why I am not making this an answer), the exponential $X^{\beta\mathbb N}$ is homeomorphic to the cartesian power $X^{\mathbb N}$ for any compact Hausdorff $X$. It follows that continuous maps ${\beta\mathbb N}\times{\beta\mathbb N}\to X$ are in one-to-one correspondence with arbitrary sequences of continuous maps $f_1,f_2,...:{\beta\mathbb N}\to X$ $\endgroup$ – მამუკა ჯიბლაძე Dec 15 '18 at 16:47
  • $\begingroup$ In other words (if the above is correct), any choice of values $f(i,j)$ on principal ultrafilters, (and these values might be arbitrary) extends uniquely to such an $f$. $\endgroup$ – მამუკა ჯიბლაძე Dec 15 '18 at 16:51
  • $\begingroup$ ...which is impossible since ${\beta\mathbb N}\times{\beta\mathbb N}$ is not homeomorphic to $\beta({\mathbb N}\times{\mathbb N})$ - so I do make a mistake somewhere, sorry... $\endgroup$ – მამუკა ჯიბლაძე Dec 15 '18 at 16:58
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    $\begingroup$ The usual extension of addition is not jointly contínuous for this reason $\endgroup$ – Benjamin Steinberg Dec 15 '18 at 21:52
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    $\begingroup$ @მამუკაჯიბლაძე Although, when $X$ is compact Hausdorff, $X^{\beta\mathbb{N}}$ and $X^{\mathbb{N}}$ are isomorphic as sets, the topologies are different. The compact-open topology on $[0,1]^{\beta\mathbb{N}}$ is uniform convergence on compact sets, which by compactness of $\beta{\mathbb{N}}$ is just uniform convergence, whereas the topology coming from $[0,1]^{\mathbb{N}}$ is pointwise convergence on $\mathbb{N}$. $\endgroup$ – Robert Furber Dec 16 '18 at 4:12
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In this paper, Dimension phenomena associated with $\beta\mathbb{N}$-spaces, Ilijas Farah proved that continuous maps from $\beta\mathbb{N}^2$ (and other powers) to $\beta\mathbb{N}$ are quite simple: there is a finite disjoint cover such that the map depends on one coordinate on each member of the cover.

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  • $\begingroup$ Thank you, KP! However, that's bitter news to me, unfortunately. Do you know whether there are similar results for operations on squares (that is, saying that operations $(\beta\mathbb{N})^2\to(\beta\mathbb{N})^2$ are also somehow simple)? $\endgroup$ – Damian Sobota Dec 16 '18 at 23:42
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    $\begingroup$ You can apply the result to each coordinate separately and take a common refinement of the two covers. Then each coordinate is simple on each member of the cover. $\endgroup$ – KP Hart Dec 17 '18 at 7:06
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    $\begingroup$ Also: Malykhin, in $\beta\mathbb{N}$ (mathscinet.ams.org/mathscinet-getitem?mr=552052), shows that if $\prod_{n\in\omega}X_n$ contains a copy of $\beta\mathbb{N}$ then one of the factors already contains a copy. $\endgroup$ – KP Hart Dec 17 '18 at 9:04

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