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A topological universal algebra of type $\Omega$ is a universal algebra $A$ of type $\Omega$ that is also a topological space, such that for any $n\!\in\!\mathbb{N}$ and any operation $\omega\!\in\!\Omega$ of arity $n$, the map $\omega\!:A^n\!\rightarrow\!A$ is continuous, where $A^n$ has the product topology.

In particular, a topological group is a group $G$ that is also a topological space, such that multiplication $\cdot\!:G^2\!\rightarrow\!G$ and inversion $^{-1}\!:G\!\rightarrow\!G$ are continuous. A topological ring is a ring $R$ that is also a topological space, such that addition $+\!:R^2\!\rightarrow\!R$ and multiplication $\cdot\!:R^2\!\rightarrow\!R$ are continuous.

We know that a left $R$-module is a universal algebra $(M,+,(r\cdot)_{r\in R})$ of type $(2,(1)_{r\in R})$, and a $R$-algebra is a universal algebra $(A,+,(r\cdot)_{r\in R},\cdot)$ of type $(2,(1)_{r\in R},2)$. Now, if $R$ is a topological ring, then in the literature, a topological left $R$-module is defined as a left $R$-module $M$ that is also a topological space, such that addition $+\!:M^2\!\rightarrow\!M$ and scalar multiplication $\cdot\!:R\!\times\!M\!\rightarrow\!M$ are continuous. This does not coincide with the definition via universal algebras.

Question: If $R$ is a topological ring, and if $M$ is a left $R$-module which is also a topological space, such that $+\!:M^2\!\rightarrow\!M$ and $r\cdot\!:M\!\rightarrow\!M$ are continuous ($\forall r\!\in\!R$), is then $\cdot\!:R\!\times\!M\!\rightarrow\!M$ continuous?

Proof: More generally, if $f\!:X\!\times\!Y\!\rightarrow\!Z$ is a map, and if for each $x\!\in\!X$, the map $f(x,\_)$ is continuous, then $f$ is continuous. To see this, note that for any open subset $U\!\subseteq\!Z$, we have $$f^{-1}(U)\!=\!\{(x,y)\!\in\!X\!\times\!Y; f(x,y)\!\in\!U\}\!=\!\bigcup_{x_0\in X}\{(x_0,y)\!\in\!X\!\times\!Y;f(x_0,y)\!\in\!U\}\!=\!\bigcup_{x_0\in X}f(x_0,\_)^{-1}(U),$$ which is a union of open sets.

Is all of the above correct?

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The claim in your proof is false. For example, take $f(x, y) = h(x)$ where $h$ is discontinuous – then $f(x, -)$ is continuous for each fixed $x$, but obviously $f$ itself is not continuous. – Zhen Lin Jun 29 '12 at 7:00
You are right, because $\{(x_0,y)\!\in\!X\!\times\!Y;f(x_0,y)\!\in\!U\}\!\neq\!f(x_0,\_)^{-1}(U)$, since $f(x_0, \_ )$ is a map $Y\!\rightarrow\!Z$, and not $X\!\times\!Y\!\rightarrow\!Z$. Anyway, are the two definitions of a topological $R$-module equivalent or not? – Leon Jun 29 '12 at 7:08
I don't see any reason for them to be equivalent. I'm sure a suitable counterexample can be concocted. – Zhen Lin Jun 29 '12 at 7:22
Equip $R = \mathbb{R}$ with its usual topology and the module $M = \mathbb{R}^2$ with the discrete topology. Since $M$ is discrete, both addition and scalar multiplication $r \cdot \colon M \to M$ are continuous. You can easily check that $R \times M \to M$ is not continuous. – t.b. Jun 29 '12 at 8:01
Thank you both, @t.b. & Zhen Lin, case closed. Too bad that the terminologies don't agree. – Leon Jun 29 '12 at 8:04

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