# Joint Continuity and Separate Continuity in Topological rings

I have been stuck with the following problem for quite some time, as I will next describe, by first introducing the definitions at hand. Here I have two definitions of joint continuity and two definitions of separate continuity. It is supposed to be easy to prove, but I am having trouble proving claim 1 and claim 2. I would be grateful for any suggestions and happy to clarify any irregularities concerning the post

Joint Continuity

J1 A binary operation $$\beta: R \times R \rightarrow R$$ defined for all $$x,y \in R$$ by $$\beta(x,y)=x * y$$ is said to be jointly continuous at $$(x,y) \in R \times R$$ if and only if for each neighbourhood $$W$$ of $$x * y$$, there exists a product of open sets $$U \times V \subseteq R \times R$$ containing $$(x,y)$$ such that $$U*V \subseteq W$$. If $$\beta$$ is jointly continuous at each point in $$R \times R$$ then $$\beta$$ is called (globally) jointly continuous.

J2 Suppose that $$X,Y$$ and $$Z$$ are topological spaces. Let $$f: X \times Y \rightarrow Z$$ be a mapping. Then $$f$$ is called jointly continuous on $$X \times Y$$ if and only if the map $$f$$ is continuous from $$X \times Y$$, equipped with the product topology $$\tau^{\times}$$.

Claim 1 In the event that $$X=Y=Z$$, we have equivalence for J1 and J2.

Separate Continuity

S1 A binary operation $$\beta: R \times R \rightarrow R$$ defined for all $$x,y \in R$$ by $$\beta(x,y)=x * y$$ is said to be separately continuous if and only if for all $$a \in R$$ and for all $$U \in \mathcal{N}_0$$ there exists $$V \in \mathcal{N}_0$$ such that $$a*V \subseteq U; V*a \subseteq U$$

S2 Suppose $$X,Y$$ and $$Z$$ are topological spaces. Let $$f: X \times Y \rightarrow Z$$ be a mapping. Then $$f$$ is separately continuous on $$X \times Y$$ if and only if for all $$(x_0,y_0) \in X \times Y$$, both functions $$x \mapsto f(x,y_0)$$ and $$y \mapsto f(x_0,y)$$ are continuous on $$X$$ and $$Y$$ respectively.

Claim 2 In the event that $$X=Y=Z$$, we have equivalence for S1 and S2.

• What is $\mathcal N_0$? Jan 9, 2021 at 6:00
• Hi Alex Ravsky. $\mathcal{N}_0$ is the collection of all neighbourhoods of $0$.
– FJ W
Jan 9, 2021 at 13:07
• Then $S1$ means a separate continuity of $\beta$ at $(0,a)$ and $(a,0)$ provided $\beta(a,0)=\beta(0,a)=0$ for each $a\in R$. Jan 9, 2021 at 13:21

Claim 1 In the event that $$X=Y=Z$$, we have equivalence for J1 and J2.
Implication $$(J1 \Rightarrow J2)$$ holds since $$U\times V\in\tau^\times$$ for each open subsets $$U$$ and $$V$$ of $$R$$. On the other hand, suppose that J2 holds. Let $$(R,\tau)$$ be a topological space, $$(x,y) \in R \times R$$ be any point, and $$W$$ be any neighbourhood of $$x*y$$. By the definition of a neighborhood, there exists an open set $$W’\ni (x,y)$$ such that $$W’\subseteq W$$. J2 implies that a set $$\beta^{-1}(W’)$$ is an open neighborhood of $$(x,y)$$. Since a family $$\{ U \times V :U,V\in\tau \}$$ is a base of the topology $$\tau^\times$$, there exist $$U,V\in\tau$$ such that $$(x,y)\in U \times V\subseteq \beta^{-1}(W’)$$. Then $$U*V=\beta(U\times V) \subseteq \beta(\beta^{-1}(W’))\subseteq W’\subseteq W.$$
• I don't think it is correct to say that $(x,y) \in W'$. Surely it should be $\beta(x,y) \in W'$