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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.

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  • $\begingroup$ What is $\mathcal N_0$? $\endgroup$ Jan 9 '21 at 6:00
  • $\begingroup$ Hi Alex Ravsky. $\mathcal{N}_0$ is the collection of all neighbourhoods of $0$. $\endgroup$
    – FJ W
    Jan 9 '21 at 13:07
  • $\begingroup$ Then $S1$ means a separate continuity of $\beta$ at $(0,a)$ and $(a,0)$ provided $\beta(a,0)=\beta(0,a)=0$ for each $a\in R$. $\endgroup$ Jan 9 '21 at 13:21
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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 $$U*V=\beta(U\times V) \subseteq \beta(\beta^{-1}(W’))\subseteq W’\subseteq W.$$

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  • $\begingroup$ I don't think it is correct to say that $(x,y) \in W'$. Surely it should be $\beta(x,y) \in W'$ $\endgroup$
    – FJ W
    Jan 20 '21 at 6:18

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