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Let $f$ denote a map from a product space $X \times Y$ to $Z$. If for every $x\in X$, the map $f(x,-)$ is continuous, and the same holds for every $y \in Y$, then is $f$ continuous in general? If not, is there any condition to be imposed to make $f$ continuous?

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A categorical reason why you would not believe this statement holds is that you are asking of a product to satisfy a coproduct property. This is why at first glance it must feel false. The product behaves well when you project it on its components, i.e. $X \times Y \to X$ ; the coproduct behaves well when you inject its components in, i.e. $X \to X \times Y$. Notice that the maps $f(x,-)$ are nothing more than composing $\pi_Y : Y \to X \times Y$ defined by $\pi_Y(y) = (x,y)$ with $f$, i.e. $f(x,-) = f \circ \pi_Y$. –  Patrick Da Silva Dec 12 '12 at 6:48
    
Interesting question. Does the converse hold by chance? (I'm guessing not.) –  Noldorin Apr 13 at 2:39

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up vote 7 down vote accepted

No, separate continuity does not in general imply joint continuity. The function

$$f:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto\begin{cases}\frac{xy}{x^2+y^2}&,\text{if }\langle x,y\rangle\ne\langle0,0\rangle\\\\0,&\text{if }\langle x,y\rangle=\langle 0,0\rangle\end{cases}$$

is continuous everywhere except at the origin. At the origin it is continuous in each variable separately, but it is not continuous.

Added: Useful keywords on which to search are separate continuity and joint continuity. Z. Piotrowski, a former colleague of mine, did quite a lot of work in this area; you’ll find many of his papers here, and in them both results and further references. Be warned, though, that the links don’t always match the text: the link to the comprehensive (if now dated) survey Separate and joint continuity is actually at number $17$, not at number $19$. As I recall, there are more results giving conditions under which the set of points of continuity contains a dense $G_\delta$-set in $X\times Y$ than there are giving conditions that guarantee that a separately continuous function is jointly continuous.

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Curiosity : Why did you use $\langle x,y \rangle$ instead of $(x,y)$ for the coordinates? This doesn't feel natural, I've never seen that before. –  Patrick Da Silva Dec 12 '12 at 6:41
    
Well, in set theory, bivariate functions are actually functions from ordered pairs to the result set, i.e. $f(x,y) := f(\langle x,y\rangle)$. –  Mario Carneiro Dec 12 '12 at 6:44
    
@Mario Carneiro : You are suggesting that $\langle x,y \rangle$ is a set-theoretic notation for the ordered pair? I know ordered pairs are defined something like $\langle x,y \rangle = \{x, \{x,y \} \}$ so I understand the need for a notation but just let me know if I am wrong. –  Patrick Da Silva Dec 12 '12 at 6:50
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@Patrick: Yes, the notation comes from set theory. I habitually use it, partly because a lot of my background is in set theory, and partly simply because I prefer it: ordinary parentheses have enough work to do! –  Brian M. Scott Dec 12 '12 at 6:51
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I'm pretty sure $\langle x,y\rangle$ is the standard notation for ordered pairs in general. The definition of the ordered pair in set theory is $\langle x,y\rangle := \{\{x\},\{x,y\}\}$, which is what I think you are thinking of. –  Mario Carneiro Dec 12 '12 at 6:52

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