Continuity of a function of product spaces

Question

Let $f: X \times Y \rightarrow Z $ such that.

$ \forall x_0 \in X , f_{x_0} : Y \rightarrow Z, y\mapsto f(x_0,y)$ is continuous.

$ \forall y_0 \in Y , f_{y_0} : X \rightarrow Z, x \mapsto f(x,y_0)$ is continuous.

How i can prove that $f$ is continuous?

Thanks in advance.

$U$ is open in $X\times Y$ if it is equal to $U_1 \times U_2$ with $U_1$ open in $X$ and $U_2$ open in $Y$. — user73577, May 16 '13 at 04:50
This is called the box topology (different from the product topology). — Martin Brandenburg, May 16 '13 at 09:06
@MartinBrandenburg. I think OP meant that these boxes form a basis for the topology, and not all sets are of this form, otherwise it wouldn't be a topology. But then it is not different from the product topology. — Stefan Hamcke, May 16 '13 at 11:21
@Martin: What the OP describes is a base for both the product and the box topology: for products of finitely many spaces they’re identical. It is not itself a topology at all (except in some trivial cases). — Brian M. Scott, May 16 '13 at 17:30
Does this answer your question? Does factor-wise continuity imply continuity? — tryst with freedom, Jul 24 '22 at 01:11

score 11 · Answer 1 · answered May 16 '13 at 08:44

It isn't. Let $f:\Bbb R\times \Bbb R\to \Bbb R$ be the function
$(x,y)\mapsto\begin{cases} 0 &\text{ , if }x=y=0\\ \frac{xy}{x^2+y^2} &\text{ , else } \end{cases}$
For any fixed $x_0$ the function $f_{x_0}$ is continuous. Since $f$ is symmetric in $x$ and $y$, $f_{y_0}$ is continuous for any $y_0$. But $f$ is not continuous in $x=y=0$. Consider the sequence $(x_n,y_n)=(1/n,1/n)$. It converges to $(0,0)$ along the diagonal and $f(x_n,y_n)=1/2$. But $f(\lim_{n\to\infty}(x_n,y_n))=f(0,0)=0$.

Continuity of a function of product spaces

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