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Maybe it is a silly question, but:

If I have a function $f$ that is continuous on $I\times J\subseteq\mathbb{R}^2$; does this imply that $f$ is continuous in $I$ and in $J$?

My intuitive answer is: Yes, of course, because if one looks at the criterion of continuity that uses series, i.e.

$$ (x_k,y_k)\to (x,y)\Rightarrow f((x_k,y_k))\to f((x,y)) $$

that implies

$$ x_k\to x\Rightarrow f(x_k)\to f(x) $$

respectively

$$ y_k\to y\Rightarrow f(y_k)\to f(y). $$

Or am I totally wrong and confused?

Regards

egreg
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    What do you mean by continuous on $I$ or $J$? The function is defined on $I \times J$. – copper.hat Oct 17 '13 at 17:56
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    You probably mean that, fixed $y\in J$, the function $x\mapsto f(x,y)$ is continuous on $I$ and similarly fixing the other variable. This is true and the reasoning you present is substantially correct, but must be fixed. – egreg Oct 17 '13 at 17:59
  • Yes, thats what I meant respectively wanted to say. Of course it is necessary to define new functions for that, because, as copper.hat said, $f$ itself is defined on $I\times J$. –  Oct 17 '13 at 18:05
  • Maybe you should also see the definitions of 'separate continuity' and 'joint continuity'. If $f$ is (jointly) continuous in $I\times J$, then $f$ is separately continuous, i.e., $f(x,\cdot)$ is continuous in $J$ for each fixed $x\in I$ and $f(\cdot,y)$ is continuous in $I$ for each fixed $y\in J$. However, the converse is not true! – bkarpuz Oct 17 '13 at 20:00
  • Do you mean "see" or "say"? –  Oct 17 '13 at 20:02

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It is easy to check that the functions $i_y: I \to I \times J$ and $j_x: J \to I \times J$ defined by $i_y(x) = (x,y)$ and $j_x(y) = (x,y)$ are continuous. Hence the functions $f \circ i_y$, $f \circ j_x$ will be contiinuous.

copper.hat
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