TITLE: Continuous Function
QUESTION [6 upvotes]: Let  $ f: \Bbb R^2 \to \Bbb R $ such that :
$ \forall _{y_0 \in \Bbb R\ }:  $ function $ x \to f(x,y_0) $ continuous function and increasing
$ \forall _{x_0 \in \Bbb R\ }:  $ function $ y \to f(x_0,y) $ continuous function 
I mean the continuity of one variable.
Prove the continuity of a function $f: \Bbb R^2 \to \Bbb R $
I know definition, but I can not do. I wanted to use the definitions of Cauchy.

REPLY [6 votes]: Let $\varepsilon > 0$. There is $\delta_1 > 0$ such that $$f(x_0, y_0) - \varepsilon/2 < f(x_0 - \delta_1, y_0) < f(x_0 + \delta_1, y_0) < f(x_0, y_0) + \varepsilon/2.$$
Then there is $\delta_2 > 0$ such that
$$f(x_0 - \delta_1, y_0) - \varepsilon/2 < f(x_0 - \delta_1, y) < f(x_0 - \delta_1, y_0) + \varepsilon/2$$
and
$$f(x_0 + \delta_1, y_0) - \varepsilon/2 < f(x_0 + \delta_1, y) < f(x_0 + \delta_1, y_0) + \varepsilon/2$$
for any $y \in (y_0 - \delta_2, y_0 + \delta_2)$.
Now for any $(x,y) \in (x_0 - \delta_1, x + \delta_1) \times (y_0 - \delta_2, y_0 + \delta_2)$ you have $|f(x,y) - f(x_0, y_0)| < \varepsilon$.