# Individually continuous function + monotonic = continious

1. Jun 3, 2010

### esvee

1. The problem statement, all variables and given/known data

Given $$f:{R^2} \to R$$.

Prove that if $$f$$ is continuous individually for each variable, and monotone in the first variable, then $$f$$ is continuous.

2. Relevant equations

3. The attempt at a solution
Well I "succeeded" to "prove" it by choosing $$\min \left( {{\varepsilon _1},{\varepsilon _2}} \right)$$ of the corresponding inequalities that follow from the definition of the limit. But I did not use the monotonicity at all, that means I got it wrong :)

Explanation:

If f is cont. relative to x:

$${\lim }\limits_{\left( {x,y} \right) \to \left( {{x_0},{y_0}} \right)} f\left( {x,y} \right) = f\left( {{x_0},y} \right) \Leftrightarrow \forall \varepsilon > 0,\exists {\delta _1} > 0,\left\| {\left( {x,y} \right) - \left( {{x_0},{y_0}} \right)} \right\| < {\delta _1} \Rightarrow \left| {f\left( {x,y} \right) - f\left( {{x_0},y} \right)} \right| < \varepsilon$$

etc. for y.

I'm sure it must be some simple inequality trick or the like, please help me out people!

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