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Individually continuous function + monotonic = continious

  1. Jun 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Given [tex]f:{R^2} \to R[/tex].

    Prove that if [tex]f[/tex] is continuous individually for each variable, and monotone in the first variable, then [tex]f[/tex] is continuous.

    2. Relevant equations

    3. The attempt at a solution
    Well I "succeeded" to "prove" it by choosing [tex]\min \left( {{\varepsilon _1},{\varepsilon _2}} \right)[/tex] 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 :)


    If f is cont. relative to x:

    [tex] {\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 [/tex]

    etc. for y.

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