Continuity of a particular scalar field I have been struggling with the problem posed and answered in this post:
Continuous Function
In particular, I am confused by the accepted answer's final assertion: if $(x,y) \in (x_0 - \delta_1, x_0 + \delta_1) \times (y_0 - \delta_2, y_0 + \delta_2)$, then $\|f(x_0,y_0) - f(x,y)\| < \epsilon$. I'm certain I'm missing something, but I do not see how this arises from the preceding observations.  
My own attempt at solution runs as follows:
For all fixed $x_0$, call the function that takes $y \to f(x_0,y)$  $f_{x_0}$; let $f_{y_0}$ be the analogous function for $y_0$.
Choose $(x_0,y_0)$, and let $\epsilon > 0$.  Since $f_{x_0}$ is continuous, we can choose $\delta_y$ such that 
\begin{align}
y &\in (y_0 - \delta_y, y_0 + \delta_y)\Rightarrow\\
f(x_0,y) &\in \left(f(x_0,y_0) - \frac{\epsilon}{2}, f(x_0,y_0)+\frac{\epsilon}{2}\right)
\end{align} 
Now consider all $f_y$ such that $y \in (y_0 - \delta_y, y_0 + \delta_y)$.  We want to find a $\delta_x$ such that 
\begin{align}
x \in (x_0 - \delta_x, x_0 + \delta_x)\Rightarrow\\
f(x,y) \in \left(f(x_0,y) - \frac{\epsilon}{2}, f(x_0,y) + \frac{\epsilon}{2}\right)
\end{align}
This is, for me, the sticking point.  For each individual $f_y$ we can find a suitable $\delta$, and it's certainly tempting to choose the infimum of the set of all such $\delta$s -- but what if said infimum is $0$?  How do we use the monotonicity of the $f_y$ to obtain $\delta_x$? 
Thanks in advance.
 A: You're right about taking the infimum:  to do this trick, you have to use only finitely many deltas, and then you can take the smallest.  I think the point here is to notice that, because each $f_y$ is increasing, you only need finitely many $\delta$s.
First of all, pick your $\delta _x$ first.  Fix $x_0$ and $y_0$, let $\epsilon >0$, and choose $\delta _x$ so that $x\in (x_0-\delta _x,x_0+\delta _x)$ implies that $f(x,y_0)\in \left( f(x_0,y_0)-\epsilon ,f(x_0,y_0)+\epsilon \right)$.
The problem you had before was that it seemed you need to now pick a $\delta$ for each $x\in (x_0-\delta _x,x_0+\delta _x)$, but in fact you do not:  you only need to choose $\delta$s for $x_0-\delta _x$ and $x_0+\delta _x$ because of the fact that each $f_y$ is increasing.  So let $\delta _1,\delta _2>0$ be such that
$$
y\in (y_0-\delta _1,y_0+\delta _1)\Rightarrow f(x_0-\delta _x,y)\in \left( f(x_0-\delta _x,y_0)-\epsilon ,f(x_0-\delta _x,y_0)+\epsilon )\right)
$$
and
$$
y\in (y_0-\delta _2,y_0+\delta _2)\Rightarrow f(x_0+\delta _x,y)\in \left( f(x_0+\delta _x,y_0)-\epsilon ,f(x_0+\delta _x,y_0)+\epsilon )\right)
$$
Then, we can take $\delta _y:=\min \{ \delta _1,\delta _2 \}$.  Then, for $x\in (x_0-\delta _x,x_0+\delta _x)$ and $y\in (y_0-\delta _y,y_0+\delta _y)$, we have, first of all, that $f(x-\delta _x,y)\leq f(x,y)\leq f(x+\delta _x,y)$.  From this, it follows that if both $f(x-\delta _x,y)$ and $f(x+\delta _x,y)$ are with (some multiple of) $\epsilon$ of $f(x_0,y_0)$, so too will be $f(x,y)$.
$$
\left| f(x+\delta _x,y)-f(x_0,y_0)\right| \leq \left| f(x+\delta _x,y)-f(x,y_0)\right| +\left| f(x,y_0)-f(x_0,y_0)\right| <\left| f(x+\delta _x,y)-f(x+\delta _x,y_0)\right| +\left| f(x+\delta _x,y_0)-f(x,y_0)\right| +\epsilon <\epsilon +\epsilon +\epsilon =3\epsilon .
$$
And similarly for $f(x-\delta _x,y)$.
