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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.

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solitaireartist
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  • The other answer that you reference uses the triangle inequality: $|f(x_0, y_0) - f(x,y)| \le |f(x_0,y) - f(x_0,y_0)| + |f(x_0,y) - f(x,y)|$, as I understand it. Each of those two terms are bounded by single-variable continuity. – Joshua Mundinger Nov 06 '14 at 23:16
  • I agree about the triangle inequality. I also agree that each of the two terms are bounded by single-variable continuity. Here's where I'm sure I'm being dense: if $y \neq y\prime$, then the terms $|f(x_0,y) - f(x,y)|$ and $|f(x_0,y\prime) - f(x,y\prime)|$ are differently bounded, so to speak. In other words, just because $|(f(x_0,y) - f(x,y)| < \frac{\epsilon}{2}$ for $x$ in some neighborhood of $x_0$, there's no guarantee that $|f(x_0,y\prime) - f(x,y\prime)| < \frac{\epsilon}{2}$ for $x$ in the same neighborhood. – solitaireartist Nov 06 '14 at 23:23
  • Or, rather, there is -- but that's because we're talking about a finite number $y$s here. In the problem, we have to contend with an infinite number of $y$s... – solitaireartist Nov 06 '14 at 23:28

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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 $f(x-\delta _x,y)$.

Jonathan Gleason
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