# When $\epsilon$ shrinks, does $\delta$ necessarily? If so, my proof makes sense. If not, can you help me fix it?

I'm working through the problems in Rudin's Principles of Mathematical analysis, chapter 4. Problem 3 is as follows:

Let $f$ be a continuous real function on a metric space $X$. Let $Z(f)$ (the zero set of $f$) be the set of all $p\in X$ at which $f(p)=0$. Prove that $Z(f)$ is closed.

Here's what I have.

Let $p$ be a limit point of $Z(f)$. Then for every neighbourhood $N_{\epsilon}(p)$ of p, there is a point $p'\in Z(f)\cap N_{\epsilon}(p)$ so that $f(p')=0$. Then for any $\epsilon>0$ there is some $\delta$ such that $|f(p)-f(p')|=|f(p)|<\delta$. Since we can take $\epsilon$ arbitrarily small, $f(p)=0$ and so $p\in Z(f)$.

I thought that looked good, but I realized I'm not sure that $\epsilon$ getting arbitrarily small implies $\delta$ getting arbitrarily small, which is what I would actually need. It seems to make sense that $\delta$ should get arbitrarily small but I'm not sure now—could you have a function which is continuous where, with $\epsilon,\delta$ as in the definition of continuity, taking $\epsilon'<\epsilon$ would require us to choose $\delta'\geq\delta$?

Maybe this is a dumb question. If my proof is incorrect, could someone give me a hint to fix it?

-

Your concern is justified. However, you started on the right foot, so the proof isn’t too hard to fix. For each $\epsilon>0$ you have a point $p_\epsilon\in N_\epsilon(p)$ such that $f(p_\epsilon)=0$. In particular, for each $n\in\Bbb Z^+$ you have a point $p_n\in N_{1/n}(p)$ such that $f(p_n)=0$. Now $\langle p_n:n\in\Bbb Z^+\rangle$ converges to $p$, and $f$ is continuous, so ... ?

-
Sorry, I'm not familiar with the notation $\langle p_n:n\in\Bbb Z^+\rangle$. Is that just some sequence of $p_n$? –  crf Nov 20 '12 at 8:57
@crf: Yes; it’s a standard notation for the sequence $\langle p_1,p_2,\dots\rangle$. The statement that it converges to $p$ could also be expressed as $$\lim_{n\to\infty}p_n=p\;.$$ –  Brian M. Scott Nov 20 '12 at 8:58
Thanks, I got it! –  crf Nov 20 '12 at 20:29
@crf: Excellent! You’re welcome. –  Brian M. Scott Nov 20 '12 at 20:37

You are on the right track, but your $\epsilon-\delta$ manipulations are a bit wonky.

Given any $\epsilon > 0$, there exists $\delta > 0$ such that for all $p'$ satisfying $d(p,p') < \delta$, $|f(p) - f(p')| < \epsilon$. Since $p$ is a limit point of $Z(f)$, there is always such a $p'$ which lies in $Z(f)$, so $f(p') = 0$, and we get $|f(p)| < \epsilon$ for any $\epsilon > 0$.

-