# Compactness and epsilon delta

I have trouble in understanding some mathematical statement.

Suppose there is a non-vanishing function $g(z)$ on $\partial D(P,r)$. Then the compactness of $\partial D(P,r)$ implies that there is an $\epsilon>0$ such that $|g(z) | > \epsilon$ on $\partial D(P,r)$.

What i have in trouble is what "implies" means.

For compact domain, how can above statement is true?

• For posterity, it would be helpful to add details about the meaning of notation (symbols in mathematics, such as $g$, $z$, $|\ |$, $\partial$, $D$, $P$, and $r$ do not have universal meaning), and/or to trim out details irrelevant to your question. Here, for example, it appears your question boils down to "Let $f$ be a continuous, positive, real-valued function defined on a compact set $X$. Why does there exist a number $\varepsilon > 0$ such that $f(z) > \varepsilon$ for all $z$ in $X$?" Dec 12 '16 at 12:01

The statement is true if your function is continuous. Any continuous real-valued function on a compact set takes its infinum (and supremum) at some point in this set. If $g(z)$ is continuous, $|g(z)|$ is continuous as well. If there were no such $\epsilon$, your function would have arbitrarily small absolute values: there would be points $z_n\in \partial D$ so that $|g(z_n)|$ would be less than $1/n$, but still positive. But that would mean that $\inf_{\partial D}|g(z)|=0$, and by continuity of $|g|$, there must be a point $z_0\in\partial D$ so that $|g(z)|=\inf_{\partial D}|g(z)|=0$, so that $g(z)=0$, a contradiction.
If your function is not continuous, the statement is false. Consider the following function $g$ on the unit circle. If $z=x+iy$ has rational, but non-zero real part (i.e., $x$ is a non-zero rational number), set $g(z)=|x|$, in all other cases, set $g(z)=1$.
• (+1) Or more simply, $|g|$ has a minimum value $|g(z_{0})|$, which is positive by hypothesis; put $\varepsilon = \frac{1}{2}|g(z_{0})|$. Dec 12 '16 at 11:53
• (To the OP). The continuous image of a compact space is compact. A subset of $\mathbb R$ is compact iff it is closed and bounded. So if $g$ is continuous and $\inf \{|g(x)| :x\in \partial D(P,r)\}=0$ then $0$ is in the closure of $\{g(x):x\in \partial D(P,r)\}$ which is $\{g(x):x\in \partial D(P,r)\}$ itself. That would require that $g(x)=0$ for some $x\in \partial D(P,r).$ Dec 13 '16 at 10:45