Validity of allowing $\epsilon$ to vanish in Baby Rudin Theorem 3.10a

Question

In Theorem 3.10a of Rudin's PMA, we prove that $$\text{diam } \bar E = \text{diam } E$$ by

[fixing] $\epsilon>0$ and [choosing] $p \in \bar E, q \in \bar E$. By the definition of $\bar E$ there are points $p'$, $q'$, in $E$ such that $d(p,p')<\epsilon, d(q,q')<\epsilon.$ Hence

\begin{align} d(p,q) &\leq d(p,p') + d(p',q')+ d(q',q) \\ &< 2\epsilon + d(p',q') \leq 2\epsilon+\text{diam } E \end{align}

It follows that $$\text{diam } \bar E \leq 2\epsilon+ \text{diam } E$$

Since $\epsilon$ was arbitrary, the theorem is proved.

Why is this valid? To me it seems that $\text{diam } \bar E$ approaches, but never reaches, $\text{diam } E$. I get that $\epsilon$ can be arbitrarily small, but we $did$ fix $\epsilon>0$, so how can they be equal?

Answer 1

$\text{diam}\;\bar{E}$ doesn't approach a value, it has a fixed value, $x$ say.

Similarly, $\text{diam}\;E$ has a fixed value, $y$ say.

Since the inequality $0\le x-y\le 2\epsilon$ holds for all $\epsilon > 0$, it follows that $x=y$.

Answer 2

Suppose otherewise. That is, suppose that $\operatorname{diam}\overline E>\operatorname{diam}E$. Then take $\varepsilon<\dfrac{\operatorname{diam}\overline E-\operatorname{diam}E}2$. So $\varepsilon>0$ and $\operatorname{diam}\overline E>2\varepsilon+\operatorname{diam}E$, which goes against what Rudin proved.

Note that Rudin uses no passage to the limit.