Clarification of Proof of Baby Rudin Theorem 4.2

Question

I'm a bit confused about the proof in Theorem 4.2 in Baby Rudin, and could use some help filling in the intermediate steps.

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Theorem 4.2 states: $$ \lim_{x\to p} f(x) = q \iff \lim_{n\to\infty} f(p_n) = q $$ for every sequence $\{p_n\} \subset E \text{ s.t. } p_n \to p, p_n \neq p$. [*]

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The forward direction ($\implies$) is fairly straightforward, but the converse is a bit confusing. An outline of his proof by contraposition goes:

Suppose $\lim_{x\to p} f(x) \neq q$, then $$ \exists \epsilon > 0 \text{ s.t. } \forall \delta > 0, \exists x \in E \text{ with } 0 < d_X(x, p) < \delta \text{ but } d_Y(f(x), q) \geq \epsilon $$

Take $\delta_n = \frac{1}{n}$, thus we find a sequence in $E$ that satisfies [*] above but $\lim_{n\to\infty} f(p_n) =q$ is false.

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I'm assuming $\delta_n$ is supposed to be the radius of the neighborhood around $p$. But what is the sequence $p_n$ that satifies [*]? Furthermore we have not defined $f$ anywhere, how do we know $\lim_{n\to\infty} f(p_n) = q$ is false?

Answer 1

Yes, $f$ is not defined. You still know however that, for some number $\varepsilon>0$, whenever $\delta>0$ there is some point $x\in E$ such that $0<d_X(x,p)<\delta$ and $d_Y\bigl(f(x),q\bigr)\geqslant\varepsilon$. In particular, if $n\in\Bbb N$, there is some $p_n\in E$ such that $0<d_X(p_n,p)<\frac1n$ and $d_Y\bigl(f(p_n),q\bigr)\geqslant\varepsilon$. Now, it follows from the fact that you have $0<d_X(p_n,p)<\frac1n$ for each $n\in\Bbb N$ that $\lim_{n\to\infty}p_n=p$. And it follows from the fact that $d_Y\bigl(f(p_n),q\bigr)\geqslant\varepsilon$ for each $n\in\Bbb N$ that you don't have $\lim_{n\to\infty}f(p_n)=q$.