Question about the connection between limits and sequences ( Rudin 4.2 )

Question

I'll give a part of the proof for Rudin theorem $4.2$ as context. This theorem states that $$\lim_{x\to p} f(x) = q \iff \lim_{n\to \infty} f(\{p_n\})=q$$

for every sequence $\{p_n\}$ in the domain of $f$ such that $p_n \neq p$, and $\lim_{n\to \infty} p_n = p$. In the proof of the '$\implies$' direction there is a point that I don't understand.

The logic goes like this : Let $\epsilon >0$ be given. Then there exists $\delta >0$ such that $d(f(x),q)<\epsilon$, if $d(x,p)<\delta$. Also there exists $N\in \Bbb{N}$ such that $n>N$ implies $d(p_n,p)<\delta$. Thus, for $n>N$, we have $d(f(p_n), q)<\epsilon$.

I don't see how the conclusion in bold is justified. I can see why it makes sense though. Since $x\in E$, ( $E$ is some subset of the domain of $f$ ) the point $p$ is a limit point of $E$. Certainly there is a sequence in $E$ which is converging to $p$. So the behaviour of $\{p_n\}$ and $x$ with respect to the point $p$ is similar, it is to be expected that the behaviour of $f$ with respect to these two is similar. Can anyone explain how this makes sense logically? I tried to mention only the parts relevant of the proof, and if needed I can give more details.

Answer 1

I'm looking at page 84 of my copy of the third edition of Rudin's Principles of Mathematical Analysis. Quoting verbatim one sentence of the proof:

Then there exists $\delta > 0$ such that $d_Y(f(x),q) < \varepsilon$ if $x\in E$ and $0 < d_X(x,p) < \delta.$

The phrase "if $x \in E$" means that the implication $0 < d_X(x,p) < \delta \implies d_Y(f(x),q) < \varepsilon$ is true for any number $x$ that happens to be in $E.$

So if $0 \in E$ then $0 < d_X(0,p) < \delta \implies d_Y(f(0),q) < \varepsilon.$
If $1 \in E$ then $0 < d_X(1,p) < \delta \implies d_Y(f(1),q) < \varepsilon.$
If $17\pi+2 \in E$ then $0 < d_X(17\pi+2,p) < \delta \implies d_Y(f(17\pi+2),q) < \varepsilon.$

Likewise, since each $p_n$ is a number, If $p_n \in E$ then $0 < d_X(p_n,p) < \delta \implies d_Y(f(p_n),q) < \varepsilon.$ And the premises of the theorem say that $p_n \in E.$

There is no statement that $x$ must be $p_n,$ only that $x$ can be $p_n.$

The proof might be a little less confusing if it had used some symbol other than $x$ for this particular statement. The $x$ in "if $x \in E$" is quantified right there and really has no meaning outside of the sentence I quoted above. In certainly has nothing to do with the $x$ in $\lim_{x\to p}$ except for the coincidental fact that both of these symbols are numbers and are desired to take values in the set $E.$ You could change the quoted sentence to

Then there exists $\delta > 0$ such that $d_Y(f(t),q) < \varepsilon$ if $t\in E$ and $0 < d_X(t,p) < \delta$

without changing one single letter or symbol in the rest of the proof, and the meaning of the entire proof would be exactly the same as it was when Rudin wrote it.