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Baby Rudin theorem 4.2 presents an alternative definition of a limit:

Suppose $X, Y$ are metric spaces, $E \subset X$, $f: X \rightarrow Y$, $p$ is a limit point of $E$.

Then $\lim_{x \rightarrow p}f(x) = q$ $\textbf{(4)}$ if and only if $\lim_{n \rightarrow \infty}f(p_n) = q$ $\textbf{(5)}$ for every sequence $(p_n)$ in $E$ such that $p_n \neq p$, $\lim_{n\rightarrow \infty}p_n = p$ $\textbf{(6)}$.

For the proof of $\impliedby$, Rudin supposes that $\textbf{(4)}$ is false, then finds a sequence in $E$ that satisfies $\textbf{(6)}$ but not $\textbf{(5)}$. I'm not certain as to how this proves the implication, nor where the contradiction is that would close the argument.

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    $\begingroup$ It looks like a proof by contrapositive, rather than contradiction ($p\implies q $ is equivalent to $\neg q\implies \neg p$. $\endgroup$ – AnyAD Nov 12 '18 at 22:24
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That proposition has this structure:$$A\iff(B\implies C).$$In order to prove $\Longleftarrow$, Rudin proves that $\neg A\implies\neg(B\implies C)$. And, in turn, $\neg(B\implies C)$ is equivalent to $B\wedge\neg C$.

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