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.

  • 1
    $\begingroup$ 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

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 $B\wedge\neg C$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.