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Refering to the question Rudin 4.2 definition of a limit of a function, I dont have enough reputation to add a comment.

I have the same question. Is'nt the proof in $\impliedby\ $ direction a logical fallacy? If $P \implies\ Q$ then all we know more is $\neg Q \implies\ \neg P$ called modus tollens.

Further $\neg P\implies\ \neg Q$ is called "Denying the antecedent, or Inverse error or Fallacy of the inverse"

Isnt that what Rudin is doing here? So the proof should'nt start by assuming (4) is false, but that (5) is false.

Then from $\neg (5) \land (6) \implies\ \neg (4)$ or $ (5) \land \neg (6) \implies\ \neg (4)$ or (since logical and) $\neg (5) \land \neg (6) \implies\ \neg (4)$ hence a contradiction.

Endre Moen
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  • Note that $P \leftarrow Q$ is equivalent to $\lnot P \rightarrow \lnot Q$ by modus tollens. It only becomes denying the antecedent if you assume he's trying to prove $P \rightarrow Q$ in that instant, which is where your error lies. – Brian Moehring Nov 05 '19 at 18:29
  • I've always thought that making a mistake was ok, but I have to say that if it involves thinking that Rudin might personally fall into elementary logic flaws, then I retract my vote. – Arnaud Mortier Nov 05 '19 at 18:48
  • @BrianMoehring agree. $ Q \to P$ implies by modus tollens $\neg P \to \neg Q$. – Endre Moen Nov 05 '19 at 21:01

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No. The two statements claimed to be equivalent are:

  • $P$: $\lim_{x \rightarrow p}f(x) = q$
  • $Q$: for every sequence $(p_n)$ in $E$ such that $p_n \neq p$ and $\lim_{n\rightarrow \infty}p_n = p$, $\lim_{n \rightarrow \infty}f(p_n) = q$.

He shows first that $P \implies Q$ and then that $\neg P \implies \neg Q$. The latter is equivalent to $Q \implies P$, which completes the proof that $P \iff Q$.

Matthew Leingang
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  • I was confused at first by the contrapositive of $P \implies Q$ which is $\neg P \implies \neg Q$ which happens also to be modus tollens of $Q \implies P$ which is last step in the proof. Thanks! – Endre Moen Nov 05 '19 at 21:15