distinguish $\neg P\implies (Q\land\neg R)$ and $(\neg P\land Q)\implies\neg R $ in application

Question

The function $f:\mathbb{R}\to\mathbb{R} $ is continuous at $a\in\mathbb{R}$

if and only if

if $\displaystyle\lim_{n\to\infty} x_n=a$, then $\displaystyle\lim_{n\to\infty}f(x_n)=f(a).$

Denoting $P$ as the $\epsilon-\delta$ continuity definition, $Q$ as $\displaystyle\lim_{n\to\infty} x_n=a$, and $R$ as $\displaystyle\lim_{n\to\infty}f(x_n)=f(a),$ we have $$P\implies (Q\implies R)\equiv(P\land Q)\implies R$$ and $$(Q\implies R)\implies P\equiv \neg P\implies \neg(Q\implies R)\equiv\neg P\implies (Q\land \neg R).$$

However, the capture seems to be proving $$(\neg P\land Q)\implies \neg R.$$

How to distinguish $$(\neg P\land Q)\implies \neg R$$ from $$\neg P\implies (Q\land \neg R)$$ in this application?

Add-on: I know that $\neg P\implies (Q\land \neg R)$ is the true formulation and know how to obtain it. What really bugs me is the capture uses $Q, \lim_{n\to\infty}x_n=a$, not arrived this.

Answer 1

What really bugs me is the capture uses $Q, \lim_{n\to\infty}x_n=a$, not arrived this.

No, the screencap is assuming only that $P$ is false, and proving this:

$$\lnot P\implies \text{there is some }\langle x_n:n{\in}\mathbb N\rangle \text{ such that }\big(Q\land\lnot R\big);\tag1$$

equivalently (by contrapositive), it is proving that $$\Big[\text{for each }\langle x_n:n{\in}\mathbb N\rangle, \big(Q\implies R\big)\Big]\implies P.$$

the argument seems having Q in antecedent. If not, what will $$(¬P∧Q)⟹¬R\tag2$$ looks like in plain language?

Statement $(2)$ translates as “Suppose that $P$ is false but $Q$ true; we shall now show that $R$ is false” (interpreted: “Suppose that $f$ is discontinuous at $a$ and that $\langle x_n:n{\in}\mathbb N\rangle$ converges to $a;$ we shall now show that $f(x_n)$ does not converge to $f(a)”).$

The problem with this suggestion is that the sequence $\langle x_n:n{\in}\mathbb N\rangle$ in it is hanging freely: while the original theorem/definition that started your post was referring to every possible sequence $\langle x_n:n{\in}\mathbb N\rangle$ that converges to $a,$ and the screencap is referring to some sequence $\langle x_n:n{\in}\mathbb N\rangle$ coverging to $a,$ in statement $(2),$ a particular sequence appears to be referred to, yet it is unidentified.

The function $f:\mathbb{R}\to\mathbb{R} $ is continuous at $a\in\mathbb{R}$

if and only if

if $\displaystyle\lim_{n\to\infty} x_n=a$, then $\displaystyle\lim_{n\to\infty}f(x_n)=f(a).$

To be clear: you've probably transcribed the theorem/definition incorrectly: $$\text{for each }\langle x_n:n{\in}\mathbb N\rangle$$ ought to be inserted at the beginning of its third line.

Answer 2

No, the logical form of the argument is correct and does what it is supposed to do. First, I try to adapt the logical form of the argument to your formalization. Then, I show that your formalization is missing something important, the quantifiers, to give a full account for the logical structure of the proof argument.

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The proof you cited shows that if the condition about limits is fulfilled (the $Q \implies R$ part of your formalization, i.e. for every sequence $\{x_n\}_{n\in \mathbb{N}}$, if $\lim_{n \to \infty} x_n = a$ then $\lim_{n \to \infty} f(x_n) = f(a)$), then $f$ is continuous in $a$ (the $P$ part of your formalization). The proof uses the technique called "by contrapositive": it supposes that the function $f$ is not continuous at $a$ (i.e. it assumes $\lnot P$) and then it shows that the condition about limits is not fulfilled, i.e. it shows that $Q \land \lnot R$ (which is equivalent to $\lnot (Q \implies R)$).

Indeed, the proof argument shows that there exists a sequence $\{x_n\}_{n\in \mathbb{N}}$ such that $\lim_{n \to \infty} x_n = a$ (i.e. $Q$ holds) but it is not true that $\lim_{n \to \infty} f(x_n) = f(a)$ (i.e. $\lnot R$ holds).

Note that this part of the proof you quoted is maybe badly worded and a bit confusing. You do not suppose to have a sequence $\{x_n\}_{n \in \mathbb{N}}$ such that $\lim_{n \to \infty} x_n = a$ (the phrase "Also $\lim_{n \to \infty} c_n = c$" is misleading there, because the sequence $\{c_n\}_{n \in \mathbb{N}}$ is not defined yet). Instead, you construct a sequence $\{x_n\}_{n \in \mathbb{N}}$ for which it turns out that $\lim_{n \to \infty} x_n = a$. How? Since $f$ is not continuous in $a$ by hypothesis, we know that $$\text{there is $\varepsilon \!>\! 0$ such that, for all $\delta \!>\! 0$, $|x - a| \!<\! \delta$ and $|f(x) - f(a)| \!\geq\! \varepsilon$ for some $x$}\tag{*}$$ For every $n \in \mathbb{N}^{>0}$, let $x_n$ be the real number such that $|x_n - a| < \frac{1}{n}$ and $|f(x_n) - f(a)| \geq \varepsilon$. Such a $x_n$ exists thanks to $(\text{*})$, by taking $\delta = \frac{1}{n}$. From that definition of $\{x_n\}_{n \in \mathbb{N}}$, it is immediate to see that $\lim_{n \to \infty} x_n = a$ and that $\lim_{n \to \infty} f(x_n) \neq f(a)$.

See also the proof quoted here to have an explanation of the argument that, in my opinion, is better than the one you quoted in the OP.

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What is missing in your formalization of the logical form of the argument? The universal quantification about sequences. I guess this can partially be the source of your misunderstanding.

Indeed, a more faithful formalization of the proof argument would be of the form: we assume $\lnot P$ and we show that this implies that $\lnot \forall s (Q(s) \implies R(s))$, which amounts to show that $\exists s (Q(s) \land \lnot R(s))$. Here, by $Q(s)$ I mean that the sequence $s$ converges to $a$, and by $R(s)$ I mean that the limit of $f$ applied to each element of the sequence $s$ is $f(a)$.

Said differently, in order to prove that the function $f$ is continuous in $a$ (i.e. to prove that $P$ holds), it is sufficient to prove that for every sequence $s$, if $Q(s)$ then $R(s)$. By contrapositive, it amounts to show that if $\lnot P$ does not hold, then there exists some sequence $s$ such that $Q(s)$ and $\lnot R(s)$ hold.