Intuition. Equivalence of Characterization of Limits and Continuity (Abbott p106 t4.2.3, p110 t4.3.2)

Question

What are the intuitions of these equivalences? Not questioning about proofs or any rigour.
I question both equivalences jointly because they look similar.

And Are there any figures?

Abbott p 106 THeorem 4.2.3. (Sequential Characterization of Limits) Let $f : D → R$ and let $c$ be a limit point of D. c doesn't have to $ ∈ D$. TFAE.

(i) $\lim_{x \to c} f(x) = L$
(ii) All sequences $\{ x_n \} \subseteq D$, $x_n → c$ as $n → ∞$
implies $\{f(x_n)\}$ converges to $L$

Abbott p110 THeorem 4.3.2. Let $f : D → R$ and let $c$ be a limit point of D with $c ∈ D$. TFAE.

(i) f is continuous at c.
(ii) All sequences $\{ x_n \} \subseteq D$, $x_n → c$ as $n → ∞$
implies $\{f(x_n)\}$ converges to $f(c)$.
(iii) $\lim_{x \to c} f(x) = f(c)$.

https://math.la.asu.edu/~dajones/class/371/ch4.pdf p 1 sur 14

Answer 1

The intuition of (i)$\>\Rightarrow\>$(ii), i.e., $$\biggl(\lim_{x\to c}f(x)=L\quad \wedge\quad \lim_{n\to\infty} x_n=c\biggr)\qquad\Longrightarrow \qquad \lim_{n\to\infty} f(x_n)=L$$ is obvious, and the same is true for (i) $\>\Rightarrow\>$ (ii), i.e., $$\biggl(x\mapsto f(x)\ {\rm continuous\ at}\ c\quad \wedge\quad \lim_{n\to\infty} x_n=c\biggr)\qquad\Longrightarrow \qquad \lim_{n\to\infty} f(x_n)=f(c)\ .$$ What is not obvious (and is false in some topological spaces) is the fact that the "converse" also holds, i.e., $$\lim_{n\to\infty} f(x_n)=L\quad {\rm for\ all\ sequences}\quad x_n\to c\qquad\Longrightarrow \qquad \lim_{x\to c} f(x)=L\ ,$$ resp., $$\lim_{n\to\infty} f(x_n)=f(c)\ {\rm for\ all\ sequences}\ x_n\to c\qquad\Longrightarrow \qquad x\mapsto f(x)\ {\rm is\ continuous\ at}\ c\ .$$ So the first two implications are immediate consequences of the definition of limit, resp., continuity, but their "converses" are statements about the "fine structure" of ${\mathbb R}$ (or of any metric space, for that matter). These "converses" allow to treat convergence and continuity in metric spaces completely in terms of sequences instead of in terms of neighborhoods.

E.g., the last implication in the above list allows to prove the continuity of $f$ at $c$ by testing all imaginable sequences $x_n\to c\>$ (a huge infinity of sequences!) instead of working with distances $|x-c|$ and estimates for $|f(x)-f(c)|$ directly. Of course nobody ever tests all these sequences. In fact a convergence or continuity proof in terms of sequences can easily be converted into a direct proof.

Therefore let me close this answer with the following personal remark: In my view using sequences to better understand (or even define, as some authors do) limits, resp., continuity of functions is complicating matters unnecessarily. It adds at least two $\forall$'s and one $\exists$ in a nested sequence of quantifiers, which is difficult enough to swallow for many in its simplest form.

Answer 2

I will give some disconnected ideas (well, they are not totally disconnected). Hope you find'em useful.

• If $f$ is continuous and $x$ approaches $c$, then $f(x)$ approaches $f(c)$, no matter which values use $x$ to approach $c$.
• If $f$ is not continuous $x$ can approach $c$ in a way that $f(x)$ approaches $a$ and in another way $f(x)$ approaches $b\neq a$. Consider a lightbulb that is on $1$ second, off $1/2$ seconds, on $1/4$ seconds, off $1/8$ seconds and so on. If you close and open your eyes in the same way, you'll always see the lightbulb on and your limit is "on". But if your partner does just the opposite thing, he/she'll tell that the limit is "off". The guy that is making this wierd experiment has his eyes open, so he knows that there is no limit at all.
• A similar but simpler thing may occur if $f$ has different side limits. The guy that approaches from the left obtains a limit and the one that approaches from the right obtains a different one.

Answer 3

Note that the equivalences (for a metric space) depend on the axiom of countable choice, which is accepted by most but is strictly stronger than induction. If you didn't notice that, it's because most textbooks gloss over that, most incorrectly stating that it is induction. This affects the implication from the sequential criterion to the epsilon-delta criterion. The proof should go like this:

If $f(x) \not\to f(c)$ as $x \to c$:

  Let $ε>0$ such that:

    For any $δ>0$:

      $|f(x)-f(c)| > ε$ for some $x$ such that $|x-c|<δ$

  By the axiom of countable choice let $( x_n : n \in \mathbb{N} )$ be a sequence such that:

    For any $n \in \mathbb{N}$:

      $|x_n-c| < \frac{1}{n}$ and $|f(x_n)-f(c)| > ε$

  Then $x_n \to c$ as $n \to \infty$ because $\frac{1}{n} \to 0$ as $n \to \infty$

  Also $f(x_n) \not\to f(c)$ as $n \to \infty$

Therefore if $f(x_n) \to f(c)$ as $n \to \infty$ for any sequence $( x_n : n \in \mathbb{N} )$ such that $x_n \to c$ as $n \to \infty$:

  $f(x) \to f(c)$ as $x \to c$

Note

Induction does not work because it can only obtain any finite prefix of the sequence because each element requires an instantiation of an existential quantifier. Likewise Bolzano-Weierstrass theorem requires something even stronger than the axiom of countable choice, because each element in the convergent subsequence is dependent on the previous element, and the standard proof of it uses the axiom of dependent choice.