Sequential definition of continuity: What does "all sequences mean I've been exposed to both the classic and sequential definition of continuity. The sequential definition is the following:

A function $f: A \to \mathbb R$ is continuous at a point $a \in A$ if for each sequence $(x_n)$ in $A$ such that $x_n \to a$, we have $f(x_n) \to f(a)$.

I've watched some videos on this and also read other questions, but I'm not fully onboard with this definition yet.
What's clouding my understanding is "... if for each sequence $(x_n)$ in $A$ such that $x_n \to a$ ...". In particular:

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*What does "each sequence" (in the interval $A$) actually mean? My understanding is that we want to ensure there are no jumps/gaps in the graph of $f$, thus we'd need to check the output of $f$ for every real number on the $x$-axis to give a point on the $y$-axis - which is of course not possible - but how does "every sequence in $A$" cover all real numbers in the interval on the $x$-axis?

*When I look at a graph of for example the topologist's sine function, and we can focus on the interval $(-1, 1)$ and we let $a=0$ - then I intuitively think about two sequences on the $x$-axis: the sequence of real numbers approaching $0$ from the left and one approaching $0$ from the right (just "following" the $x$-axis in my mind towards $0$.) And, intuitively, I want to verify that both tend or not tend to $f(a)$ according to the definition. But - $f(0)$ is not even defined, since that'd be a division by $0$. How does this then fit with the definition?



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*The phrase "for each sequence in $A$" means that you can take any sequence $x_1,x_2,x_3, \dots$ so that all the $x_i$'s are in $A$, and if this sequence converges to $a$, then the sequence $f(x_1),f(x_2),f(x_3),\dots$ should converge to $f(a)$.

For example, the function $f(x) = 2x^2+3$ is continuous at $x=0$. For example, let $A$ be the interval $[-1,1]$. One possible sequence that converges to $0$ is the sequence $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots, \frac{1}{n}, \dots$. If you evaluate the function at these values, you get the sequence $3+\frac{1}{2}, 3 + \frac{2}{9}, 3 + \frac{1}{8}, \dots, 3+\frac{2}{n^2}, \dots$. This sequence certainly converges to $3$ (which is $f(0)$).
But of course that isn't the only sequence which converges to $0$. You could also have the sequence $\frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \dots, \frac{1}{n^2}$, and this also converges to $0$. Evaluating the function $f$ at these values gives the sequence $3 + \frac{2}{n^4}$, which again converges to $3$.
And then of course there are lots of other sequences which converge to $0$ (I only gave two examples). So the sequential definition of continuity in this example says that for every possible sequence which converges to zero, the corresponding sequence $f(x_i)$ (where you plug the values of the sequence in to the function) converges to $3$.


*How does the topologist's sine curve fit with this definition?

This function is not continuous at $a=0$. The reason is that we can find two different sequences which both converge to $0$, but where the limits of sequences $f(x_i)$ are different. For example, take for the first sequence
$$ \frac{1}{\pi}, \frac{1}{2\pi}, \frac{1}{3\pi}, \frac{1}{4\pi}, \dots, \frac{1}{n\pi}, \dots $$
This sequence certainly converges to $0$.
Since the function is $f(x) = \sin(1/x)$, the corresponding $f(x_i)$ sequence
is $\sin(1/(1/n\pi)) = \sin(n\pi) = 0$. In other words, this sequence is $0,0,0, \dots$. The limit of this sequence is zero.
On the other hand, consider the sequence
$$ \frac{1}{\pi/2}, \frac{1}{2\pi + \pi/2}, \frac{1}{4\pi+\pi2}, \dots, \frac{1}{2n\pi + \pi/2}, \dots $$
Again, this sequence also converges to $0$. But this time when we evaluate the function at these points, we get $\sin(2n\pi + \pi/2) = \sin(\pi/2) = 1$. So the sequence is just $1,1,1,1, \dots$, and so the limit is $1$ (and not $0$!).
