Doubts: Discontinuity Criterion and motivation behind it Discontinuity Criterion Let $A \in \mathbb{R}$, let $f : A \to \mathbb{R}$, and let $ c \in A$. Then f is discontinuous at c if and only if there exists a sequence $(x_n)$ in A such that $(x_n)$ converges to c, but the sequence $(f(x_n))$ does not converge to $f(c)$
I'm having trouble understanding this definition. With the divergence criteria, we pick out any sequence that converges to c such that it isn't equal to c. However, over here, it says that the sequence is in A. What does that mean? How does that change the way I pick out sequences. Moreover, I don't think I get the motivation behind the sequential criteria. Can anyone give me a geometric interpretation(or any other for that matter) of this so I can see how functions and their convergence and continuity is related to that of sequences?
Thank you.
 A: I'll try to give a good example (as intuitive as I can) to help you understand this definition.

Define the function $f:[0,2]\rightarrow\mathbb{R}$ 
$$
f(x)=
\begin{cases} 
      1,& 0\leq x< 1 \\
1000,&x=1\\
      3,& 1< x\leq 2\\
\end{cases}
$$
And note that $f$ is not continuous at $x=1$. 
Now, define a sequence $(x_{n})_{n=1}^{\infty}\subseteq[0,2] $ by
$$x_{n}=1-\dfrac{1}{n}$$
for all positive integers $n\in\mathbb{N}$. Then you can easily verify that $\lim\limits_{n\rightarrow \infty}x_{n}=1$. But now, observe that when you take the limit of $f$ evaluated at each $x_{n}$ you can see that
$$\lim\limits_{n\rightarrow\infty}\bigl\{f(x_{n})\bigr\}=\lim\limits_{n\rightarrow\infty}\Bigl\{f\bigl(1-\frac{1}{n}\bigr)\Bigr\}$$
For me personally, I think that it helps to think of the actual terms in this sequence. The first few are
$$\Bigl\{f(0),f\Bigl(\frac{1}{2}\Bigr),f\Bigl(\frac{2}{3}\Bigr),f\Bigl(\frac{3}{4}\Bigr),\dots\Bigr\}=
\Bigl\{1,1,1,1,\dots\Bigr\}$$
So $\lim\limits_{n\rightarrow\infty}x_{n}=1$, but clearly $\lim\limits_{n\rightarrow\infty}\bigl\{f(x_{n})\bigr\}=1\neq 1000$. 

Now, why do we care? Why not just negate the $\epsilon-\delta$ definition of continuity to characterize discontinuity? The answer to this is that it is often much easier to show a function is not continuous at a point using this Sequential Criterion for discontinuity than it is to use the negation of the $\epsilon-\delta$ definition. 
If you want to convince yourself, try this as an exercise. Consider the function 
$$g(x)=
\begin{cases}
\sin\bigl(1/x\bigr),&x\neq 0\\
0,& x=0
\end{cases}
$$
on any interval containing $0$ and try to produce a single $\epsilon > 0$ such that for every $\delta>0$, there exists $x\in(-\delta,\delta)$ with $0<|x-0|<\delta$ and $|g(x)-g(0)|$. And after a minute or two, just make a sequence $(x_{n})_{n=1}^{\infty}$ which converges to $0$, such that $g(x_{n})$ doesn't converge to $g(0)$, and you are done.

The Sequential Criterion for Discontinuity also constitutes a beautifully short proof that Dirichlet's Function defined by 
$$g(x)=
\begin{cases}
0, &x\in\mathbb{R}\setminus\mathbb{Q}\\
1, &x\in\mathbb{Q}\\
\end{cases}
$$
Is discontinuous at every rational number in $\mathbb{Q}$ (Just construct a sequence of rational numbers which converge to an irrational number). 
NOTE. Discontinuity of $g$ can also be proved using the Archimedean Property.

Hope this helps, let me know if you have any questions
