Need explaining on aspect of a proof So the textbook I'm reading just stated three definitions of continuity:
a) f is continuous at a (using neighborhoods)
b) the epsilon-delta definition
c) If $x_n$ is a any sequence of elements of $D(f)$ which converges to a, then the sequence $(f(x_n))$ converges to $f(a)$
Then it proceeded to show that the three statements are equivalent. I'm having trouble with understanding the part of the proof where c) implies a).
The portion of the proof goes as follows:
We shall argue that if condition a) does not hold, then condition c) does not hold. If a) fails, then there exists a neighborhood $V_0$ of $f(a)$ such that for any neighborhood $U$ of $a$, there is an element $x_U$ belonging to $D(f)\cap U$ but such that $f(x_U)$ does not belong to $V_0$. For each natural number consider the neighborhood $U_n$ of $a$ defined by $U_n=\{x\in \mathbb{R^p}:\|x-a\|<\frac{1}{n}\}$;from the preceding sentence, for each $n$ in $\mathbb{N}$ there is an element $x_n$ belong to $D(f)\cap U$ but such that $f(x_n)$ does not belong to $v_0$. The sequence $(X_n)$ just constructed belongs to $D(f)$ and converges to a, yet none of the elements of the sequence $(f(x_n))$ belong to the neighborhood $V_0$ of $f(a)$. Hence we have constructed a sequence for which the condition c) does not hold. This shows that part c) implies a).$\square$
Would the proof have been any different if $U_n$ been defined differently? Or not defined at all?
 A: The sequence of nested balls with radii $1/n$ is a standard argument technique.  It can be replaced by any nested sequence of balls or even any nested sequence of open sets as long as their diameters go to zero.  Thus, for any $\epsilon > 0$, there is eventually an $N \in \mathbb{N}$ such that for all $n > N$, the $n^\text{th}$ nested ball or nested open set has radius less than $\epsilon$.  In this way, the nested sequence of sets is doing the job of bridging between the sequence and the $\epsilon$ in parts (b) and (c) of the theorem.
A: The essential conditions to $U_n$ are:

  
*
  
*Each $U_n$ is a neighborhood of $a$.
  
*Every neighborhood of $a$ contains at least one $U_n$ (for some $n\in\Bbb N$).
  
*$U_n\supseteq U_{n+1}$ for all $n$'s.
  

Note that if a collection $U_i$ of subsets satisfies properties 1. and 2., it is called a basis of neighborhoods around $a$, and a critical thing is that $\Bbb R^N$ has countable basis around each of its points.
So, 1.-3. together guarantee that, if $x_n\in U_n$ then $x_n\to a$ which is the key part of the proof.
