Question about continuous function in terms of limits of sequences I am reading about continuous function, in this site http://en.wikipedia.org/wiki/Continuous_function specifically the section "Definition in terms of limits of sequences". My question is, let be $c\in \mathbb{R}$ an arbitrary element belonging to domain of $f$, where $f$ is in a closet and bounded set. Are there always a sequence $x_n$  which converges to $c$?.
thanks by your replies.
 A: There is always the constant sequence $x_n = c$ for all $n$.  In the extreme case that the domain of the function is $\{c\}$, there isn't anything else.
In general there is a dichotomy to be made according to whether $c$ is an isolated point of the domain $X$ of $f$.  (Let's assume that $X$ is a subset of the real numbers $\mathbb{R}$, which seems to be the context of the question.  In fact, with only mild notational change, this answer works in the context of arbitrary metric spaces.)  We say that a point $c \in X$ is isolated if there is a $\delta > 0$ such that if $y \in X$ and $|x-y| < \delta$, then $y= c$. 
For instance, if $f(x) = \sqrt{x^2(x-1)}$, the natural domain is $\{0\} \cup [1,\infty)$ and $0$ is an isolated point.  
Now, for $X \subset \mathbb{R}$ and $c \in X$, the following are equivalent:  
(i) $c$ is an isolated point of $X$.
(ii) Every sequence $\{x_n\}$ in $X$ which converges to $c$ is eventually constant: that is, $x_n = c$ for all sufficiently large $n$.
(iii) Every function $f: X \rightarrow \mathbb{R}$ is continuous at $c$. 
In other words, an isolated point is something of a trivial case.  
