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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.

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Your question doesn't make much sense. You were asking about continuity and then you switched to closed sets. If $f$ a function or a subset? –  Asaf Karagila Jul 19 '12 at 22:30
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What about $x_n = c$ for all $n$? –  Dylan Moreland Jul 19 '12 at 22:42
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Why is $f$ in the closet? I thought we as a society were past that... –  Quinn Culver Jul 19 '12 at 23:24

1 Answer 1

up vote 1 down vote accepted

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.

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