Doubts about definition of open sets in "Understanding Analysis" by Stephen Abbott

In the book "Understanding Analysis" by Stephen Abbott, the author defines an open set as:

A set $O \subseteq \mathbb{R}$ is open if for all points $a \in O$ there exists an $\epsilon$-neightborhood $V_\epsilon (a) \subseteq O$.

Here the $\epsilon$-neighborhood is defined as: $$V_\epsilon(a) = \{ x \in \mathbb{R} \mid |x-a| < \epsilon \}$$

Now, according to the above definitions, if we consider the closed interval $[c,d]$, then it appears to me that $[c,d]$ is not an open set (because for the points $c$ and $d$ we cannot find an $\epsilon$-neighborhood such that $V_\epsilon(c) \subseteq [c,d]$ nor $V_\epsilon(d) \subseteq [c,d]$). However, I know that for any metric space $X$ (and $[c,d]$ is a metric space), the whole set $X$ is open in $X$. Thus, something has gone wrong in the above reasoning. Can anybody shed some light on this?

You changed your metric space from $\mathbb{R}$ to $X$. So your definition of $V_\epsilon(a)$ needs to change, too:
$$V_\epsilon(a) = \{ x \in X \mid |x-a| < \epsilon \}$$
• Thanks, that is what I thought but I wasn't sure. Do you know why Stephen Abbott defines $\epsilon$-neighborhoods like that, or do you think it is a (sloppy) mistake on his part? Sep 16, 2014 at 23:19
• @Hunter, what's the mistake? In the definition you quote only defines "open" for subsets of $\mathbb R$, in which case his definition of $\epsilon$-neighborhood is fine. Sep 16, 2014 at 23:30