3
$\begingroup$

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?

$\endgroup$

1 Answer 1

4
$\begingroup$

You changed your metric space from $\mathbb{R}$ to $X$. So your definition of $V_\epsilon(a)$ needs to change, too:

$$V_\epsilon(a) = \{ x \in X \mid |x-a| < \epsilon \}$$

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – Hunter
    Sep 16, 2014 at 23:19
  • $\begingroup$ It's a standard definition in analysis, proven equivalent to the definition in topology for a given metric space topology. $\endgroup$ Sep 16, 2014 at 23:29
  • 3
    $\begingroup$ @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. $\endgroup$ Sep 16, 2014 at 23:30
  • $\begingroup$ @SantiagoCanez ahh of course, I can't believe I didn't get that. Thanks! $\endgroup$
    – Hunter
    Sep 16, 2014 at 23:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .