Open Sets

Recall the definition of the Epsilon-Neighborhood of $x$ :

ε

-neighborhood of

a

$V_{ε} (a) = (a - ε, a + ε) = {x \in R | | x - a | < ε}$

open set

Let $U \subseteq R$ . Then $U$ is open if:

\forall x \in U \exists ε > 0 (V_{ε} (x) \subseteq U)

ie: each $x \in U$ has some $ε$ -neighborhood contained in $U$ .

$(a, b)$ is open.

Since choose $ε = min (| x - a |, | x - b |)$ implies that $V_{ε} (x) \subseteq (a, b)$ .

By constrast if we considered even a one-sided interval like $(a, b]$ then it is NOT open:

Notice if $x = b$ then any $\varepsilon > 0$ would have some element in $V_\varepsilon(b)$ be outside the interval.