Answer to Question #159250 in Real Analysis for Sylvia

Question

Show that every open interval is an open set

Answer 1

Let $(i,j)$ be an open interval. Let $a \in (i, j)$ . We show that, we can find $\epsilon > 0 \ni$ $N(a, \epsilon) \sub (i,j)$

Since $a \in (i,j)$ we have that $i<a<j$ this implies that $a-i > 0$ and $j-a > 0$

Set $\epsilon = min (a-i, j-a)$

Suppose further that $b \in N(a, \epsilon)$ this implies that $|b-a| < \epsilon$

$- \epsilon < b - a< \epsilon \\ a - \epsilon < b < a + \epsilon$

And since $\epsilon$ is as defined we have that $a - \epsilon \geq i \\ a + \epsilon \leq j$

This implies that $b \in (i, j)$

Hence $N(a, \epsilon ) \sub (i,j)$