Find max $\{\epsilon : N(x;\epsilon) \subseteq S\}$, the largest $\epsilon$ such that the neighborhood centered at $x$ of radius $\epsilon$ is contained in $S$. That is, state the largest possible radius of an open interval centered at $x$ and contained in $S$.
$x=7.2$
$S=[1,4) \cup (4,9]$
Based upon what I know so far, the largest possible radius of an open interval centered at $x$ and contained in $S$ would be (I imagine) the boundary points, which would be my $x-\epsilon$ and $x+\epsilon$, where $\epsilon$ is equal to my boundary points.
Am I way off here?
I'm looking for the largest radius of an open interval centered at $x$ and contained in $S$, I want an $\epsilon$ that does not extend past my boundary points $\{4,9\}$
Since my boundary points are in $S$, I will use them to determine the largest radius.
Therefore I need to meet the following two conditions:
(1) $x-\epsilon\ge4$
(2) $x+\epsilon\le9$
Testing boundary (1) and setting $x=7.2$ I have $7.2-\epsilon \ge 4$ or $\epsilon \le 7.2-4=3.2$
If I use $\epsilon=3.2$ then $x-\epsilon=7.2-3.2=4$, which is in the set; however, $x+\epsilon=7.2+3.2=10.4$, which is not in the set. So this boundary point does not work.
Testing boundary (2) and setting $x=7.2$ I have $7.2+\epsilon \le 9$ or $\epsilon \le 9-7.2 = 1.8$
If I use $\epsilon=1.8$ then $x-\epsilon=7.2-1.8=5.4$, which is in the set and $x+\epsilon=7.2+1.8=9$, which is also in the set.
So $1.8$ is the largest $epsilon$ in $S$ that meets the requirements.