Answer:
See proof below
Step-by-step explanation:
We will use properties of inequalities during the proof.
Let [tex]y\in (x-\epsilon,x+\epsilon)[/tex]. then we have that [tex]x-\epsilon<y<x+\epsilon[/tex]. Hence, it makes sense to define the positive number delta as [tex]\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}[/tex] (the inequality guarantees that these numbers are positive).
Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that [tex](y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)[/tex], and if we prove this, we are done. To prove it, let [tex]z\in (y-\delta,y+\delta)[/tex], then [tex]y-\delta<z<y+\delta[/tex]. First, [tex]\delta \leq y-(x-\epsilon)[/tex] then [tex]-\delta \geq -y+x-\epsilon[/tex] hence [tex]z>y-\delta \geq x-\epsilon[/tex]
On the other hand, [tex]\delta \leq x+\epsilon-y[/tex] then [tex]z<y+\delta\leq x+\epsilon[/tex] hence [tex]z<x+\epsilon[/tex]. Combining the inequalities, we have that [tex]x-\epsilon<z<x+\epsilon[/tex], therefore [tex](y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)[/tex] as required.
