Question

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in other words, if y (x - epsilon, x epsilon), then there is a delta > 0 such that (y - delta , y delta ) (x - epsilon, x epsilon).

Answer 1

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$. then we have that $x-\epsilon$. Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (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 $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$, and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$, then $y-\delta$. First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$. Combining the inequalities, we have that  $x-\epsilon$, therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.