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Tpy6a [65]
3 years ago
8

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).
Mathematics
1 answer:
kaheart [24]3 years ago
3 0

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.  

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