Proving the Openness of Neighborhoods in \Re with Metric \rho(x,y) = |x-y|

[rjw5002]

Homework Statement

Consider \Re with metric \rho (x,y) = |x-y|. Verify for all x \in \Re and for any \epsilon > 0, (x-\epsilon, x+\epsilon) is an open neighborhood for x.

Homework Equations

Neighborhood/Ball of p is a set Nr(p) consisting of all q s.t. d(p,q)<r for some r>0.

The Attempt at a Solution

Take \alpha > 0, \alpha < \epsilon. Take \rho(x, x-\alpha) = |x-(x- \alpha )| = \alpha < \epsilon.
and
Take \rho(x, x+\alpha) = |x-(x+\alpha)| = \alpha < \epsilon.
Therefore, any positive \alpha < \epsilon is in N\epsilon(x).

I initially misplaced this thread, and was told that this shows that (x-\epsilon, x+\epsilon) is a neighborhood but it was not an "open neighborhood." But there is a theorem (2.19 in Rudin) that says: "Every neighborhood is an open set." I must be misinterpreting this theorem then?

(x - \epsilon) is a limit point of the set, but (x - \epsilon) \notin N\epsilon (x), and (x + \epsilon) is a limit point of the set, but (x + \epsilon) \notin N\epsilon (x). Therefore the set is open.

Will this complete the proof?

 

[MathematicalPhysicist]

don't see what there is to prove here.
one definition of open set in R is that it contains an open ball, well obviously this interval is an open ball is it not?

 

[NateTG]

Different authors will sometimes use different definitions for particular terms. That said, loop quantum gravity has pointed out that the set you're dealing with is a basis element, thus clearly open.