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.