- #1

#### rjw5002

**Question:**

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

**Relevant Definitions:**

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

**Attempt at solution:**

Take [tex]\alpha[/tex] > 0, [tex]\alpha[/tex] < [tex]\epsilon[/tex]. Take [tex]\rho[/tex](x, x-[tex]\alpha[/tex]) = |x-(x- [tex]\alpha[/tex] )| = [tex]\alpha[/tex] < [tex]\epsilon[/tex].

and

Take [tex]\rho[/tex](x, x+[tex]\alpha[/tex]) = |x-(x+[tex]\alpha[/tex])| = [tex]\alpha[/tex] < [tex]\epsilon[/tex].

Therefore, any positive [tex]\alpha[/tex] < [tex]\epsilon[/tex] is in N[tex]\epsilon[/tex](x).

So, I think that this proof is ok, but I also feel that it is missing something.

Thanks in advance for any comments or suggestions.