Does the Proof Verify that an Open Ball is a Neighborhood in Metric Spaces?

[rjw5002]

First of all, sorry if the notation is hard to read, I'm still getting used to this text entry.

Question:
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.

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 $$\alpha$$ > 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).

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.

 

[Saketh]

You've shown that $(x-\epsilon, x+\epsilon)$ is a "neighborhood" of radius $\epsilon$ about $x$, but you haven't shown that it's open.

 

[rjw5002]

Ok, 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.

Does this complete the proof?

 

[rjw5002]

Sorry for misplacing this post, this belongs in Calculus and Beyond

 

[Saketh]

Rudin defines an open set as one which contains only interior points -- that is, points which have neighborhoods contained within the set. Prove that neighborhoods are open, given this definition. (Rudin does it for you in the theorem you mentioned, but it would be helpful for you to reproduce it yourself.)

Does the logic you just provided show that the neighborhood fits this definition?