A Question Involving Metrics and Topologies

So, my professor asked me to prove a theorem, and I only have a question about PART of it.

Theorem: Let $\displaystyle X$ be a set and let $\displaystyle \mu$ be a metric on $\displaystyle X$. The set $\displaystyle B = \{\emptyset\} \cup \{B_r(x) : x \in X, r \in \mathbb{R}\}$ forms a basis for a topology on $\displaystyle X$.

My question is this: Is it necessary to union the empty set with the open balls? I don't understand why this is necessary.

Re: A Question Involving Metrics and Topologies

Quote:

Originally Posted by

**Aryth** So, my professor asked me to prove a theorem, and I only have a question about PART of it.

Theorem: Let $\displaystyle X$ be a set and let $\displaystyle \mu$ be a metric on $\displaystyle X$. The set $\displaystyle B = \{\emptyset\} \cup \{B_r(x) : x \in X, r \in \mathbb{R}\}$ forms a basis for a topology on $\displaystyle X$.

My question is this: Is it necessary to union the empty set with the open balls? I don't understand why this is necessary.

Without going into detail here is a quick answer. A topology on a set contains the empty set. In this case, because by definition balls are not empty it is necessary to include the empty set in the definition of the basis.

Re: A Question Involving Metrics and Topologies

Actually, the only instance that I can think of where you would need the empty set as part of the basis would be if the set $\displaystyle X$ consisted of only one point. So long as the set contains at least two points, metrizability will ensure that there exists two disjoint non-empty open sets, and the empty set would not be necessary to create either (either through unions or finite intersections). So, any basis for a space with at least two points will pick up the empty set through finite intersections of disjoint open sets. Seems to me that the single point set is the only exception. (Unless you consider the empty set to be metrizable...)

Re: A Question Involving Metrics and Topologies