How to show something forms a base topologically speaking Suppose $(X,d)$ is a metric space. Consider the set of all open balls in $(X,d)$ as well as $\emptyset$. How would I show that together they form a base topologically speaking? I mean it seems rather straightforward. You have a collection of open sets (or balls) along with $\emptyset$ which is also open. Then if you take the union you should form a base right? Although I would like to mathematically write it down.

Let $X$ be a set. A collection $\mathcal{B}$ of subsets of $X$ is a base for some (then uniquely determined) topology iff 


*

*$\cup \mathcal{B} = X$ 

*For all $B_1, B_2 \in \mathcal{B}$, for all $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$, such that $x \in B_3 \subset B_1 \cap B_2$


If then $(X,d)$ is a metric space and $\mathcal{B} = \{B(x, r): x \in X, r > 0 \}$, the collection of all open balls, does fulfill these conditions.
To check 1. we only need to notice that $x \in B(x,1) \in \mathcal{B}$, for every $x \in X$.
For 2. pick $B(p, r), B(q, s) \in \mathcal{B}$, and $x \in B(p,r) \cap B(q, s)$. Then $d(p, x) < r$ and $d(q, x) < s$. Then setting $ t = \min(r - d(p,x), s - d(q,x)) > 0$. Then $B(x, t) \subset B(p,r) \cap B(q, s)$, by the triangle inequality (check this!)
