Proving the Metric Space Topology is a Topology

Let $(X, d)$ be a metric space. Define $\tau = \left \{ u \subseteq X | \exists S \subseteq B \ such\ that \ u = \bigcup_{s\in S}s \right \}$, where $B$ is the set of all open balls in $X$. Show that $\tau$ is a topology on $X$.

I can show easily that $\phi \in \tau$ and that $X \in \tau$.

I've tried to prove that if $A\subseteq \tau$, then $\bigcup_{a\in A}a \in \tau$, but I'm not sure if it is correct and would appreciate verification. Here is my proof:

Let $A\subseteq \tau$ We want to find $S\subseteq B$ such that $\bigcup_{a \in A}a=\bigcup_{s \in S}s$.

From the definition of $\tau$, we know that for each $a \in A$ there exists $S_a \in B$ such that $a=\bigcup_{s\in S_a}s$. Choose $S = \bigcup_{a \in A}S_a$ This is a subset of $B$ as each $S_a$ is a subset of $B$

Let $x\in \bigcup_{a \in A}a$, then $x \in a_i$ for at least some $a_i \in A$.

But, $\bigcup_{s \in S}s=(\bigcup_{s \in S_{a_i}}s)\cup(\bigcup_{s \in comp(S_{a_i})}s)=(a_i)\cup(\bigcup_{s \in comp(S_{a_i})}s)$ so $a_i \subseteq \bigcup_{s \in S}s$, thus $\bigcup_{a \in A}a \subseteq\bigcup_{s \in S}$

Now let $x\in \bigcup_{s \in S}s$, then $x \in b$ for at least one $b \in S$

Recall, $S = \bigcup_{a \in A}S_a$, so $b\in S_{a_i}$, for some $a_i\in A$ But, $a_i=\bigcup_{s\in S_{a_i}}s$, so $b \subseteq a$ and thus $b \subseteq \bigcup_{a \in A}a$ so $x \in \bigcup_{a \in A}a$. Thus, $\bigcup_{a \in A}a \supseteq\bigcup_{s \in S}$

So we have shown $\bigcup_{a \in A}a =\bigcup_{s \in S}$, as required.

Finally I'm trying to prove that if $u_1,u_2 \in \tau$, then $(u_1 \cap u_2)\in \tau$. I can't seem to figure it out, but here is my progress so far:

As $u_1,u_2 \in \tau$, there exists $S_1,S_2\subseteq B$ such that $u_1 = \bigcup_{s\in S_1}s$, and $u_2 = \bigcup_{s\in S_2}s$.

We need to find $S\subseteq B$ such that $(\bigcup_{s\in S_1}s) \cap (\bigcup_{s\in S_2}s)=\bigcup_{s\in S}s$

Intuitively I want $S$ to consist of an open ball centred on each point in $u_1 \cap u_2$ of small enough radius that the ball is contained entirely in $u_1 \cap u_2$, but I have been unable to translate this idea into maths.

Any help with the verification of part 3 or any ideas/help for the proof of part 3 would be greatly appreciated!

• is this literally this long! – MAN-MADE Aug 4 '17 at 8:54
• I know right!! Thats why I'm looking for verification, I feel like im missing something much simpler. – CoffeeCrow Aug 4 '17 at 8:59
• I think what you need to show is $B_d(x,\epsilon)=\{y\mid d(x,y)<\epsilon\}$ are basis elements. – MAN-MADE Aug 4 '17 at 9:05
• He is already showing that the open balls are basis elements. That's what $B$ is, it is the set of open balls. @MANMAID – Lee Mosher Aug 4 '17 at 14:31
• Before the place where you get stuck, I don't see anything particularly wrong with the proof. The reason it seems so long is that you are being very, very careful about set theoretic conclusions which more experienced writers would glide over with barely a notice, for instance in your sentence "This is a subset of $B$ as each $S_a$ is a subset of $B$". Your proof would look simpler if you also glided over these places, although then it would contain the same simple set theoretic jumps that everyone else's proof contains. – Lee Mosher Aug 4 '17 at 14:31

I don't like your choice of notation. It's better to use upper cases for sets. Having said that let's observe that $$\tau=\{U\subseteq X|\ \forall x\in U:\exists B\in\mathcal{B}:x\in B\subseteq U\}$$, where $\mathcal{B}$ is the family of all open balls in $(X,d)$. Now all you need to prove is that $\mathcal{B}$ is a base for $\tau$, i.e.

1. For each $x\in X$ , there exists $B\in\mathcal{B}$ such that $x\in B$;
2. For each $B_1,B_2\in\mathcal{B},$ if $x\in B_1\cap B_2,$ then there exists $B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$.

$U_1,U_2\in \tau\implies U_1\cap U_2\in \tau$ is immediate now.

Your notation could be better.

$O$ is open iff $$O = \bigcup \{B(x_i, r_i): i \in I(O)\}$$

for some set of centres $x_i$ and radii $r_i$ for some index set $i \in I(O)$.

Now, if we write $U$ and $V$ as such unions for index sets $I(U), I(V)$.

For every $y \in U \cap V$ we have $i_1 \in I(U)$ and $x_{i_1}, r_{i_1}$ and $i_2 \in I(V)$ and $x_{i_2}, r_{i_2}$ such that

$$y \in B(x_{i_1}, r_{i_1}) \cap B(x_{i_2}, r_{i_2})$$

Now, the triangle inequality says that if we take $r(y)=\min(r_{i_1}-d(y, x_{i_1}, r_{i_2}-d(y, x_{i_2})$, we have

$$B(y, r(y) \subseteq B(x_{i_1}, r_{i_1}) \cap B(x_{i_2}, r_{i_2})$$

so that

$$U \cap V = \bigcup \{B(y, r(y): y \in U \cap V\}$$

which is open as a union of balls. Proving we have a (local) base with the balls is easier though. But I thought I'd give a proof along your lines.

As to unions, we write $O_j, j \in J$ as a union as in the beginning with index set $I(O_j)$. Then with the index set $I = \bigcup_j I(O_j)$ we can write

$$\bigcup O_j = \bigcup_{i,j} \{B(x_i, r_i) :i \in I(O_j) \}$$