# Topology generated and metric topology.

Definition 1. If $$X$$ is a set, a basis for a topology on $$X$$ is a collection $$\mathcal{B}$$ of subset of $$X$$ such that

$$(i_B)\quad$$ For each $$x\in X$$, there is at least $$B\in\mathcal{B}$$ such that $$x\in B$$;

$$(ii_B)\quad$$ If $$x$$ belongs to the intersection of two basis element $$B_1$$ an $$B_2$$, there is a basis element $$B_3$$ containing $$x$$ such that $$B_3\subseteq B_1\cap B_2$$.

If $$\mathcal{B}$$ satisfies $$(i_B)$$ and $$(ii_B)$$, then we define the topology $$\mathcal{T}_{\mathcal{B}}$$ generated by $$\mathcal{B}$$ as follows: a subset $$U$$ of $$X$$ is said open in $$X$$ if for each $$x\in U$$, there is a basis element $$B\in\mathcal{B}$$ such that $$x\in B$$ and $$B\subseteq U.$$ We observe that $$\mathcal{B}\subseteq\mathcal{T}_{\mathcal{B}}.$$

When we say that $$\mathcal{B}$$ is a basis for a specific topology $$\mathcal{T}$$ we are saying that $$\mathcal{B}$$ is a basis for a topology on $$X$$ and $$\mathcal{T}=\mathcal{T}_{\mathcal{B}}.$$ Moreover, I just proved that $$\mathcal{T}_{\mathcal{B}}$$ equals the collection of all unions of elements of $$\mathcal{B}$$.

Viceversa, If we start from a set $$X$$ and a basis $$\mathcal{B}$$ for a topology on $$X$$, then the collection $$\mathcal{T}_{\mathcal{B}}'=\bigg\{\bigcup{\mathcal{C}\;\bigg|\;\mathcal{C}\subseteq \mathcal{B}}\bigg\}$$ is a topology on $$X$$ and $$\mathcal{T}_{\mathcal{B}}=\mathcal{T}_{\mathcal{B}}'$$.

We denote with $$\mathcal{B}_d$$, where $$d$$ is a metric on $$X$$, the family of all open ball of center $$x$$ and radius $$r>0$$, that is $$\big\{B(x,r)\;\big|\;x\in X, r>0\big\},$$ where $$B(x,r)=\{y\in X\;|\; d(x,y)< r\}.$$

Definition 2. If $$d$$ is a metric on a set $$X$$, then $$\mathcal{B}_d$$ is a basis for a topology on $$X$$, called the metric topology $$\mathcal{T}_d.$$

This definition means that $$\mathcal{B}_d$$ is a basis on $$X$$ and $$\mathcal{T}_d=\mathcal{T}_{\mathcal{B_d}}$$, that is if $$U\in\mathcal{T}_d$$, exists $$r_1>0$$ such that $$x\in B(x_1,r_1)\subseteq U.$$ I've already proved that $$\mathcal{B}_d$$ check $$(i_B)$$ and $$(ii_b)$$ and that exists $$r>0$$ such that $$B(x,r)\subseteq B(x_1,r_1)\subseteq U$$. Then $$(U\in\mathcal{T}_d) \iff(\forall y\in U, \exists r>0\;\text{such that}\; B(y,r)\subseteq U).$$ We observe that the open balls are actually open set, because $$\mathcal{B}_d\subseteq \mathcal{T}_d.$$

Proposition 1. If $$U\in\mathcal{T}_d$$, then $$U$$ is unions of open balls. That is $$\mathcal{T}_d$$ equals the collection of all unions of open balls.

Proof. Let $$\mathcal{C}\subseteq\mathcal{B_d}$$, since $$\mathcal{T}_d$$ is a topology $$\bigcup\mathcal{C}\in\mathcal{T}_d$$.Viceversa, let $$U\in\mathcal{T}_d$$ then for all $$x\in U$$ exists $$r_x>0$$ such that $$B(x,r_x)\subseteq U$$. Therefore $$U=\bigcup_{x\in U} B(x,r_x).$$

Now, we define the topology metric $$d$$ on a set $$X$$ as $$\mathcal{T}_d'=\bigg\{\bigcup\mathcal{C}\;\bigg|\;\mathcal{C}\subseteq\mathcal{B}_d\bigg\}.$$ I've just proved that $$\mathcal{T}_d'$$ is a topology on $$X$$.

Proposition 2. $$\mathcal{T}_d'=\mathcal{T}_d$$

Proof. Since $$\mathcal{B}_d\subseteq\mathcal{T}_d$$ and $$\mathcal{T}_d$$ is a topology if $$\mathcal{C}\subseteq\mathcal{B}_d\subseteq\mathcal{T}_d$$ we have that $$\bigcup\mathcal{C}\in\mathcal{T}_d$$, therefore $$\mathcal{T}_d'\subseteq \mathcal{T}_d$$. Obviously $$\mathcal{B}_d\subseteq\mathcal{T}_d'$$, indeed $$B(x,r)=\bigcup\{B(x,r)\}$$; we must prove that $$\mathcal{T}_d\subseteq\mathcal{T}_d'$$. Let $$U\in\mathcal{T}_d$$ then for all $$x\in U$$ eists $$r>0$$ such that $$B(x,r)\subseteq U$$, then $$U\subseteq\bigcup_{x\in U}B(x,r)\subseteq U$$ therefore $$U\in\mathcal{T}_d'$$.

Question. Is this reasoning correct, or can we say more about the generated topology and the metric topology?

Thanks!

• Bases are useful tools. E.g. if a space has a countable base then it has a countable dense subset. And if a metrizable space has a countable dense subset then it has a countable base. – DanielWainfleet Mar 18 at 5:56

What you have done is mostly correct and only a little part of it is connected to a metric at all, really.

If we have a collection of subsets of $$X$$, $$\mathcal{B}$$, that obeys your axioms $$(i)_B$$ and $$(ii)_B$$, we can define a topology from $$\mathcal{B}$$ in two, essentially equivalent ways:

$$\mathcal{T}_1(\mathcal{B})$$ can be defined as

$$\mathcal{T}_1(\mathcal{B}) = \{O \subseteq X: \forall x \in O: \exists: B_x \in \mathcal{B}: x \in B_x \subseteq O\}$$

as you do in the beginning. We can check this is indeed a topology using these two axioms. It is inspired by the most standard way to define a topology from a metric: if $$A$$ is a subset of $$X$$ and $$x \in A$$ we call $$x$$ an interior point of $$A$$, if there is some $$r>0$$ such that $$B(x,r) \subseteq A$$. A set is then called open iff all its points are interior points.

We note that this is exactly the topology we get when we use $$\mathcal{B}_d = \{B(x,r): x \in X, r>0\}$$ as a base, which we can do after we have checked (as you claim to have done) that $$\mathcal{B}_d$$ obeys $$(i)_B$$ and $$(ii)_B$$.

The second way to define a topology from the base $$\mathcal{B}$$ is

$$\mathcal{T}_2(\mathcal{B}) = \{\bigcup \mathcal{C}: \mathcal{C} \subseteq \mathcal{B}\}$$

and this is the exactly same topology as $$\mathcal{T}_1(\mathcal{B})$$:

If $$O \in \mathcal{T}_1(\mathcal{B})$$ then for each $$x \in O$$ pick $$B_x \in \mathcal{B}$$ as promised by the definition of $$\mathcal{T}_1(\mathcal{B})$$ and note that $$\bigcup \{B_x: x \in O\} = O$$ each $$x \in O$$ is in its "own" $$B_x$$, hence in the left hand union, and all sets in the left hand union are subsets of $$O$$ hence so is their union (for the other inclusion). Then using $$\mathcal{C} = \{B_x: x \in O\}\subseteq \mathcal{B}$$ we see that $$O \in \mathcal{T}_2(\mathcal{B})$$.

OTOH, if $$O \in \mathcal{T}_2(\mathcal{B})$$ write $$O = \bigcup \mathcal{C}$$ for some $$\mathcal{C} \subseteq \mathcal{B}$$. If now $$x \in O$$ this means that there is some $$C\in \mathcal{C} \subseteq \mathcal{B}$$ such that $$x \in C$$ (definition of union) and as clearly $$C \subseteq \bigcup \mathcal{C}=O$$ we have shown (as $$x \in O$$ was arbitrary) that $$O \in \mathcal{T}_1(\mathcal{B})$$, showing equality of these topologies.

Note that this holds regardless of whether the base comes from a metric or not.

So the fact that the induced topology on a metric space is the set of unions of open balls, is a combination of the fact that $$\mathcal{B}_d$$ obeys the two base properties (the only place where we use the metric axioms (and not even their full force)) plus the general fact I descibed above on how the two ways of generating a topology from a base ($$\mathcal{T}_1(\mathcal{B})$$ vs $$\mathcal{T}_2(\mathcal{B})$$) are equivalent in general.

You might recognise some or all of the arguments you used yourself. I just want to point out the separation of these two facts.