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?


  • $\begingroup$ 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. $\endgroup$ – 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.


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