194: Set of Subsets with Whole Set and Empty Set Constitutes Subbasis

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that set of subsets with whole set and empty set constitutes subbasis

---

Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of subbasis.
• The reader admits a criterion for any collection of open sets to be a basis.

---

Target Context

• The reader will have a description and a proof of the proposition that for any set, any set of subsets with the whole set and the empty set constitutes a subbasis.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any set, $S$, any set, $S^{\prime }$, of subsets of $S$ with the whole set and the empty set constitutes a subbasis.

---

2: Proof

Let us show that the set of open sets defined as that each open set is any union of any finite intersections of subsets from $S^{\prime }$ constitutes a topology. The whole set and the empty set are obviously included.

Let us think of any union of open sets, ${\cup }_{\alpha }{\cup }_{\beta }{\cap }_i{S}_{\alpha ,\beta ,i}$, where $\{\alpha \}$ is a possibly uncountable index set, $\{\alpha ,\beta \}$ is a possibly uncountable index set for each fixed $\alpha$, $\{\alpha ,\beta ,i\}$ is a finite index set for each fixed $\alpha ,\beta$, and $S_{\alpha ,\beta ,i}$ is a subset from $S^{\prime }$. ${\cup }_{\alpha }{\cup }_{\beta }{\cap }_i{S}_{\alpha ,\beta ,i}={\cup }_{\alpha ,\beta }{\cap }_i{S}_{\alpha ,\beta ,i}$, which is a union of finite intersections of subsets from $S^{\prime }$, open.

As for finite intersections of open sets, let us 1st think of intersections of 2 open sets. Let us think of any intersection of 2 open sets, $({\cup }_{\alpha }{\cap }_i{S}_{\alpha ,i})\cap ({\cup }_{\beta }{\cap }_j{S}_{\beta ,j})$ where $\{\alpha \}$ and $\{\beta \}$ are possibly uncountable index sets, $\{\alpha ,i\}$ and $\{\beta ,j\}$ are finite index sets for each fixed $\alpha$ and $\beta$, and $S_{\alpha ,i}$ and $S_{\beta ,j}$ are subsets from $S^{\prime }$. Let us express $A_{\alpha }:={\cap }_i{S}_{\alpha ,i}$ and $B_{\beta }:={\cap }_j{S}_{\beta ,j}$ for brevity. $({\cup }_{\alpha }{A}_{\alpha })\cap ({\cup }_{\beta }{B}_{\beta })={\cup }_{\alpha }(A_{\alpha }\cap {\cup }_{\beta }{B}_{\beta })$, because for any $p\in ({\cup }_{\alpha }{A}_{\alpha })\cap ({\cup }_{\beta }{B}_{\beta })$, $p\in {\cup }_{\alpha }{A}_{\alpha }$ and $p\in {\cup }_{\beta }{B}_{\beta }$, $p\in A_{\alpha }$ for an $\alpha$ and $p\in B_{\beta }$ for a $\beta$, $p\in A_{\alpha }$ for an $\alpha$ and $p\in {\cup }_{\beta }{B}_{\beta }$, $p\in A_{\alpha }\cap {\cup }_{\beta }{B}_{\beta }$ for an $\alpha$, so, $p\in {\cup }_{\alpha }(A_{\alpha }\cap {\cup }_{\beta }{B}_{\beta })$; for any $p\in {\cup }_{\alpha }(A_{\alpha }\cap {\cup }_{\beta }{B}_{\beta })$, $p\in A_{\alpha }\cap {\cup }_{\beta }{B}_{\beta }$ for an $\alpha$, $p\in A_{\alpha }$ for an $\alpha$ and $p\in B_{\beta }$ for a $\beta$, $p\in {\cup }_{\alpha }{A}_{\alpha }$ and $p\in {\cup }_{\beta }{B}_{\beta }$, so, $p\in ({\cup }_{\alpha }{A}_{\alpha })\cap ({\cup }_{\beta }{B}_{\beta })$. $A_{\alpha }\cap {\cup }_{\beta }{B}_{\beta }={\cup }_{\beta }(A_{\alpha }\cap B_{\beta })$, because for any $p\in A_{\alpha }\cap {\cup }_{\beta }{B}_{\beta }$, $p\in A_{\alpha }$ and $p\in {\cup }_{\beta }{B}_{\beta }$, $p\in A_{\alpha }$ and $p\in B_{\beta }$ for a $\beta$, $p\in A_{\alpha }\cap B_{\beta }$ for a $\beta$, so, $p\in {\cup }_{\beta }(A_{\alpha }\cap B_{\beta })$; for any $p\in {\cup }_{\beta }(A_{\alpha }\cap B_{\beta })$, $p\in A_{\alpha }\cap B_{\beta }$ for a $\beta$, $p\in A_{\alpha }$ and $p\in B_{\beta }$ for a $\beta$, $p\in A_{\alpha }$ and $p\in {\cup }_{\beta }{B}_{\beta }$, so, $p\in A_{\alpha }\cap {\cup }_{\beta }{B}_{\beta }$. So, $({\cup }_{\alpha }{A}_{\alpha })\cap ({\cup }_{\beta }{B}_{\beta })={\cup }_{\alpha }({\cup }_{\beta }(A_{\alpha }\cap B_{\beta }))={\cup }_{\alpha \beta }{A}_{\alpha }\cap B_{\beta }={\cup }_{\alpha \beta }({\cap }_i{S}_{\alpha ,i})\cap {\cap }_j{S}_{\beta ,j}$)\), which is a union of finite intersections of subsets from $S^{\prime }$, open. Obviously, by induction, any finite intersection of open sets is open.

So, the set of open sets defined by $S^{\prime }$ is a topology.

The set of all the finite intersections of $S^{\prime }$ constitutes a basis for the topology, because it satisfies a criterion for any collection of open sets to be a basis: for any topological space, T, any set of open sets, $\{B_{\alpha }\}$, is a basis of T, if and only if each open set on T is the union of some elements of the set.

So, $S^{\prime }$ is a subbasis.

---

3: Note

Although the whole set and the empty set are prevalently claimed to be included in the topology based on the conventions that the intersection of nothing is the whole set and the union of nothing is the empty set, those conventions do not seem natural (especially the former one), and just requiring that $S^{\prime }$ contains the whole set and the empty set seems more natural and more clear with no problem.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>