# Equivalence of the definitions of the Subbasis of a Topology [duplicate]

I came to know about two definitions of a subbasis of a topology.

1. A collection of subsets of $X$ viz. $S$ is said to be a subbasis of a topology $T$ on $X$ if $S$ covers $X$ and the finite intersections of elements of the subbasis form the basis elements (H.L.Royden).
2. If we consider the all the topologies on $X$ containing $S$, then their intersection is also a topology on $X$ and it is the smallest topology containing $S$. Then $S$ is said to be a subbasis of the smallest topology, and the topology is said to be generated by $S$.

Now I want to prove their equivalency, i.e. $S$ is a subbasis of a topology $T$ iff $T$ is generated by $S$. I have tried in the way:

First let $T$ is generated by $S$. Then by $2^{nd}$ definition $T$ is a topology containing $S$. Let $B$ be the set of all finite intersections of elements of $S$.Then $S$ is a subset of $B$ and $B$ is a subset of $T$ (as $T$ is a topology). LET $T_1$ be the collection of possible unions of elements of $B$, we will show that $T = T_1$

After this I am not able to proceed. Also I am unable to sketch the converse part. If something is wrong, please let me know, and also I hope to get help. Thank You.

## marked as duplicate by Carsten S, heropup, C. Falcon, Shailesh, Zain PatelMay 3 '17 at 1:14

• Ya.. that is as – hiren_garai May 2 '17 at 16:34
• That is also done by me. – hiren_garai May 2 '17 at 16:34
• And you should not have asked two identical questions. I suggest that you delete the other one. – Carsten S May 2 '17 at 16:40
• Can you tell me how to delete a question here plz. – hiren_garai May 2 '17 at 16:44
• If you look at a question of your own you should find a "delete" link below the tags. – Carsten S May 2 '17 at 16:47

If $B$ is defined as the the collection of all finite intersections of elements of $S$ then it can be shown (and that is not really difficult) that the collection of unions of elements of $B$ (let's denote it by $B^{\bigcup}$) is a topology.

Denote $\mathcal T$ as the collection of all topologies that contain $S$ as a subcollection and notice that:

• $B^{\bigcup}\subseteq\tau$ for every $\tau\in\mathcal T$. This because $S\subseteq\tau\implies B\subseteq\tau\implies B^{\bigcup}\subseteq\tau$.
• $B^{\bigcup}\in\mathcal T$.

These two facts together imply that $B^{\bigcup}=\bigcap\mathcal T$.

The way I would show this is a little different. It goes as follows:

Let $\mathcal{O}'$ be the set of subsets of $X$ which are unions of finite intersections of sets of $S$. We first show that this defines a topology on $X$ and then that this topology is equal to the topology generated by $S$. This shows equivalency of the two definitions.

Showing that this defines a topology is straightforward, the only subtlety is how to get the empty set and the space $X$. The empty set is defined by the empty union of elements of $S$ and $X$ is defined as the empty intersection of elements of $S$. Then of course you have to show the other properties (arbitrary unions and finite intersections).

Now we want to show $\mathcal{O}' = \mathcal{O}(S)$ where the latter denotes the topology generated by $S$.

The relation $S \subset \mathcal{O}'$ is obvious because we defined $\mathcal{O}'$ to be the set of subsets that are unions of finite intersections of sets of $S$ which of course includes $S$ itself. But $\mathcal{O}(S)$ is the intersection of all topologies containing $S$ so we must also have $\mathcal{O}(S) \subset \mathcal{O}'$.

The other direction is also not hard to see: Take an open set $O \in \mathcal{O}'$. By definition, $O$ is the union of finite intersection of sets in $S$. But since $S\subseteq \mathcal{O}(S)$ we must have $O \in \mathcal{O}(S)$. Thus we have shown

$$\mathcal{O}' = \mathcal{O}(S)$$

Or in words: The topology generated by $S$ is equal to the topology generated by unions of finite intersections of sets of $S$.

The definitions are not equivalent. Royden has restricted himself to a (common) subtype of the general definition as presented in 2, which is more general.

To start with an example: take $$X = \{0,1\}$$ in the indiscrete topology $$\mathscr{T}_i = \{\emptyset,X\}$$. Then by the definition $$\mathcal{S} = \emptyset$$ is a subbase for this, as the smallest topology on $$X$$ that contains the empty set (and as all sets contain the empty set, this is just the smallest topology on $$X$$ full stop) is the indiscrete one (all topologies must contain $$\emptyset$$ and $$X$$). But $$\emptyset$$ is clearly no cover of $$X$$. In fact any subset of $$\mathscr{T}_i$$ is a subbase for the same reason, but I took the most extreme example.

Royden's definition is misleading in another way: he says "the base for $$X$$" instead of a base for the topology on $$X$$. (A set by no means has a unique basis, in general, except in trivial cases).

His definition refers to the following general fact

Proposition

Let $$X$$ be any set (no topology yet) If $$\mathcal{S} \subseteq \mathscr{P}(X)$$ then the smallest topology $$\mathcal{T}_{S}$$ on $$X$$ that contains $$\mathcal{S}$$ as a subset (i.e. the intersection of al such topologies) is given by $$\mathcal{T}_S = (\mathcal{S}^{\cap, <\infty})^{\cup}$$

Here for a collection $$\mathcal{A}$$ of subsets of $$X$$ we define $$\mathcal{A}^{\cap, < \infty} = \{\bigcap \mathcal{A}': \mathcal{A}' \subseteq \mathcal{A} \text{ finite }\}, \text{ where } \bigcap \emptyset = X$$

And $$\mathcal{A}^\cup = \{ \bigcup \mathcal{A}': \mathcal{A}' \subseteq \mathcal{A}\}$$

For the final part of the first definition, see the discussion here, e.g.

The proof is not hard: any topology $$\mathscr{T}$$ that contains $$\mathcal{S}$$ contains $$\mathcal{S}^{\cap, < \infty}$$ as well, as topologies are closed under finite intersections, and then it contains $$(\mathcal{S}^{\cap, <\infty})^{\cup}$$ as well, as topologies are closed under all unions. So $$\mathscr{T}_S \subseteq \mathscr{T}$$. This shows the minimality, one only needs to show that $$(\mathcal{S}^{\cap, <\infty})^{\cup}$$ is indeed a topology. It also contains all members $$S$$ of $$\mathcal{S}$$ as $$S = \bigcup\{ \bigcap\{S\}\}$$, and $$\{S\} \subseteq \mathcal{S}$$ is finite.

It is a topology as well: it contains $$X$$ as $$X = \bigcup \{ \bigcap \emptyset \}$$ and $$\emptyset = \bigcup \emptyset$$ is in it as well. The collection is trivially closed under all unions (a union of unions is a union), while it's closed under finite intersections as

$$\left(\bigcup_{i \in I} \bigcap \mathcal{A}_i\right) \cap \left(\bigcup_{j \in J} \bigcap \mathcal{B}_j\right) = \bigcup_{(i,j) \in I \times J} \{ \bigcap \left(\mathcal{A}_i \cup \mathcal{ B}_j)\right) \}$$

Where all $$\mathcal{A}_i$$ and $$\mathcal{B}_j$$ are finite subsets of $$\mathcal{S}$$, and then applying induction. (Or doing the equivalent for finite intersections of the above formula, which is a bit messier).

Alternatively we can note that $$\mathcal{S}^{\cap, <\infty}$$ satisfies the requirements for a base for a topology : it covers $$X$$ by the empty intersection clause; and it's closed under finite intersections.

So indeed in the general case $$\mathcal{S}^{\cap, < \infty}$$ is a base for $$\mathscr{T}_S$$, and covers $$X$$, provided we use the empty intersection definition. So with that proviso the equivalence does hold.

First, it is not necessary to require that a subbase $S$ covers $X$.

Your beginning of a proof is correct, and it can be continued like this: $T_1$ is a topology (simple check), $T_1 \subseteq T$ and $S \subseteq T_1$. Now $T$ is the intersection of all topologies containing $S$, among which is $T_1$. Hence $T \subseteq T_1$. So far for 2 $\to$ 1.

As to 1 $\to$ 2, let $S \subseteq T$ be a collection such that the set $B$ of finite intersections of elements of $S$ is a basis for $X$, i.e., each member of $T$ is a union of members of $B$. Suppose $T_1$ is the topology generated by $S$. Then $S \subseteq T_1$ and as $T_1$ is a topology, we conclude that, first $B \subseteq T_1$, and next that $T \subseteq T_1$. Hence $T$ is the smallest topology with subset $S$.

• Your first remark needs some explanation. It might be that in "Royden" they only look at finite non-empty intersections. In that case covering of $X$ by elements of $S$ is necessary. – drhab Apr 27 '17 at 9:51
• It is beneficial in topology to let the intersection of an empty collection of subsets of a set X be equal to X. Royden seems to be an exception, perhaps for didactic reasons. Under his conditions, the suggested equivalence is not quite true. – Marcel Van de Vel Apr 27 '17 at 12:29
• $T_1$ is a topology is fine but how $T_1$ subset T ?? Will you elaborate? Thank you. – hiren_garai Apr 28 '17 at 5:46