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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 Patel May 3 '17 at 1:14

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


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 $S$.

  • $\begingroup$ 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. $\endgroup$ – drhab Apr 27 '17 at 9:51
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    $\begingroup$ 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. $\endgroup$ – Marcel Van de Vel Apr 27 '17 at 12:29
  • $\begingroup$ $T_1$ is a topology is fine but how $T_1$ subset T ?? Will you elaborate? Thank you. $\endgroup$ – hiren_garai Apr 28 '17 at 5:46

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