# Very confused about the definition of subbasis for a topology in Munkres's book

In Munkres's Topology textbook, it says,

Definition

A subbasis S for a topology on X is a collection of subsets of X whose union equals X.

The topology generated by the subbasis S is defined to be the collection T of all unions of finite intersections of elements of S.

I am having a very hard time understanding the first sentence. Let's say $$X = \{1, 2, 3\}$$. Then let $$A = \{\{1\}, \{2\}, \{3\}\}$$, or $$A = \{\{1, 2\}, \{3\}\}$$. Either way, $$A$$ is a collection of subsets of $$X$$, and the union of the elements in $$A$$ is $$X$$, but $$A$$ does not seem like a subbasis here...

What did I misunderstand here? What should be the correct interpretation of "a collection of subsets of X whose union equals X"?

Edit:

So turned out $$A$$ is a subbasis in both cases, thanks @Surb for your comment.

But here is what I don't understand (which made me mistakenly think A may not even be subbasis in the first place). If $$A = \{\{1\}, \{2\}, \{3\}\}$$, then wouldn't "the collection T of all unions of finite intersections of elements of $$A$$" be {$$\varnothing$$}, since any intersection of any two elements in $$A$$ is empty? But since T is a topology, T must also contain $$X$$, but, in this case, it does not?

• Both work and will provide different topologies on $X$.
– Surb
Mar 9, 2021 at 20:34
• Why do you say $A$ does not seem like a subbasis here? Mar 9, 2021 at 20:38
• @Tanner I added my thoughts in the post. Thanks!
– Erin
Mar 9, 2021 at 21:02
• Note in response to edit: the intersection of two elements of $A$ is not necessarily empty, because for example $\{1\}\cap\{1\}=\{1\}$ Mar 9, 2021 at 21:04
• Ah, that clarifies everything. Thank you! @Tanner
– Erin
Mar 9, 2021 at 21:08

The first "subbase" is {{1}, {2}, {3}}. The set of all possible unions are {{1}, {2}, {3}, {1,2}, {1, 3}, {2, 3}, {1, 2, 3}}. The set of all possible intersections is {{1}, {2}, {3}, {}}.

The set of all possible unions and intersections is {{}, {1}, {2}, {3}, {1,2}, {1, 3}, {2, 3}, {1, 2, 3}}. That is a topology, in fact it contains all subsets, the "discrete topology" for {1, 2, 3}.

The second subbase is {{1, 2}, {3}}.

The set of all possible unions is {{1, 2}, {3}, {1, 2, 3}}.

The set of all possible intersections is {{1, 2}, {3}, {}}.

The set of all unions and intersections is {{}, {3}, {1, 2}, {1, 2, 3}}. Again, that is a topology for {1, 2, 3}.

If you could not see that, are you clear what a "topology" for a set is? A topology for a set, X, is a collection of subsets or X that has four properties:

1. It contains X itself.
2. It contains the empty set, {}.
3. It contains all unions of its sets.
4. It contains all finite intersections of its sets.

What you’re missing is that a finite subcollection of $$A$$ can be a collection with only one element; in your example, for instance, $$\big\{\{1\}\big\}$$ is a finite subcollection of $$A$$.

If $$\mathscr{C}$$ is any family of subsets of a set $$X$$, by definition

$$\bigcap\mathscr{C}=\{x\in X:\exists C\in\mathscr{C}(x\in C)\}\,;$$

$$\bigcap\mathscr{C}$$ is the set of all things that belong to at least one member of $$\mathscr{C}$$. In your example $$\big\{\{1\}\big\}$$ is a finite subset of $$A$$, and and its intersection is the set of all members of $$\{1,2,3\}$$ that belong to some member of the collection $$\big\{\{1\}\big\}$$. The only member of that collection is $$\{1\}$$, and the only thing that belongs to it is $$1$$, so $$\bigcap\big\{\{1\}\big\}=\{1\}$$.

More generally, if $$S$$ is any set, $$\bigcap\{S\}=S$$: the only member of the collection $$\{S\}$$ is $$S$$, and the set of its members is just $$S$$ itself.

Munkres has a slightly modified definition of subbase, from my perspective.

For any family $$\mathcal{S}$$ of subsets of $$X$$ there is a unique smallest topology $$\mathcal{T}(\mathcal{S})$$ that contains $$\mathcal{S}$$, which we can define abstractly as the intersection of all topologies on $$X$$ that contain $$\mathcal{S}$$; we at least have one such topology (the discrete one, which is just all subsets of $$X$$) and any intersection of topologies is again a topology.

But there is also a less abstract way to define $$\mathcal{T}(\mathcal{S})$$ which is quite natural once, you have seen bases of topologies:

Let $$\mathcal{S}'$$ be the set of all intersections of finite subfamilies of $$\mathcal{S}$$. (if $$\mathcal{N} \subseteq \mathcal{S} \subseteq \mathscr{P}(X)$$, we define $$\bigcap \mathcal{S} = \{x \in X\mid\forall O \in \mathcal{N}: x \in O\}\tag{1}$$

as usual in set theory and only take those intersections for finite subfamilies $$\mathcal{N}$$. Now topologies are always closed under finite intersections so if $$\mathcal{T}$$ is any topology that contains $$\mathcal{S}$$, it will also contain $$\mathcal{S}'$$, defined by all finite intersections from $$\mathcal{S}$$ in the above sense.

By the notion of "void truth" it holds that $$\bigcap \emptyset = X$$ as we have a universal quantifier over the $$\mathcal{N} = \emptyset$$ in $$(1)$$. So we get "for free" that $$X \in \mathcal{S}$$, as the empty subfamily is surely finite. Also $$\mathcal{S} \subseteq \mathcal{S}'$$ as when $$S \in \mathcal{S}$$, $$\{S\}$$ is also a finite subfamily of $$\mathcal{S}$$ and $$\bigcap \{S\} = S \in \mathcal{S}'$$ by the definition. But of course also $$S_1, S_2 \in \mathcal{S}$$ implies $$S_1 \cap S_2= \bigcap \{S_1,S_2\} \in \mathcal{S}'$$ and likewise for intersections of $$3$$,$$4$$ etc. members. All of these must be in any topology containing $$\mathcal{S}$$, so in $$\mathcal{T}(\mathcal{S})$$ in particular.

Anyway, we then recall the criteria that Munkres gives for a family $$\mathcal{B}$$ to be a base for some topology:

1. $$\bigcup \mathcal{B} = X$$
2. $$\forall B_1, B_2 \in \mathcal{B}: \forall x \in B_1 \cap B_2 \exists B_3 \in \mathcal{B}: x \in B_3 \subseteq B_1 \cap B_2$$

and if these conditions are satisfied, the set of all unions of subfamilies $$\mathcal{B}' \subseteq \mathcal{B}$$, where $$\bigcup \mathcal{B}' = \{x \in X\mid \exists B \in \mathcal{B}: x \in B\}\tag{2}$$ forms the unique minimal topology that has $$\mathcal{B}$$ as a base.

Now we easily check that our $$\mathcal{S}'$$ defined from $$\mathcal{S}$$ indeed satisfies 1. and 2. That 1. holds follows from $$X \in \mathcal{S}'$$ trivially. That 2. holds is because $$\mathcal{S}'$$ is closed under finite intersections: given $$B_1 = \bigcap \mathcal{N}_1$$ and $$B_2 = \mathcal{N}_2$$, with $$\mathcal{N}_1, \mathcal{N}_2 \subseteq \mathcal{S}$$ finite, we just pick for any $$x$$ in their intersection $$B_3 = \bigcap (\mathcal{N}_1 \cup \mathcal{N}_2) \in \mathcal{S}'$$ (as a union of two finite sets is finite).

So $$\mathcal{S}'$$ forms a base for $$\mathcal{T}(\mathcal{S})$$ by minimality and the topology is just the set of all unions of (finite intersections from $$\mathcal{S}$$).

This $$\mathcal{S}$$ is traditonally called the subbase for this minimal topology $$\mathcal{T}(\mathcal{S})$$ generated by $$\mathcal{S}$$, so just another word for generating set: we just specify (as minimally as possible, usually) a collection of sets that has to be open to define a topology. This happens quite a lot in topology. The order topology is one example : we just specify that all sets of the form $$\{x\mid x < a\}$$ and $$\{x\mid x > a\}$$ where $$a \in X$$ have to be open, so these form a subbase for the order topology. (But not yet a base).

Munkres almost does the same thing but seems not to want to rely on the void truth convention to ensure that $$X \in \mathcal{S}'$$, but just demands as a condition that $$\bigcup \mathcal{S} = X$$ so that 1. is also satisfied for the base $$\mathcal{S}' \supseteq \mathcal{S}$$. The rest of the arguments then stay the same.

In practice this doesn't make much difference as standard subbases automatically satisfy that condition (the order topology subbase on $$X$$ only does if $$|X| \ge 2$$) and the set $$\{p_i^{-1}[O]\mid i \in I, O \subseteq X_i \text{ open }\}$$ for the product topology on $$\prod_{i \in I} X_i$$ etc.

The main reason why subbases got their separate name is that the Alaxander subbase lemma (or theorem) can be used to prove compactness of a space: $$X$$ is compact iff every open cover with elements from $$\mathcal{S}$$ has a finite subcover. This allows for efficient proofs of Tychonoff's theorem and other applications.

SO for me a subbase can be any family, but Munkres wants to avoid logical discussions and demands the union be $$X$$. But the basic idea stays the same: specify a small set that generates the topology.