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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"?

Thank you in advance!


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?

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

3 Answers 3

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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.
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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.

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  • $\begingroup$ Hi professor Scott, could I ask your assistance here, please? $\endgroup$ Mar 9, 2021 at 21:22
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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.

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