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James Munkres defines a subbasis $\mathcal S$ for a topology on a set $X$ as a collection of subsets of $X$ whose union equals $X$. Then the topology generated by the subbasis $\mathcal S$ is defined to be the collection $\mathcal T$ of all unions of finite intersections of elements of $\mathcal S$. To prove that $\mathcal T$ is indeed a topology, one only needs to show that the collection $\mathcal B$ of all finite intersections of elements of $\mathcal S$ is a basis.

My question is what if $\mathcal S$ is a partition of $X$? Then any finite intersection of the elements of $\mathcal S$ is either $\emptyset$ or the elements of $\mathcal S$. It is clear that $\emptyset$ is not enough to generate a topology. Hence the topology is essentially generated by all the unions of the elements of $\mathcal S$, which is in fact the property of a basis rather than a subbasis. Hence, if my argument is correct, then I can conclude that if elements of $\mathcal S$ form a partition of $X$, then there is no difference between basis and subbasis. Is this right, please? Thank you!

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What about $S \cap S$, for $S \in \mathcal{S}$? –  Nick Mar 2 at 5:09
    
@Nick Good point. Thank you. I will update my question. –  20824 Mar 2 at 5:13
    
@Nick Just wondering whether $S\cap S$ is valid since there is only one $S$ in $\mathcal S$. Can you intersect with yourself? –  20824 Mar 2 at 5:26
    
Why is there only one $S$ in $\mathcal{S}$? –  Nick Mar 2 at 5:27
    
@Nick Because by set theory convention, all elements in a set is unique. For example, $\{1,1,1\}=\{1\}$. –  20824 Mar 2 at 5:32

1 Answer 1

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You are quite right: if $\mathcal{S}$ forms a partition of the set $X$ (union all of $X$, pairwise disjoint), they actually form a base for a topology $\mathcal{T}$ on $X$, which are exactly the unions of members of $\mathcal{S}$.

Recall the characterization of base $\mathcal{B}$: $\cup \mathcal{B} = X$ and for every $B_1, B_2 \in \mathcal{B}$, for all $x \in B_1 \cap B_2$ there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subset B_1 \cap B_2$. If $\mathcal{B}$ forms a partition, the first is automatic, and the second condition can only be satisfied when $B_1 = B_2$, as all other intersections are empty (and so the for all $x \in B_1 \cap B_2$ condition is trivially true), and then taking $B_3 = B_1 = B_2$ is enough. So a partition is a base for a topology .

In fact any collection of subsets of $X$ forms a subbase and the statement that the finite intersections of members of $\mathcal{S}$ form a base remains true. But by convention/logic we include the empty intersection (the intersection of zero members of $\mathcal{S}$), which equals $X$, so the union to $X$ is then automatically fulfilled. Of course all 1-set intersections from $\mathcal{S}$, i.e. all members of $\mathcal{S}$ are also in this, etc. Maybe Munkres wanted to avoid the whole empty intersection explanation, and his condition makes it unnecessary (but makes a subbase more restrictive).

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