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How is the empty set generated by the arbitrary union of half open intervals of the form $[a,b), a<b, a,b\in R$.

I can't come up with a union.

Jhon Doe
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  • https://math.stackexchange.com/questions/151924/unnecessary-property-in-definition-of-topological-space – cqfd Jan 21 '19 at 12:07

2 Answers2

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Let $B$ be the set of half-open intervals, open at the top. A set $S$ is open in the Sorgenfrey topology iff $S=\cup C$ for some $C\subset B.$ We have $\phi\subset B$ so $\cup\phi$ is open. And $\cup \phi=\phi$.

When we say "$S$ is a union of (some things)" we mean there is a set $B$ of "some things" and there exists $C\subset B$ such that $S=\cup C.$ Unless stated otherwise, this does not exclude the case $C=\phi.$

BTW. The only way that $\phi=\cup C,$ for any set $C,$ is that $C=\phi$ or $C=\{\phi\}.$

DanielWainfleet
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It is not. When people talk about "the topology generated by" they always implicitely put both $\emptyset$ and the whole space in the topology. Regardles of how you generate it. This simply follows from the definition of "the topology generated by" being the intersection of all topologies containing given collection.

For example say you have $X=\{1,2\}$ and you define $\tau$ to be the topology generated by $\{\{2\}\}$. We know that $\tau=\{\emptyset, \{2\}, X\}$ even though neither $\emptyset$ nor $X$ can be recreated from $\{2\}$ via union/intersection operations.

In other words when you read "every open subset is a union of half intervals" it means "every nonempty open subset" in this particular case. Note that whole $\mathbb{R}$ can be easily written as a union of half open intervals. But again: it is just this particular case, it is not true in general.

freakish
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  • Here $\mathscr{B}={[a,b), a<b, a,b\in \Bbb R}$ is the basis of the topology, right? Then we can say that empty set is the union of empty collection of sets of $\mathscr B$. See here for a proper treatment. – cqfd Jan 21 '19 at 13:21
  • @ThomasShelby you first need topology to talk about basis. But I suppose we can take the union of empty subcollection. Even though it is confusing imo. And you are still left with the problem of whole space. – freakish Jan 21 '19 at 13:25
  • Sorry, I meant the topology generated by $\mathscr B$. But by definition, a basis should cover the whole space,right? – cqfd Jan 21 '19 at 13:38