2

There are two different definitions for a subbase of a topological space that are not equivalent: Let $(X,\mathcal{T})$ be a topological space.

1) $\gamma \subset \mathcal{T}$ is a subbase of $\mathcal{T}$ if the collection of all finite intersections of elements of $\gamma$ together with the set $X$, forms a basis for $\mathcal{T}$.

2) $\gamma \subset \mathcal{T}$ is a subbase of $\mathcal{T}$ if additionally $\gamma$ covers $X$, i.e. $\bigcup\limits_{U \in \gamma} U=X$.

theorem of alexander: If every open cover by elements from $\gamma$ has a finite subcover, then the space is compact.

question: Does the theorem holds even for the 1) defintion?

considerations: If $\gamma$ doesn't cover $X$, then for each open cover $\mathcal{U}\subset \mathcal{T}$ of $X$ there holds $X \in \mathcal{U}$ (since each open set $U \in \mathcal{U}$ can be expressed as union of finite many intersections of elements of $\gamma$) and hence $X$ is a finite subcover of itself. So $X$ is compact.

  • If Alexander's theorem is true when using definition (1), then it is automatically true when using definition (2). On the other hand, suppose Alexander's theorem is true when using definition (2). If $\gamma$ covers $X$, then the theorem holds by hypothesis; otherwise, $\gamma$ doesn't cover $X$, and one can quickly show that $X$ is compact (as you note in your "considerations"), so the theorem's conclusion (and hence the theorem) holds. Therefore, the theorem holds true for either definition. – WillG Jun 03 '24 at 18:58
  • So I agree with the conclusions of both answers below, although I find their explanations somewhat unclear. – WillG Jun 03 '24 at 19:00

2 Answers2

1

Yes because with the addition of X, $\gamma$ will cover X.
The 2nd definition is much preferred to the unusual, weaker 1st definition.

William Elliot
  • 18,025
  • 1
    One could also point out that by convention one may consider the intersection the empty family (which is clearly finite) of subsets of $X$ to be equal to $X$. – Mirko Oct 31 '19 at 12:44
  • @WilliamElliot Im not sure wether I understand your answer the right way. In the 1) definition $X$ doesnt have to be an element of $\gamma$ –  Oct 31 '19 at 12:57
  • @Mirko yes, for this definition the proof of alexanders lemma is clear. –  Oct 31 '19 at 12:59
  • 1
    @chrisMainly If $\gamma_0$ is as in definition 1, and $\gamma=\gamma_0\cup{X}$, then $\gamma_0$ and $\gamma$ generate the same topology. We do not get "more sets" as intersections of finite subfamilies of $\gamma$. If $\mathcal F$ is a non-empty finite subfamily of $\gamma_0$ then $\cap\mathcal F=\cap\bigl(\mathcal F\cup{X}\bigr)$. – Mirko Oct 31 '19 at 13:12
  • ok that make sense, thank you for you affort! –  Oct 31 '19 at 13:24
1

In my definition a subbase $\mathcal{S}$ for a topology $\mathcal{T}$ is a subfamily of $\mathcal{T}$ such that the smallest topology on $X$ that contains $\mathcal{S}$ equals $\mathcal{T}$.

So $\emptyset$ is actually a subbase for the indiscrete (trivial) topology on $X$ in that definition.

So a subbase for $\mathcal{T}$ need not contain a cover of $X$, as Munkres demands in his text (it is the only popular textbook that defines a subbase this way), but what is true is that a base for $\mathcal{T}$ is given by all finite intersections from $\mathcal{S}$ under the proviso that the empty intersection (the intersection of $0$ members of $\mathcal{S}$ equals $X$, the void intersection "convention") and of course the intersecton of a one set family is that one set in the family etc. All these count as finite intersections)

Alexander's subbase lemma holds both for my definition of subbase as well as the Munkres one. You can always just add $X$ for free to any of "my" subbases and get a valid "Munkres" subbase. This adds no new interesting covers to check.

Henno Brandsma
  • 250,824