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Need some help here, so is every smaller collection of open sets a basis for topology? Let's say for example, a topology, (T) which contains the following stuffs: {{empty set},{0},{N}}, where N is a set of natural numbers. Is this also a basis for topology T ?

Math_Is_Fun
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    For example, the collection of all open intervals in $\mathbb R$ forms a base for the standard topology on $\mathbb R$; every open set can be written as a union of elements of the base – J. W. Tanner Feb 03 '20 at 03:19
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    Not every smaller collection of open sets are a basis; some collections of open sets (satisfying certain properties) will be bases for their corresponding topologies. In your example, not, the set $N$ is not a basis for the topology, because it does not satisfy those properties that I am sure your book will introduce you to in short order, after it is done with the introduction. – Arturo Magidin Feb 03 '20 at 03:20
  • This answer is probably too advanced for you now, but possibly some of the expository parts could be of use to you. – Dave L. Renfro Feb 03 '20 at 07:01

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No, not every collection of subsets $\mathcal{B}$ of $X$ can serve as a basis for a topology on $X$. Probably your text will introduce these shortly after this introduction, but in anticipation of those, the following two are necessary and sufficient:

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

The second condition would be easily satisfied if $\mathcal{B}$ is closed under finite intersections (but that is not a condition, just an easy way to satisfy it, as we can take $B_3 = B_1 \cap B_2$ in that case). Open intervals in ordered spaces are an example of this, while the balls in a metric space generally will just obey 2, and are not closed under intersections, e.g.

The family $\{\emptyset, \{0\}, \Bbb N\}$ is already a topology on its own, and not generated by a strictly smaller base.

Henno Brandsma
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