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Let $(S,+,*, \leq)$ be a totally ordered field.

For any set $X$ let $\mathcal B \subseteq \mathcal {P}(X)$ be called a basis on $X$ iff:

  1. $X \subseteq \bigcup\mathcal B$
  2. $\forall a,b \in \mathcal B:\exists \mathcal A \subseteq \mathcal B:a \, \cap\,b=\bigcup \mathcal A $

Let any subsets of $S$ of the form ${\{x:x

How to prove that the $I$ is a basis on $S$?

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    Have you tried to prove the defining properties of a basis? – 2017-01-21
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    don't you need $X=\bigcup \mathcal B$? – 2017-01-21
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    @JorgeFernándezHidalgo Apparently not - https://proofwiki.org/wiki/Definition:Basis_(Topology). – 2017-01-21
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    @JorgeFernándezHidalgo equality with $X$ is automatic as already all $a \in \mathcal{B} \subseteq X$ by definition. – 2017-01-21

1 Answers 1

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As to 1: let $p \in X$.

We must assume $X$ has at least two elements so we have some $q \neq p$, or all the open intervals are empty, and we have no base. 3 cases:

  1. $p =\max(X)$. Then $p \in \{x: x > q\}\in \mathcal{B}$.
  2. $p = \min(X)$. Then $p \in \{x: x < q\}\in \mathcal{B}$.
  3. In the other case, there are $a,b \in X$ with $a < p, p < b$. Then $p \in \{x: a < x < b\} \in \mathcal{B}$.

So $X \subseteq \bigcup \mathcal{B}$.

For 2, you have to distinguish all the cases of two different intervals intersecting each other, where you can assume WLOG that the intersection is non-empty, or we take $\mathcal{A} = \emptyset$. It's a bit tedious case checking again.