# Axioms for a topology I found this theorem in a book. I am not sure how I would prove it. To start with, I can't see why those conditions imply why the sets are open, nor even why they aren't closed (as for a finite set, for e.g., these conditions would work for closed sets I think). Can anyone give me a hand in proving this? Thanks!

The definition of topology in the book (which btw is "Modern Differential Geometry for Physicists" by Chris J Isham) is: • Hmm, that's often the definition of "topology on $X$," so you'll need to give the book's definition of "topology." – Thomas Andrews Aug 6 '14 at 18:33
• What was the definition of a topology given in the book? – vociferous_rutabaga Aug 6 '14 at 18:33
• What is your definition of topology (usually $\tau1$, $\tau2$ and $\tau3$ are the axioms for a topology). – Luiz Cordeiro Aug 6 '14 at 18:33
• What this book states as a "theorem" is exactly what most books state as the definition of a topology. So, you'll have to supply us with more context before your question makes any sense. In particular, I do not know what this book means by a "neighborhood space". – Lee Mosher Aug 6 '14 at 18:33
• It's clear from the previous paragraph that it is defining "open" via something it is calling a "neighborhood space." – Thomas Andrews Aug 6 '14 at 18:34

There are several things to prove. So assume $\tau$, a family of subsets of $X$, satisfies the three axioms. Then we define for every $x \in X$: $\mathcal{B}(x) = \{ O \in \tau: x \in O \}$. Claim 1: this is a neighbourhood space, with neighbourhood filters generated by $\mathcal{B}(x)$.
As $X \in \mathcal{B}(x)$ for all $x$ (as $X$ is in $\tau$), all sets $\mathcal{B}(x)$ are non-empty. Also, by definition, all members of $\mathcal{B}(x)$ contain $x$. When $O_1, O_2 \in \mathcal{B}(x)$, then $x \in O_1 \cap O_2$, and $O_1, O_2 \in \tau$, so $O_1 \cap O_2 \in \mathcal{B}(x)$, so that the sets $\mathcal{B}(x)$ indeed form a filter base at each point. So $\mathcal{N}(x) = \{N: \exists O \in \mathcal{B}(x): O \subset N \}$ is the neighbourhood filter at $x$.
We then define open sets as always: $O \subset X$ is open, iff for all $x \in O$, $O \in \mathcal{N}(x)$ ($O$ is a neighbourhood of each of its points), so (as the $\mathcal{B}(x)$ form a filter base) for all $x \in O$, $\exists O_x \in \tau$ such that $O_x \subset O$. But then $O = \cup \{O_x: x \in O \}$, for these $O_x$ (as each $x \in O$ is in its "own" $O_x$, so $O$ is a subset of the union, while all $O_x \subset O$, so the other inclusion also holds). But all $O_x$ are in $\tau$ so $O \in \tau$ by the third axiom. So $O$ open in the neighbourhood space implies $O \in \tau$. But also, if $O \in \tau$, for every $x \in O$, $O \in \mathcal{B}(x)$ by definition, so indeed $O$ is a neighbourhood of each of its points and $O$ is open in the neighbourhood space. So $\tau$ is exactly the collection of open sets of the neighbourhood space definded by the filter bases $\mathcal{B}(x)$.
• That was an awesomely clear explanation, thanks a lot! Just to confirm, in the second paragraph you made use of the third axiom ($\tau 3$) didn't you? And in the third, of the second ($\tau 2$) (in the orden in my picture), no? – guillefix Aug 6 '14 at 23:50
• Yes, finite intersections show open sets containing $x$ are a filter base, unions show that the two ideas of open (nbhd space vs being in $\tau$) are the same. – Henno Brandsma Aug 7 '14 at 3:51