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Suppose that we have a topological space $(X,\tau)$, a subset $A\subseteq X$ is a neighborhood of $x \in X$ if there exists an open subset $U_x$, containing $x$ such that $U_x \subseteq A$. Also, suppose that the collection of all neighborhoods of $x \in X$ in $(X,\tau)$ is denoted by $\mathcal{N}_x$.

Each $\mathcal{N}_x$ has the following interesting properties, which I managed to prove.

  1. $\forall A \in \mathcal{N}_x$, $x \in A$
  2. $\forall A,B \in \mathcal{N}_x$, $A \cap B \in \mathcal{N}_x$
  3. $\forall A \subseteq X$ and $\forall B \in \mathcal{N}_x$, if $B \subseteq A$, then $A \in \mathcal{N}_x$
  4. $\forall A \in \mathcal{N}_x$, there exists $N \in \mathcal{N}_x$ such that, $\forall y \in N$, $A \in \mathcal{N}_y$

Let us denote the collection of subsets of $X$ satisfying the four above-mentioned properties by $\mathcal{U}_x$.

If we define $\tau_{\mathcal{N}} = \{G \subseteq X \:{:}\: \forall x \in G, G \in \mathcal{U}_x \}$, then it can easily be shown that $\tau_{\mathcal{N}}$ is a topology on $X$. Moreover, $\tau_{\mathcal{N}} = \{G \subseteq X \:{:}\: \forall x \in G, \exists U_x \in \mathcal{U_x}, U_x \subseteq G \}$ is also an equivalent expression.

My struggle starts from proving that $\mathcal{U}_x = \mathcal{N}_x$ where $\mathcal{N}_x$ is the collection of all neighborhoods of $x \in X$ in $(X,\tau_{\mathcal{N}})$. One way is easy because each $\mathcal{N}_x$ satisfies the four conditions of $\mathcal{U}_x$.

For the other way, suppose $A \in \mathcal{U}_x$, we need to show that $A \in \mathcal{N}_x$. Firstly, there exists $N \in \mathcal{U}_x$ such that, $\forall y \in N$, $A \in \mathcal{U}_y$, which implies that $N \subseteq A$. Since $x \in N$, all we need to show is that $N \in \tau_{\mathcal{N}}$, which means that I need to show that either

  1. $\forall z \in N$, $N \in \mathcal{U}_z$, or
  2. $\forall z \in N$, $\exists U_z \in \mathcal{U_z}$ such that $U_z \subseteq N$

I spent hours but could not make any progress to show that $N \in \tau_{\mathcal{N}}$. Any help will be greatly appreciated.

James
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    Note that Kelley and Willard differ in the exact axioms here: Kelley uses $4'$: $$\forall A \in \mathcal{N}_x: \exists M \in \mathcal{N}_x: (M \subseteq N) \land (\forall y \in M: M \in \mathcal{N}_y)$$ This is more explicit in its intention: "every nbhd of $x$ contains an open neighbourhood of $x$". I initially learnt "your" version, IIRC, and its analogue proof is slightly more subtle, I think. What text are you using? Willard? – Henno Brandsma May 29 '20 at 15:19
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    Note that you cannot show in general that $N$ as you use is open, e.g. $A=[-2,2]$, $N=[-1,1]$ is a valid instance for $x=0$ (in $\Bbb R$, usual topology). But $A^\ast = (-2,2)$ (as I defined in my answer) would work. – Henno Brandsma May 29 '20 at 16:35
  • I used Willard. I will definitely have a look at Kelley's too. I agree with you regarding the subtlety. Thank you so much. – James May 29 '20 at 17:07
  • You’re welcome. Most texts seem to use such neighbourhood (base) characterisations as exercises. In Hausdorff’s book ( sort of the first topology book) he uses neighbourhoods as the primitive notion ( inspired on metrics?) and defines open from it. So it’s an old alternative axiom system. – Henno Brandsma May 29 '20 at 17:18
  • Hausdorff's old axioms are found here. He uses neighbourhood for what I would call "open neighbourhood", so he has no filter (in $\mathscr{P}(X)$) as you have, but his "coherence axiom" (an axiom to connect systems at different points, our 4 or 4' or his 3 in the link) is a bit simpler. He only considers what we now call Hausdorff spaces by the linked axiom 4. Hence the current name of this separation axiom. – Henno Brandsma May 29 '20 at 23:08

1 Answers1

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This is a subtle matter and I already devoted a pretty comprehensive answer to it here, where you can also double check your proof that $\tau$ is a topology.

The basic idea is for $A \subseteq X$ to define

$$A^\ast = \{y \in X: A \in \mathcal{N}_y\}$$

(a neighbourhood version of the interior) and note that the $N\in \mathcal{N}_x$ promised by 4. to exist for $A$ is a subset of $A^\ast$, so $A^\ast \in \mathcal{N}_x$ too. Then show that $A^\ast \in \tau$ and so $A \in \mathcal{U}_x$.

To see that $A^\ast \in \tau$: let $z \in A^\ast$ be arbitrary. By definition, $A \in \mathcal{N}(z)$. Apply axiom 4. to $A$ to get an $N \in \mathcal{N}(z)$ such that $\forall y \in N: A \in \mathcal{N}(y)$, so by definition $N \subseteq A^\ast$ and so $A^\ast \in \mathcal{N}(z)$ by the enlargement property. Hence $\forall z \in N^\ast: A^\ast \in \mathcal{N}(z)$ which means that $A^\ast$ is open, as claimed.

Henno Brandsma
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  • Thank you for your help. I will have a look at your previous answer. In my question, I had to show that $A \in \mathcal{N}_x$, but in your answer it seems like showing that $A \in \mathcal{U}_x$. – James May 29 '20 at 14:51
  • @James You have two inclusions. $\mathcal{U}_x \subseteq \mathcal{N}_x$ and reversly. You asked for the reverse: Let $A \in \mathcal{N}_x$. Then $A^\ast \subseteq A$ and $x \in A^\ast$ and so knowing $A^\ast \in \tau$ then ensures $A \in \mathcal{U}_x$. – Henno Brandsma May 29 '20 at 15:00
  • @James If however $A \in \mathcal{U}_x$ we know that $O \in \tau$ exists with $x \in O \subseteq A$. Then the definition of $\tau$ says $O \in \mathcal{N}_x$ and so $A \in \mathcal{N}_x$ by 3. So that inclusion is immediate. – Henno Brandsma May 29 '20 at 15:02
  • Wow using the neighbourhood version of the interior perfectly answers my question. Thank you so much for such a beautiful yet insightful idea and help. However, I still believe $\mathcal{N}x$ and $\mathcal{U}_x$ are interchanged in your explanation. $\mathcal{U}_x$ is a collection of subsets satisfying 4 above conditions and $\mathcal{N}_x$ is the collection of neighbourhoods of $x$ in $(X,\tau{\mathcal{N}})$. So $\mathcal{N}_x \subseteq \mathcal{U}_x$ was an obvious part for me as every neighbourhood system of $x$ meets the above four requirements. Kindly let me know if I'm still confused. – James May 29 '20 at 17:03
  • @James no, you’re not. In my head the N-versions are pregiven and the U-versions are the ones from a topology. It’s confusing, that’s why I defined the T and N operators in the more extensive older answer. – Henno Brandsma May 29 '20 at 17:22