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If $X$ is a topological space, we say that $U$ is a neighborhood of $x$ if there exists an open set $V$ such that $x\in V\subseteq U$.

Let $X\neq\emptyset$ and suppose that for every $x\in X$ there exists $\mathcal{V}(x)$ a family of subsets of $X$ that satisfy:

a) If $U\in\mathcal{V}(x)$, then $x\in U$.

b) If $U,V\in\mathcal{V}(x)$ then $U\cap V\in\mathcal{V}(x)$.

c) If $U\in\mathcal{V}(x)$ then there exists $V\in\mathcal{V}(x)$ such that for every $y\in V$ we have $U\in\mathcal{V}(y)$.

d) If $U\subseteq V$ and $U\in\mathcal{V}(x)$ then $V\in\mathcal{V}(x)$.

Now, we define $\tau$ as the family of all $G\subseteq X$ such that for every $x\in G$ there exists $U\in\mathcal{V}(x)$ with $U\subseteq G$.

I already proved $\tau$ is a topology over $X$, and that if $x\in X$ then every neighborhood of $x$ is an element of $\mathcal{V}(x)$.

Now, I want to prove that if $U\in\mathcal{V}(x)$ then $U$ is a neighborhood of $x$. I tried to do this: Using c), we can find $V\in\mathcal{V}(x)$ such that for every $y\in V$ we have that $U\in\mathcal{V}(y)$. Then, if $y\in V$, since $U\in\mathcal{V}(y)$ then $y\in U$, i.e., $V\subseteq U$. Can we prove now that $V\in\tau$? I don't know if it is possible.

Thanks.

Talexius
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    You can't prove $V \in \tau$ if you don't constrain $V$ somewhat. You could, for example have $V$ be the union of an open set and an isolated point in the interior of $U$, that would generally not belong to $\tau$. Try $V = { y \in U : U \in \mathcal{V}(y)}$. – Daniel Fischer Sep 09 '13 at 17:50
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    By the way, you can define $\tau$ is the family of all $G$ such that for every $x\in G$ we have $G\in\mathcal V(x)$ (follows from (d)). – Stefan Hamcke Sep 09 '13 at 18:02
  • That is pretty much only half of the story: $\mathcal{V}\to\tau\to\mathcal{V}$, $\tau\to\mathcal{V}\to\tau$ Checking one of these relations only shows injectivity resp. surjective of your identification only. As an applied example how it can actually fail is the Cayley transform: $U_A:=\frac{A-i}{A+i}$ – C-star-W-star Dec 01 '15 at 17:22

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Define $V=\{y\mid U\in\mathcal V(y)\}$. It is clear that $V\subseteq U,$ and we want to show that $V$ is open, i.e. that $V\in\mathcal V(y)$ for each $y\in V$.
So take $y\in V$. Then $U\in\mathcal V(y)$ and by (c) there is a $W\in\mathcal V(y)$ such that $U\in\mathcal V(z)$ for each $z\in W$. But this implies that $W\subseteq V$ by definition of $V.$ By (d) it then follows that $V\in\mathcal V(y).$

Did you know that condition (b) and (d) can be subsumed by saying that the $\mathcal V(x)$ are so-called filters? Also have a look at Going from a fundamental system of neighborhoods to a topology and vice versa to see how we can go from a system of filter bases to a topology.

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Stefan Hamcke
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