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This problem is probably closely related to a bunch of similar questions touched upon elsewhere (see for example Going from a fundamental system of neighborhoods to a topology and vice versa and others), yet I was not able to distill any satisfactory solution to it from them. Specifically: suppose we are given a set $X$ and a function $\sigma$ which associates with every $x \in X$ a nonempty system $\sigma(x)$ of subsets of $X$ such that $\sigma(x)$ enjoys the following properties for every $x \in X$:

  1. $x \in U$ whenever $U \in \sigma(x);$
  2. if $U,V$ are arbitrary members of $\sigma(x),$ then there is a third one, say $W \in \sigma (x),$ with $W \subset U \cap V;$
  3. to every $U \in \sigma(x)$ there is a $V \in \sigma(x)$ such that for every $y \in V$ there is $W \in \sigma(y)$ such that $W \subset U.$

Now, as the alternative terminology for $\sigma(x)$ dubbing it a base of a filter of neighbourhoods seems to suggest, the system $\mathcal{N}(x)$ defined for every $x \in X$ as $\mathcal{N}(x):= \{U \subset X | \exists V \in \sigma(x): V \subset U\}$ should be precisely the system of neighbourhoods of certain topology $\tau$ on $X$ which in turn should be equal, according to a common sense, to the system of subsets $U \subset X$ which are neighbourhoods of all of their points: $U \in \tau \iff \forall x \in X:(x \in U \Rightarrow U \in \mathcal{N}(x)).$

It is really not hard to check that $\tau$ is really a topology on $X.$ The weird thing about this is that you actually don't need the third property of the system $\sigma(x)$ to prove it. My suspicion was that the third property should be used somehow to show that in fact the system of neighbourhoods $\tau(x)$ of an arbitrary point $x \in X$ in the topology $\tau$ should be equal to $\mathcal{N}(x)$ at the same point.

Now, I am able to show that if $U \in \tau(x),$ then $U \in \mathcal{N}(x)$ (still no third property needed), but I am totally stuck on the other inclusion: if $U \in \mathcal{N}(x),$ then $U \in \tau(x).$ Please note, that a set $U \subset X$ is called a neighbourhood of a given point $x \in X$ (hence $U \in \tau(x)$), if there is an open set $V$ such that $x \in V \subset U.$ Any suggestions? Thank you!

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If $U\in\mathcal{N}(x)$, there exists $U'\in\sigma(x)$ such that $U'\subset U$. Call $U''=\{y\in U'|\exists U_y\in\sigma(y),U_y\subset U'\}$. I claim that $U''$ is open in $\tau$, and hence $U\in\tau(x)$.

Clearly, $x\in U''$, we only need to show that $U''$ is a neighbourhood of $y$ for every $y\in U''$. By definition of $U''$, there exists $U_y\in\sigma(y)$, $U_y\subset U'$. Let $V_y\subset U_y$ be the the subset given by the third property.

By definition of $V_y$, for every $z\in V_y$ there exists $W_z\in\sigma(z)$ with $W_z\subset U_y\subset U'$: this is exactly the condition for $z$ to belong to $U''$. Since $z$ is generic, we have proved that $V_y\subset U''$, which implies that $U''$ is a neighbourhood of $y$. Since $y$ is generic, $U''$ is open, as desired.

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  • $\begingroup$ @GB: I don't understand two things. First, you write "In particular, $z \in U^{\prime \prime},$ and so $V_y \subset U^{\prime \prime}.$" What "particular" $z$ do you mean and why, then, $V_y \subset U^{\prime \prime}$? I can't see it. Second, you are supposed to show $U \in \tau(x)$ not $U^{\prime \prime} \in \tau(x).$ Does that former follow somehow from the latter? Thanks! $\endgroup$ – Jorge.Squared Oct 15 '14 at 13:02
  • $\begingroup$ $z$ is a generic point of $V_y$. Saying that $V_y\subset U''$ means exactly that, for every $z\in V_y$, $z\in U''$. What I've proved is that $U''$ is an open set of $\tau$ such that $x\in U''$. Moreover, $U''\subset U$, and hence $U\in\tau(x)$. $\endgroup$ – Giulio Bresciani Oct 15 '14 at 13:19
  • $\begingroup$ @GB: So what you are saying is the following: pick an arbitrary $z \in V_y.$ Then $z \in U^{\prime \prime}.$ I am asking: why so? Still can't see it. The rest is now clear. $\endgroup$ – Jorge.Squared Oct 15 '14 at 13:28
  • $\begingroup$ I've rewritten the answer, I hope it's clearer! $\endgroup$ – Giulio Bresciani Oct 15 '14 at 13:41
  • $\begingroup$ $z\in U''$ because $W_z\in\sigma(z)$ and $W_z\subset U'$. $\endgroup$ – Giulio Bresciani Oct 15 '14 at 14:16

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