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Question: Let $\mathcal{N}$ be a neighborhood system. Prove the neighborhood system induced by the topology induced by $\mathcal{N}$ is $\mathcal{N}$ itself.

Definition: A neighborhood system $\mathcal{N}:X\rightarrow \mathcal{P}(\mathcal{P}(X))$ is a function that satisfy

  1. $\mathcal{N}(x)$ is non-empty
  2. $\forall N \in \mathcal{N}(x), x \in N$
  3. $N \in \mathcal{N}(x)$ and $N \subseteq M \implies M \in \mathcal{N}(x)$
  4. If $ N, M \in \mathcal{N}(x) $ then $ N \cap M \in \mathcal{N}(x) $.
  5. For every $ N \in \mathcal{N}(x) $, there exists $ M \in \mathcal{N}(x) $ such that $ M \subseteq N $ and for every $ y \in M $, $ N \in \mathcal{N}(y) $.

Definition: Neighborhoods system $\mathcal{N}$ induces a topology $\mathfrak{O}$ by defining open sets as the sets $O$ such that $O$ is a neighborhood around each point in $O$.

Definition: A topology $\mathfrak{O}$ induces a neighborhood system by defining neighborhoods around $x$ as the sets containing an open set containing $x$.


I have proved $\forall x,\mathcal{M}(x)\subseteq \mathcal{N} (x)$, where $\mathcal{M}$ is the neighborhood system double induced by $\mathcal{N}$. But I don't know how to prove the other way around. Thanks for any help. I have stucked here for a week already.

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1 Answer 1

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Suppose $N \in \cal N(x)$ and let $U = \{y \in N \mid N \in \cal N(y)\}$.

Lemma: $U$ is open in the topology induced by $\cal N$, that is $U \in \cal N(y)$ for every $y \in U$.

Proof: Suppose $y \in U$. Then $N \in \cal N(y)$, so (by property 5) there exits $M \in \cal N(y)$ such that $M \subseteq N$ and $N \in \cal N(z)$ for every $z \in M$. Now $M \subseteq U$, because for every $z \in M$ we have $z \in N$ and $N \in \cal N(z)$, so (by property 3) we have $U \in \cal N(y)$.

So $U$ in open in the topology induced by $\cal N$, and clearly $x \in U$ and $U \subseteq N$, so $N$ is a neighborhood of $x$ in the topology induced by $\cal N$.

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