Regarding Problem 5 on p. 8 of Bredon's Topology and Geometry

Question

In page 8, Section 2: Topological Spaces of Bredon's Topology and Geometry, he presents the 5th question as follows:

Suppose that $S$ is a set and that we are given, for each $x \in S$, a collection $\mathfrak{N}(x)$ of subsets of $S$ satisfying:

1. $N \in \mathfrak{N}(x) \implies x \in N$,
2. $N,M \in \mathfrak{N}(x) \implies \exists P\in\mathfrak{N}(x)$ s.t. $P\subseteq(N \cap M)$, and
3. $x \in S \implies \mathfrak{N}(x) \neq \varnothing$.

Then show that there is a unique topology on $S$ such that $\mathfrak{N}(x)$ is a neighborhood basis at $x$, for each $x \in S$. (Thus a topology can be defined by the specification of such a collection of neighborhoods at each point.)

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Now, what I have attempted the first time was to take $\{\mathfrak{N}(x) | x \in S\}$ as a subbasis (acting upon the comment in the parentheses), show that its topology (let $\tau$) makes each $\mathfrak{N}(x)$ a neighborhood basis at $x$ and it is the unique topology that does this. But, in this attempt, I realized I needed to describe $\tau$ and while attempting I couldn't conclude that $\tau$ makes each $\mathfrak{N}(x)$ a neighborhood basis at $x$ for each $x$ as I couldn't show an open set $U$ containing $x$ with $U \subseteq N \in \mathfrak{N}(x)$.

Next I thought that since neighborhoods do not need to be open sets themselves, it is more natural to take \begin{equation}\tag{Eq. 1} \bigcup\limits_{x \in S} \left \{\mathfrak{N}(x) \right \} = \{\mathfrak{N}(x_1) \} \cup \{ \mathfrak{N}(x_2)\} \cup \dots \end{equation} as a subbasis, i.e. union of elements of $\mathfrak{N}(x)$ for each $x \in S$.

Here I think I proved that $\mathfrak{N}(x)$ is a neighborhood basis at $x$ for each $x \in S$, let $x \in S$ be given, then:

• All elements in $\mathfrak{N}(x)$ are open and they contain $x$, so they are neighborhoods of $x$ by default,
• Given any neighborhood of $x$, let $K$, I have to show that for some $N \in \mathfrak{N}(x)$, $N \subseteq K$. Since $K$ is a n'hood of $x$, there exists an open set $U$ s.t. $x \in U \subseteq K$. Since $U$ is open wrt to the topology we have generated, $U$ is arbitrary union of finite intersections of elements of Eq.1. And since we can get open sets containing $x$ specifically from intersecting $N$'s from the same $\mathfrak{N}(x)$ (not like intersecting $N \in \mathfrak{N}(x_1)$ and $M \in \mathfrak{N}(x_2)$ where $x_1 \neq x_2$ and $x$ is an element of both), for some $N \in \mathfrak{N}(x)$, $N \subseteq U$, thus $N \subseteq K$. Hence $K$ contains an element from the neighborhood basis.

I am sorry if the second point is not very precise. But is my reasoning sound and valid and how do I go on and prove existence of such a topology? Can I somehow prove that it is the finest/coarsest topology instantiates the condition, does that imply uniqueness?

Thank you in advance for any help.

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Attempt at a Proof Upon @Thorgott 's Comment

To make $\mathfrak{N}(x)$ a n'hood basis at $x$, we must have the following, every set in $\mathfrak{N}(x)$ is a n'hood of $x$, and any n'hood of of $x$ contains a set from $\mathfrak{N}(x)$. Maybe, we can define an open set as accepting the sets in $\mathfrak{N}(x)$ as neighborhoods of $x$ and later check that this definition of open sets verifies the second condition of $\mathfrak{N}(x)$ being a n'hood basis. Then, let: \begin{equation} \text{A subset } U \subseteq S \text{ is open if } \mathfrak{N}(x) \subseteq U \quad \forall x \in U. \end{equation} Observe that since $U$ contains a n'hood for every one of its points, it is a n'hood for its points. Hence we satisfy the openness characterization by neighborhoods. This way, we get the topology, denote it by $\tau$. Now the difficult part. We must show that: (1) $\tau$ is a well-defined topology, and (2) $\tau$ makes $\mathfrak{N}(x)$ a n'hood basis at $x$ for each $x \in S$, and (3) $\tau$ is unique.

For (1), it is easy to see that countable intersection of open sets is again open. The space itself contains $\mathfrak{N}(x)$ for all $x \in S$, thus it is open, and $\varnothing$ is vacuously open because $\mathfrak{N}() \subseteq \varnothing$. For arbitrary unions of open sets, since we are collecting over $\mathfrak{N}(x)$'s we'll still have openness. Thus $\tau$ is a well-defined topology.

For (2), we should check the second condition for $\mathfrak{N}(x)$ to be a n'hood basis, as the first condition is already given by definition. I.e. given a n'hood $N$ of a point $x \in S$, can we find an element $K$ in $\mathfrak{N}(x)$ such that $K \subseteq N$? If $N$ is a n'hood, then it contains an open set, let $U$, such that $x$ is contained within. By def'n of our openness, $U$ contains $\mathfrak{N}(x)$, and since $\mathfrak{N}(x) \neq \varnothing$, we definitely have such a set $K$ such that $K \subseteq N$. Thus $\mathfrak{N}(x)$ is a n'hood basis at $x$ for each $x \in S$.

For (3), why is that $\tau$ is unique? Because every open set is uniquely determined by its points, as its points uniquely determine $\mathfrak{N}(x)$. I believe this unique determination is mainly due to the first and second properties of definition of $\mathfrak{N}(x)$, as given $x \in S$, we can find non-empty sets in $\mathfrak{N}(x)$ such that they are subsets of intersections of sets in $\mathfrak{N}(x)$. And this property holds only for $k$ if $\mathfrak{N}(x = k)$. Thus uniqueness.

2nd Attempt

Motivated by the neighborhood characterization of openness, define a set $U$ as open if \begin{equation} \forall x \in U \; \; \exists N \in \mathfrak{N}(x) \text{ s.t. } N \subseteq U. \end{equation}

Then, is $\mathfrak{N}(x)$ a neighborhood basis at $x$, given arbitrary $x \in S$?

• For all $N \in \mathfrak{N}(x)$, from our openness definition, $N$ are open as $N \subseteq N$. Then the sets $N \in \mathfrak{N}(x)$ are neighborhoods of $x$. This seems counter-intuitive as neighborhoods need not be open. But here I am.
• Given a neighborhood $K$ of $x$, by definition of a neighborhood, there is an open set $U \subseteq K$ with $x \in U$, then $U$ contains an element from $\mathfrak{N}(x)$, so does $K$.

Hence the set $\mathfrak{N}(x)$ is a neighborhood basis at $x$. Next, we check whether this topology is well-defined or not.

• $\varnothing$ is vacuously open. $S$ is also trivially open.
• Given $U, V$ open, is $U \cap V$ open? Let $x \in U \cap V$, then $\exists N \in \mathfrak{N}(x)$ with $N \subseteq U$ and $\exists M \in \mathfrak{N}(x)$ with $M \subseteq V$. From property (2) given in the problem, $\exists P \subseteq (N \cap M)$, such that $P \subseteq U\cap V$ and $P \in \mathfrak{N}(x)$. Thus intersections are open.
• Unions are trivially open.

I do not know how this topology is uniquely determined.

Answer 1

Your 2nd Attempt is great. Indeed, you have written down the correct definition. If you recall that a set in a topology is open if and only if it contains a neighborhood of any of its points, you see that any topology satisfying that $\mathfrak{N}(x)$ is a neighborhood basis of $x$ for any $x\in X$ has to be the topology you've written down. This establishes uniqueness. The arguments in your bullet points are all correct, except for the very first bullet point. I recommend as a bonus to make it explicit where you have used the assumption 3..

Now, your argument in the first bullet point is incorrect. To verify that $N\in\mathfrak{N}(x)$ were open as you claim, you need to verify that, for any $y\in N$, there is an $N_y\in\mathfrak{N}(y)$ s.t. $N_y\subseteq N$. For $y=x$, you can take $N_y=N$ as you do, but for any other $y$, there is no guarantee you can find such an $N_y$. Indeed, there is actually no guarantee that $N$ is a neighborhood of $x$ at all: The exercise is incorrect. (I apologize for not pointing this out earlier.)

For an easy counter-example, consider $X=\{x,y,z\}$ and take $\mathfrak{N}(x)=\{\{x,y\}\}$ and $\mathfrak{N}(y)=\mathfrak{N}(z)=\{\{x,y,z\}\}$. This trivially satisfies 1.-3., but there is no topology on $X$ with these neighborhood bases. Indeed, in such a topology, the only open set containing $y$ or $z$ would be $X$ itself, so $\{x,y\}$ cannot be open, but since $\{x\}$ itself cannot be open either, $\{x,y\}$ is not a neighborhood of $x$.

To fix the Exercise, you have to introduce another axiom relating the neighborhood bases at different points of $X$ (it is hopefully intuitive that, to specify a topology, you need to have some sort of compatibility between the neighborhood bases at various points). This is discussed here.