i am reading the book "Analysis Now" by Gert K. Pedersen. I am very new in the field of topology and i am having trouble proving the following theorem.

Suppose that to every point $x$ in a set X we have assigned a nonempty family $U(x)$ of subsets of X, s.t.:

$(i)x \in A,\forall A \in U(x)$

$(ii) A, B \in U(x) \Rightarrow \exists C \in U(x):C\subset A\cap B$

$(iii)A \in U(x) \Rightarrow \forall y\in A, \exists B \in U(y): B\subset A$

Show that if $\tau$ is the weakest topology containing all $U(x)$, $x\in X$, then $U(x)$ is a neighborhood basis for $x \in \tau ,\forall x \in$ X.

I have written the following sketch of the prove, but i am afraid that is completely wrong:

Since $\tau$ is the weakest topology that contains all $U(x)$ then we can say that(can we????) $$\tau=\{\emptyset,U(x),\cap^n U_i(x),\cup U_i(x) \},$$ then if $A\in\tau$ then it must be either the empty set, the set $U(x)$,$\cap^n U_i(x)$ or $\cup U_i(x)$ (if A is one of the trivial sets the prove is trivial.)

Let $A=\cap^n U_i(x)$ then if follows from $(ii)$ that $\mathcal{O}(x) \subseteq U(x)$

Let $A=\cup U_i(x)$then if follows from $(iii)$ that $\mathcal{O}(x) \subseteq U(x)$

therefore U(x) is a neighboord basis for x in $\tau$.

Thanks for the attention!

  • 1
    $\begingroup$ It's indeed not perfect. We also have to take the unions of the intersections, and so on.. Also, we won't get $\mathcal O(x) \subseteq U(x)$. $\endgroup$
    – Berci
    Feb 25 '20 at 23:45
  • 1
    $\begingroup$ There is an alternative description for $\tau$ that is more convenient, see my answer. $\endgroup$ Feb 26 '20 at 0:01

Let $\tau_w$ be that weakest topology (which always exists for any collection we specify) and let $\mathcal{O}(x)$ be the set of all open sets in $\tau_w$ that contain $x$. You want to show that $U(x)$ is a base for $\mathcal{O}(x)$.

First note that we can define a topology $\tau$ directly as

$$O \in \tau \iff \forall x \in O: \exists U \in U(x): U \subseteq O\tag{1}$$

That this defines a topology is clear (check the definitions, we need (ii) for intersections) and all $U(x)$, $x \in X$ form subsets of $\tau$ (axiom (iii) plays a role there). So $\tau_w \subseteq \tau$. If $O \in \tau$ we can use (1) to write $O$ as a union of elements from $\bigcup \{U(x): x \in X\} \subseteq \tau_w$, so $\tau \subseteq \tau_w$ as a topology is closed under all unions. So $\tau$ defines the same topology as $\tau_w$ does and we have a more convenient description of $\tau_w$ by (1).

For $O \in \mathcal{O}(x)$ it's clear by (1) that there is some $U \in U(x)$ with $U \subseteq O$, so indeed $U(x)$ is a base for $\mathcal{O}(x)$.


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