# Topology according to Hausdorff

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!

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