Definition: Let $(X, \mathcal{T})$ be a topological space. For $x \in X$, we call $V \subseteq X$ neighborhood of $x$ if there exists $G \in \mathcal{T}$ such that $x \in G \subseteq V$. We write $\mathcal{V}(x)$ for the sets of all neighborhoods of $x$.

Theorem: Suppose $X$ is a topological space and $x \in X$, then $\mathcal{V}(x)$ satisfies:

$$(V_1): X \in \mathcal{V}(x)$$ $$(V_2): \forall V \in \mathcal{V}(x): x \in V$$ $$(V_3): \forall V,W \in \mathcal{V}(x): V \cap W \in \mathcal{V}(x)$$ $$(V_4): V \in \mathcal{V}(x), V \subseteq W \implies W \in \mathcal{V}(x)$$ $$(V_5): \forall V \in \mathcal{V}(x): \exists W \in \mathcal{V}(x): \forall y \in W: V \in \mathcal{V}(y)$$

Conversely, if a family $(\mathcal{V}(x))_{x \in X}$ satisfies these $5$ conditions, then there exists a unique topology on $X$ such that for all $x \in X$, $\mathcal{V}(x)$ is the set of all neighborhoods of $x$.

Proof (my attempt): I'm stuck at the converse. I'll show what I tried.

Define $$\psi: \{\mathcal{T}\mid \mathcal{T} \mathrm{ \ topology \ on \ X}\} \to \{\mathcal{V}:=(\mathcal{V}(x))_{x \in X} \mid \mathcal{V} \mathrm{\ satisfies \ (V_1)-(V_5) }\}$$ $$\mathcal{T} \mapsto(\{A \mid \exists G \in \mathcal{T}: x \in G \subseteq A\})_{x \in X}$$

I want to show that this is a bijection, because then for every $\mathcal{V}$ satisfying the 5 given conditions, there is a unique topology we look for.

To do this, I defined what I believe is the inverse function:

$$\phi: (\mathcal{V}(x))_{x \in X} \mathrm{\ satisfying \ the \ conditions}\mapsto \{A \mid \forall a \in A: A \in \mathcal{V}(a)\}$$

Using the properties $(V_1)-(V_3)-(V_4)$, I managed to show that this is well defined: the image is a topology.

Now, one has to check that $\phi \circ \psi = 1$ and $\psi \circ \phi = 1$.

The first one of these equalities worked out.

For the second one, I ran into troubles:

I have to prove, that if $(\mathcal{V}(x))_x$ is a family satisfying those $5$ conditions, then $\psi \circ \phi((\mathcal{V}(x))_x) = (\{B \subseteq X \mid \exists G \in \{A \subseteq X \mid \forall a \in A: A \in \mathcal{V}(a)\}: x \in G \subseteq B\})_x= (\mathcal{V}(x))_x$

where I'm not sure about the last equality.

In particular, I want to prove that for a fixed $x \in X$:

$$\{B \subseteq X \mid \exists G \in \{A \subseteq X \mid \forall a \in A: A \in \mathcal{V}(a)\}: x \in G \subseteq B\}= \mathcal{V}(x)$$

and the $\supseteq$ part is giving me a lot of troubles.

This is what I did:

Let $B \in \mathcal{V}(x)$. Then, by $(V_5)$, there exists $W \in \mathcal{V}(x)$ such that $a \in W$ implies that $B \in \mathcal{V}(a)$. From this, it follows by $(V_2)$ that $W \subseteq B$. Because $W \in \mathcal{V(x)}$, it also follows that $x \in W \subseteq B$ so we are done when we can prove that

$$\forall a \in W: W \in \mathcal{V}(a)$$

but I did not manage to prove this. Can anyone help me to prove this?


Your claim $\forall a\in W : W\in\mathcal{V}(a)$ is not true.

For example, let $X=\mathbb{R}$ and for each $x\in X$ set $$ \mathcal{V}(x) = \{ A\subset\mathbb{R} \mid \exists\,\epsilon>0 \text{ such that } (x-\epsilon,x+\epsilon)\subset A \} $$ Then it is easy to check the five conditions ($V_1$)-($V_5$) hold.

Now let us follow your proof. Let $x=0$ and choose $B=(-2,2)\in\mathcal{V}(0)$. Then $W=[-1,1]\in\mathcal{V}(0)$ satisfies $\forall\,a\in W \Rightarrow B\in\mathcal{V}(a)$. Notice that when $a=1\in W$, $B=(-2,2)\in\mathcal{V}(1)$ but $W\notin\mathcal{V}(1)$. It means that your claim is not true.

A comment for proof

In my opinion, the fifth condition ($V_5$) should be as follows:

($V_5$) : $\forall\,V\in\mathcal{V}(x) : \exists\,W\in\mathcal{V}(x) : \forall y\in W : W,V\in\mathcal{V}(y)$

Then the proof will be much easier.

  • $\begingroup$ Thanks for your answer. The claim I made in my proof was wrong. $\endgroup$
    – user370967
    Feb 23 '18 at 14:56

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