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I need to show the following statement

Let $\mathcal{B}\subset P(X)$ be a set of subsets of a set $X$, such that $\bigcup_{U\in \mathcal{B}}U =X$ then the following are equivalent

$i)$ there exists a topology on $X$ with basis $\mathcal{B}$.

$ii)$ for each $U,V \in \mathcal{B}$ such that $U\cap V\neq \emptyset$ and each $x\in U\cap V$ there is a $W\in \mathcal{B}$ such that $x\in W\subset U\cap V$.

I think I could show the implication $i) \implies ii)$.

$"i)\implies ii)"$

Let $\mathcal{O}$ be a topology on $X$ with basis $\mathcal{B}$. Let $U,V \in \mathcal{B}$ such that $U \cap V \neq \emptyset$, then, since the intersection of two open sets is open, $U \cap V$ is again open. Let us take now $x\in U\cap V$. Since $U \cap V$ is open there exists some $\epsilon \gt 0$ such that $U(x,\epsilon) \subset U\cap V$, where $U(x,\epsilon)$ is the ball of radius $\epsilon$ with center $x$. Define $W:= U(x,\epsilon)$, then $x\in W \subset U\cap V.$

I'm trying to show the part $"ii) \implies i)"$, but I don't really know where to start. Could you please give me a hand?

Edit (after the hint of Stefan Hamcke)

for the implication $"ii)\implies i)"$ I tried to show that

$$\cal O=\left\{\bigcup S\ \middle|\ S\subseteq B\right\}$$

is a topology in this way. If I interpreted it correctly we have $\mathcal{B} \subseteq \mathcal{B} \implies \mathcal{B} \in \mathcal{O}$. We need to check the three axioms of a topology:

$i)$ $X \in \mathcal{O}$ since by assumption $\bigcup_{U\in \mathcal{B}} =X$. I still can not figure out why the empty set is in $\mathcal{O}$. Is it sufficient to say that it is in $\mathcal{O}$ just by saying that we don't take any union of subsets of $\mathcal{B}$?

$ii)$ let $I$ be an index set and $\mathcal{O}_i \in \mathcal{O}$ for every $i \in I$. Then by definition of $\mathcal{O}$ there we can write every $\mathcal{O}_i$ as $$\mathcal{O}_i = \bigcup_{S_i \in \mathcal{B}}S_i$$ where the $S_i \subseteq \mathcal{B}$ for every $i$. Hence $\bigcup_{S_i \in \mathcal{B}}S_i \subseteq \mathcal{B}$ since the union of subsets of $\mathcal{B}$ will always be a subset of $\mathcal{B}$. Therefore

$$\bigcup_{i\in I}\bigcup_{S_i \in \mathcal{B}}S_i\in \mathcal{O}$$

and the second axiom of a topology is satisfied.

$iii)$ Let $I:=\{1,\dots,n\}, n\in \Bbb{N}_{>0}$ be a finite index set and let $\mathcal{O}_i \in \mathcal{O}$ for every $i \in I$ we want that $O:=\bigcap_{i\in I}\mathcal{O}_i \in \mathcal{O}$. Since every $\mathcal{O}_i$ is the union of subsets $S_i$ of $\mathcal{B}$ we have that $O$ can be rewritten as

$$O=\bigcap_{i\in I}\bigcup_{S_i\in \mathcal{B}}S_i$$

Now we can have two cases: $1)$ the intersection is empty but the emtpyset is in $\mathcal{O}$. $2)$ the intersection is not empty, but then there exists $x\in \bigcap_{i\in I}\bigcup_{S_i\in \mathcal{B}}S_i$ and by definition of a basis there is a set $W \subseteq \mathcal{B}$ such that $x \in W$. We can find such subset for each $x$ in the intersection and therefore $O$ will be the union of all these subsets which are subsets of $\mathcal{B}$ and thus we can conclude that $O \in \mathcal{O}$

Is it correct or am i completely wrong?

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  • $\begingroup$ You start by writing that $\cal B\subseteq B\implies B\in O$, but we don't expect $\cal B$ to be an element of $\cal O$. Maybe there is some confusion about what the expression $\bigcup \cal S$ means. When we have a set $\cal S$ (or we can say a family) of subsets of $X$, then $$⋃\mathcal S,\quad⋃\{S\mid S\in\mathcal S\},\quad⋃_{S\in\mathcal S}S$$ all mean the same thing, the union of the sets that are elements of $\cal S$. $\endgroup$ – Stefan Hamcke Sep 20 '14 at 21:22
  • $\begingroup$ In $ii)$, you should rather say something like $$\mathcal O_i=⋃\mathcal S_i$$ where $\cal S_i$ is some subset of $\cal B$. That's because $\cal O_i$ is the union of some elements of $\cal B$, and we can call this subfamily $\cal S_i$. So we get $$⋃_{i\in I}\mathcal O_i=⋃_{i\in I}⋃\mathcal S_i=⋃⋃_{i\in I}\mathcal S_i$$ and since $⋃_{i\in I}\mathcal S_i$ is a subfamily of $\cal B$, this is an element of $\cal O$. $\endgroup$ – Stefan Hamcke Sep 20 '14 at 21:39
  • $\begingroup$ I have to admit that switching between the different kinds of writing the union is a bit confusing. If you are uncomfortable with that, you may replace the term $⋃\mathcal S$ with $⋃_{S\in\mathcal S}S$ whenever you encounter it. $\endgroup$ – Stefan Hamcke Sep 20 '14 at 21:43
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You want to find a topology with basis $\cal B$. You could define $$\cal O=\left\{\bigcup S\ \middle|\ S\subseteq B\right\}$$ the family of all unions of elements of $\cal B$. In particular $\cal B\subseteq O$. Try to show that $\cal O$ is a topology.

Your proof for the other direction looks good, except for the part where you take a radius $\epsilon>0$ as that's only possible in a metric space. But you can simply replace $U(x,\epsilon)$ by a basis element $W$ since the property of being a basis means precisely that there exists a $W\in\cal B$ such that $x\in W\subseteq U\cap V$

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  • $\begingroup$ +1 thank you for your help! I'll try to show that $\mathcal{O}$ is a topology and if i can i'll accept your answer! Thank you very much! $\endgroup$ – Bman72 Sep 19 '14 at 16:00
  • $\begingroup$ I tried to show that $\mathcal{O}$ is a topology. I've edited my original answer. Could you please check if what i've done is correct? $\endgroup$ – Bman72 Sep 19 '14 at 17:49
  • $\begingroup$ Thank you very much once again :) have a nice day! $\endgroup$ – Bman72 Sep 21 '14 at 9:43

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