TITLE: $***B***$ is a basis of $T$ $\Leftrightarrow$ for all $x \in X$ and for all neighborhoods $U$ of $x$ exists $B \in$ $***B***$, s.t $x \in B \subset U$
QUESTION [0 upvotes]: $ X = (X,T)$ is topological space and B $\subset T$ is set of subsets of X. Then the following is true:
B is a basis of $T$ $\Leftrightarrow$ for all $x \in X$ and for all neighboorhoods $U$ of $x$ exists $B \in$ B, s.t $x \in B \subset U$
My idea:
($\Rightarrow$) Let B be a basis of $T$. We know, that $T_{B} = T$, where $T_{B}$ is induced topology from B. Let $x \in U \subseteq X$ , where $U$ is a neighbourhood of $x$. By the definition of $T_{B}$ for all $x \in U$ exist $B \in$ B, s.t $x \in B \subseteq U$
($\Leftarrow$)
And here I have the problem with back direction.
Can you please check, if I proved right direction correctly.
Give me please a hint about the strategy of proving back direction.
EDIT:
$T_B$ = {$U \subseteq X, \forall x \in U, \exists b \in B,s.t: x\in b \subseteq U$}
Neighbourhood: U is neighbourhood of $x \in X \Leftrightarrow \exists O \in T, s.t: x \in O \subseteq U$, where $U \subseteq X$ and $X = (X,T)$ topological space

REPLY [1 votes]: I'll assume that $\mathcal{B}$ is a base for $\mathcal{T}$ iff $\mathcal{T}_{\mathcal{B}} = \mathcal{T}$, where the latter is defined in the OP.
Then the statement is almost trivially true:
Left to right: 
So we know that $\mathcal{T}_{\mathcal{B}} = \mathcal{T}$.
Suppose $x \in X$ and let $U$ be a neighbourhood of $x$. Then there is an open set $O \in \mathcal{T}$ such that $x \in O \subseteq U$ by the definition of a neighbourhood.
As $O \in \mathcal{T}_{\mathcal{B}}$ and $x \in O$, there exists some $B \in \mathcal{B}$ such that $x \in B \subseteq O$. Clearly this $B \in \mathcal{B}$ is as required, as $x \in B \subseteq O \subseteq U$.
This is basically the OP's proof (which was fine for this direction).
Right to left:
Suppose we have the condition. We want to show that $\mathcal{T}_{\mathcal{B}} = \mathcal{T}$. Two inclusions:
Let $O \in \mathcal{T}$, then for any $x \in O$ we note that $O$ is a neighbourhood of $x$, so that our condition on the right says we have $B \in \mathcal{B}$ with $x \in B \subseteq O$. As this holds for all $X \in O$ we know that $O \in \mathcal{T}_{\mathcal{B}}$. 
Let $O \in \mathcal{T}_{\mathcal{B}}$. For every $x \in O$ we pick the required $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq O$ (this just uses the definition of $\mathcal{T}_{\mathcal{B}}$). Then $O = \cup \{B_x: x \in O\}$ (every $B_x$ is contained in $O$; hence so is their union, and every $x \in O$ is in its own $B_x$ hence in the union as well; this shows two inclusions) and so $O$ is a union of members of $\mathcal{B} \subseteq \mathcal{T}$ and a topology is closed under uniosn so $O \in \mathcal{T}$, as required.
Note that we only need the condition for one inclusion in $\mathcal{T}_{\mathcal{B}} = \mathcal{T}$.
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