Show $\mathcal{B}$ is a topology basis on $X$ Let $(X,\mathcal{T})$ be a topological space. Denote by $C(X)$ the space of all continuous functions $f:X\rightarrow \mathbb{R}$. For $f\in C(X)$ we define 
$$V(f):=\{x\in X:f(x)\neq 0\}.$$
Show that


*

*The collection $\mathcal{B}:=\{V(f):f\in C(X)\}$ is a topology basis on $X$.

*If $X$ is normal, then $\mathcal{T}=\mathcal{T}(\mathcal{B})$.


Well, for 1, I can show that for every $x\in X$, there must be a continuous function such that $f(x)\neq 0$, so $x\in V(f)$ and $V(f)\in\mathcal{B}$. But how can I show that if $x\in V(f_1)\cap V(f_2)$ with $V(f_1),V(f_2)\in\mathcal{B}$, there exists a $B\in\mathcal{B}$ such that $B\subset V(f_1)\cap V(f_2)$ and $x\in B$? 
Also, I'm not sure how I can show the second thing as well. All these notations confuse me a bit, so I hope someone can help! 

If $x\in V(f_1)\cap V(f_2)$, then $x\in V(f_1f_2)\subset V(f_1)\cap V(f_2)$.  This finishes your first question.  
For your second question:  Clearly $V(f)\in\mathcal T$ for all $f\in C(X)$, and thus $\mathcal T(\mathcal B)\subset\mathcal T$.  To show that $\mathcal{T}=\mathcal{T}(\mathcal{B})$, we need to show that if $U\in\mathcal T$ and $x\in U$ then there is some $f\in C(X)$ such that $x\in V(f)\subset U$.  But since $X$ is normal, we can apply Urysohn's lemma to the closed sets $\{x\}$ and $U^c$ to obtain $f\in C(X)$ such that $f(x)=1$ and $f(y)=0$ for all $y\in U^c$.  This is precisely the $f$ we are after.
