Prove that this function space is a basis. Let $X=[0,1]^{[0,1]}$, the set of all functions $f : [0,1] \to [0,1]$. Given a subset $A \subseteq [0,1]$, let $U(A) = \{f \in X : f(x) = 0 \forall x \in A\}$.
Show that $\mathcal{B} := \{U(A) : A\subseteq [0,1]\}$ is a basis for a topology on $X$.
I’m having difficulty understanding what this question is asking. The set $X$ consists of functions. 
I know that to show that $\mathcal{B}$ is a basis we need to show that for for every $x \in X$ there is at least one basis element that contains it (is every element of $x$ in this sense a function?  ) And we need to show that for every two elements of $\mathcal{B}$ there exists an element of $\mathcal{B}$ that is a subset to their intersection. 

If $A_1 $ and $A_2$ are subsets of $[0,1]$ then $$UA_1 \cap UA_2=\{f: f(x) =0 \forall x \in A_1 \text {and} f(x) =0 \forall x \in A_2\}$$ $$= \{f: f(x) =0 \forall x \in A_1\cup A_2\}=U(A_1 \cup A_2).$$ Hence the family $(UA)_{A \subset [0,1]}$ is a basis for some topology. 
