If $f, g : X \to Y$ are functions such that $f(a) = g(a)$ for all $a \in A$, then $f(x) = g(x)$ for all $x \in A'$ 
Let $(X, T)$ be a topological space, let $(Y, U)$ be Hausdorff space, and let $A$ be nonempty subset of $X$. If $f, g : X\to Y$ are continuous functions such that $f(a) = g(a)$ for all $a\in A$, then $f(x) = g(x)$ for all $x \in  A'$ (with $A'$ denoting the closure of $A$).

How do I prove this?
 A: Take a series $x_n \to x$, such that $x_n \in A$ for all $n$.
By applying $f(x_n)=g(x_n)$ you should be able to proceed. (note that $f, g$ are continuous. What does that mean about series continuity?)
A: The approach I'd take is to use proof by contradiction by assuming there was some $x \in A' - A$ such that $f(x) \neq g(x)$, and the following facts:

*

*There are disjoint open sets $U_f$ and $U_g$ of $Y$ such that $f(x) \in U_f$ and $g(x) \in U_g$.

*Every open set containing $x$ intersects $A$.

If you want a full answer, hover over the text below:

 Since $f$ and $g$ are continuous, the sets $F = f^{-1}(U_f)$ and $G = g^{-1}(U_g)$ are both open. Since both sets contain $x$, the intersection $F\cap G$ (which is itself open and non-empty) intersects the set $A$. Thus there is some element $a \in A\cap F\cap G$. Applying the maps $f$ and $g$ to $a$, we observe that $f(a) \in U_f$, and $g(a) \in U_g$. Since $U_f$ and $U_g$ do not intersect, we have that $f(a) \neq g(a)$, which is a contradiction.

