Unable to prove an exercise in Continuous functions in Topology I am self studying Topology from C. Wayne Patty and I am unable to solve the following question in exercise 1.7
Adding image->
I tried by assuming a sequence $x_n$ $ \epsilon $ A which converges to x . I got f($x_n$) = g($x_n$ ) but  I am not able to move forward.
Please give some hint. No need to fully answer it.
 A: The proof writes itself if you use a proof from contradiction:
Suppose, for a contradiction, that $f(x) \neq g(x)$ for some (now fixed) $x \in \overline{A}$.
Then as $Y$ is Hausdorff, there are open, disjoint sets $U,V$ in $Y$ such that $f(x) \in U$ and $g(x) \in V$.
As $f$ is continuous at $x$, there is some open neighbourhood $U_x$ of $x$ such that $$f[U_x] \subseteq U\tag{1}$$
As $g$ is continuous at $x$, there is some open neighbourhood $V_x$ of $x$ such that $$g[V_x] \subseteq V\tag{2}$$
Now $U_x \cap V_x$ is an open neighbourhood of $x$ and as $x \in \overline{A}$, there exists some point $a \in (U_x \cap V_x) \cap A$.
$(1)$ implies (as $a \in U_x$) that $f(a) \in U$. Also, $(2)$ implies that $g(a) \in V$. But then $$f(a) = g(a) \in U \cap V$$
contradicts the disjointness of $U$ and $V$. This contradiction shows that or initial assumption was false and so $f(x)=g(x)$ for all $x \in \overline{A}$.
A: Hint
As you’re dealing with general topological spaces, you can’t use continuity criteria based on sequences.
Use the continuity criteria based on the fact that inverse image of open subsets are open subsets.
Then proceed by contradiction using Hausdorff criteria.
A: That style of argument has a chance of working if you know that every point in $\bar{A}$ is a limit of points in $A$. This is true if $X$ is a metric space, for instance, but not in general.
As requested I will give hints only.
The approach here depends in part on what you already know. If you know that the diagonal of $Y \times Y$, i.e. the set $\Delta_Y = \{(y,y) \mid y \in Y \}$, is a closed subset of $Y \times Y$ because Y$ is Hausdorff, that will give you a nice method of solution.
Otherwise, to give a more direct proof, I would start by assuming $f(x) \ne g(x)$ for some $x \in \bar{A}$, and apply the fact that $Y$ is Hausdorff to separate the points $f(x)$ and $g(x)$.
