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.
Best Answer
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 $x \in \overline{A}$.