I have difficulty with the following exercise from Introduction
to Topology (by Tej Bahadur Singh)(Exercise 9 on p. 36):
Let $f: X\to Y$ be a function between topological spaces, and assume
that $A\cup B= X$, where $A-B\subseteq A^\circ$, and $B-A\subseteq B^\circ$. If $f|_{A}$ and $f|_{B}$ (endowed with the relative
topologies) are continuous, show that $f$ is continuous.
I tried to use the following facts (from the book mentioned above):
Definition (locally finite). A family $\{A_i\}$ of subsets of a space $X$ is called locally finite if each point of $X$ has a neighborhood $U$ such that $U\cap A_i\neq \varnothing$ for at most finitely many indices $i$.
(1) Let $\{U_\alpha\}$ be a family of open subsets of a space $X$ with $X = \bigcup_\alpha U_\alpha$. Then a function $f$ from $X$ into a space $Y$ is continuous if and only if $f|_{U_\alpha}$ is continuous for each index $\alpha$. (See Exercise 8 on p. 36)
(2) If a space $X$ is the union of a locally finite family $\{A_i\}$ of closed sets, then a function $f$ from $X$ to
a space $Y$ is continuous if and only if the restriction of $f$ to each $A_i$ is continuous. (See Corollary 2.1.10 on p. 33)
Using the fact (1), we obtain that $f|_{A^\circ\cup B^\circ}$ is continuous. Clearly,
$f|_{A\cap B}$ is continuous. Since $A-B\subseteq A^\circ$ and $B-A\subseteq B^\circ$,
$A^\circ \cup B^\circ \cup (A\cap B)= X$. But $A\cap B$ is not open, so I cannot use the fact (1) again, I have not idea what to do next. Any ideas would be appreciated.
Best Answer
Of course $f$ is continuous in all points of $A^° \cup B^°$. Since $A^\circ \cup B^\circ \cup (A\cap B)= X$, it remains to show that $f$ is continuous in all points of $A \cap B$.
Let $x \in A \cap B$ and $V$ be an open neighborhood of $f(x)$ in $Y$. There exist open neighborhoods $U_A$ of $x$ in $A$ and $U_B$ of $x$ in $B$ such that $f(U_A) \subset V$ and $f(U_B) \subset V$. Choose open $W_A, W_B \subset X$ such that $W_A \cap A = U_A, W_B \cap B = U_B$. Define $W = W_A \cap W_B$. Then $W$ is an open neighborhood of $x$ in $X$. We have $f(W) \subset V$: Let $y \in W$. But $y \in A$ or $y \in B$, w.l.o.g. $y \in A$. Thus $y \in W \cap A \subset W_A \cap A = U_A$ and therefore $f(y) \in f(U_A) \subset V$.