Problem: Let $X$ and $Y$ be topological space and $R$ be a closed subset of $X \times Y$. For a compact set $A \subseteq X$, define
\[
R[A] \coloneqq \{ y \in Y: (x,y) \in R \text{ for some $x \in A$ } \}.
\]
Show that $R[A]$ is a closed subset of $Y$.
Hints: To show \( R[A] \) is closed, prove that \( Y \setminus R[A] \) is open.
Given \( y \notin R[A] \), note that for every \( x \in A \), the point \( (x,y) \notin R \).
Use the fact that \( R \) is closed to find open sets \( U_x \times V \) separating \( (x,y) \) from \( R \).
Use compactness of \( A \) to get an open set $U$ such that $U \times V \subseteq X\times Y \setminus R$.
Finally show that $V\subseteq Y \setminus R[A]$.
Solution: We need to show that the set $R[A]$ is a closed subset of $Y$. For that we will show that $Y\setminus R[A]$ is open in $Y$. Let $y \in Y \setminus R[A]$ this implies for any $x \in A$, $(x,y) \notin R$. Since $R$ is closed, so its complement must be open. Therefore, for every $x\in A$ we can find an open set $U_x \times V$ containing $(x,y)$ such that $U_x \times V \subseteq X\times Y \setminus R[A]$. Thus we get an open cover of $A$, namely
\[
\mathcal{U} = \{ U_x: x \in A \} .
\]
Since $A$ is compact, we can find a finite subcover of $\mathcal{U} $, say
\[
\{ U_i \in \mathcal{U} : i = 1,2,\dots, n \} .
\]
Set $U = \cup _i U_i$. So, $U\times V \subseteq X\times Y\setminus R$ for any $x \in A$.
We now claim that $V\subseteq Y\setminus R[A]$. Let $v \in V$. This implies
\[
\forall\ a \in A \quad (a,v) \notin R \implies v \notin R[A].
\]
Thus, $V \subseteq Y\setminus R[A]$ and hence $Y\setminus R[A]$ is open and therefore, $R[A]$ is closed.