To show that the intersection $\bigcap_{j \in J} F_j$ is closed in $T$, by definition, we must show that the complement $X \setminus \bigcap_{j \in J} F_j$ is open in $T$. Applying result R219: Difference of set and intersection equals union of differences, one has \begin{equation} X \setminus \bigcap_{j \in J} F_j = \bigcup_{j \in J} (X \setminus F_j) \end{equation} Since $F_j$ is closed in $T$ for each $j \in J$, then $X \setminus F_j$ is open in $T$ for each $j \in J$. By definition of a D86: Topology, an arbitrary union of open sets is open. Thus, $\bigcup_{j \in J} (X \setminus F_j)$ is open in $T$ and therefore $X \setminus \bigcap_{j \in J} F_j$ is open in $T$. By definition of a closed set, then, the intersection $\bigcap_{j \in J} F_j$ is closed in $T$. $\square$