I am having trouble understanding the following definition:
Let $(X,\tau)$ be a topological. $(X,\tau)$ is a $T_1$ space if all singletons are closed. Given $x,y\in X$ distinct points then there exists an open set $\mathscr{U}$ such that $\mathscr{U}\cap\{x,y\}=\{y\}$.
Question:
Why does $\mathscr{U}\cap\{x,y\}=\{y\}$ imply $\{y\}$ to be closed. Is not $\mathscr{U}\cap\{x,y\}=\{y\}$ the intersection of two open sets? What is this logic?
Thanks in advance!
If $U \cap \{x,y\} = \{y\}$ this says that $U$ is an open set of $X$ with $x \notin U, y \in U$. $U \cap \{x,y\}$ is not the intersection of two open sets, just one open set and a certain finite set (a doubleton).
If $X$ is $T_1$ (in the closed singleton sense) and $x \neq y$ are two distinct points of $X$, then $\{x\}$ is closed in $X$ so also closed in $\{x,y\}$ (as trivially $\{x\} \cap \{x,y\} = \{x\}$) and so $\{y\} = \{x,y\} \setminus \{x\}$ is open in $\{x,y\}$ as the complement of a closed set. And $\{y\}$ open in $\{x,y\}$ then means by the definition of the subspace topology that there is some open $U$ in $X$ such that $\{y\} = U \cap \{x,y\}$ as claimed.