Understanding the definion of $T_1$ space

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?

• Use \{ and \} for { and } in mathjax. – Lord Shark the Unknown Oct 19 '18 at 18:13
• It means that $\{x\}$ is closed, since its complementary is open. In fact each $y\in X\setminus \{x\}$ is in the interior of $X\setminus\{x\}$ – Blumer Oct 19 '18 at 18:15
• Consider the complementary of $S=y^c$, and try to figure out why it is open, by definition for each point $x\in S$ you shoul find an opne set$A\subset S$ such that $x\in A$ and this is exctaly what $T1$ means – ALG Oct 19 '18 at 18:17
• @ALG Why not an official answer? – Paul Frost Oct 19 '18 at 23:13

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.
• That would imply we are working in topological space $X,\tau$ where $X={x,y}$, right? – Pedro Gomes Oct 20 '18 at 13:05
• We consider $\{x,y\}$ as a subspace yes. @PedroGomes – Henno Brandsma Oct 20 '18 at 13:06
• There is something I am not quite understanding. Since $U$ by definition is opened in $X$, then $U\cap\{x,y\}=y$ implies that ${y}$ is opened in $X$, but this contradicts the fact it is $T_1$? – Pedro Gomes Oct 20 '18 at 14:12
• @PedroGomes It implies by definition that $\{y\}$ is open in $\{x,y\}$, not in $X$. Plus a set can be open and closed at the same time. All singletons closed, still means some singletons could be open too (or all of them, like in a discrete space). – Henno Brandsma Oct 20 '18 at 14:29