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 $\mathscr{U}\cap\{x,y\}=\{y\}$.


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!

  • 1
    $\begingroup$ Use \{ and \} for { and } in mathjax. $\endgroup$ – Lord Shark the Unknown Oct 19 '18 at 18:13
  • 2
    $\begingroup$ It means that $\{x\}$ is closed, since its complementary is open. In fact each $y\in X\setminus \{x\}$ is in the interior of $X\setminus\{x\}$ $\endgroup$ – Blumer Oct 19 '18 at 18:15
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    $\begingroup$ 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 $\endgroup$ – ALG Oct 19 '18 at 18:17
  • $\begingroup$ @ALG Why not an official answer? $\endgroup$ – 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.

  • $\begingroup$ That would imply we are working in topological space $X,\tau$ where $X={x,y}$, right? $\endgroup$ – Pedro Gomes Oct 20 '18 at 13:05
  • $\begingroup$ We consider $\{x,y\}$ as a subspace yes. @PedroGomes $\endgroup$ – Henno Brandsma Oct 20 '18 at 13:06
  • $\begingroup$ 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 $T_1$? $\endgroup$ – Pedro Gomes Oct 20 '18 at 14:12
  • $\begingroup$ @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). $\endgroup$ – Henno Brandsma Oct 20 '18 at 14:29
  • $\begingroup$ @PedroGomes open, not opened (abierto?). $\endgroup$ – Henno Brandsma Oct 20 '18 at 14:31

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