---- > [!definition] Definition. ([[separation of a topological space]]) > A **separation** of a [[topological space]] $X$ is a pair of disjoint, nonempty open subsets $U,V \subset X$ satisfying $X = U \sqcup V.$ > >> [!proposition] Subspace Characterization. > >If $Y \subset X$ is a [[subspace]] of $X$, then $U,V {\subset} X$ disjoint-unioning to $Y$ constitute a [[separation of a topological space|separation]] of $Y$ if and only if each is nonempty and neither contains a [[limit point]] of the other: $\overline{U} \cap V = \emptyset = U \cap \overline{V}$. > [!proof] Proof of Subspace Characterization. Endow $Y \subset X$ with the [[subspace topology]]. Suppose that nonempty $U,V$ are a [[separation of a topological space|separation]] of $Y$—— $Y=U \sqcup V$ for $U=O_{U} \cap Y$ and $V=O_{V} \cap Y$ where $O_{U}$ and $O_{V}$ are open in $X$. We claim that $O_{U} \cap Y = \overline{O_{U} \cap Y}$ ($=\overline{O_{U} } \cap Y$, see [[characterization of closure in subspace]]). Then, since $(O_{U} \cap Y) \cap V = \emptyset$ by assumption, we'll be done with this direction. To show this, we'll use [[closure is set together with limit points|closed iff contains all limit points]]. Suppos $x$ is a [[limit point]] of $U$ with $x\notin U$. Then $x \in V$ since $V=Y-U$. $V$ is open in $Y$, so we can fix $W \subset V$ s.t. $x \in W \subset V$. But $x$ is a [[limit point]] of $U$, meaning that every [[neighborhood]] of $x$ intersects $U$ nontrivially at a point other than $x$ itself. This includes $W$. But then $V$ and $U$ intersect nontrivially, a contradiction. So $x \in U$. > Conversely, suppose $U$ and $V$ are nonempty and neither contains a [[limit point]] of the other. Then $\overline{U} \cap V = \emptyset$, and since $U \subset \overline{U}$ we have $U \cap V \subset \overline{U} \cap V = \emptyset$ meaning $U \cap V=\emptyset$. \ (actually, I don't think this proof necessarily goes through if the topology isn't generated by a basis, [[TODO]]) > [!basicexample] > - Let $Y$ denote the [[subspace]] $[-1,0) \cup (0,1]$ of $\mathbb{R}$ with the [[standard topology on the real line|standard topology]]. Each of the sets $[-1,0)$ and $(0,1]$ are open in $Y$ (though not in $\mathbb{R}$) and they are very clearly disjoint and union to $Y$. So, they form a separation of $Y$. > > [!basicnonexample] Basicexample > - Let $Y$ be the [[subspace]] $[-1,1]$ of the real line. The sets $[-1,0]$ and $(0,1]$ are disjoint and nonempty, but they do not form a separation of the [[subspace]] $Y \subset \mathbb{R}$ because $\overline{(0,1]}=[0,1]$ intersections nontrivially with $[-1,0]$. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```