----- > [!proposition] Proposition. ([[every compact subspace of a Hausdorff space is closed]]) > Every [[compact]] [[subspace topology|subspace]] of a [[Hausdorff space]] is [[closed set|closed]]. > [!proposition] Corollary. > Intervals of the form $(a,b)$ or $(a,b]$ in the [[Hausdorff space]] $\mathbb{R}$ cannot be [[compact]]. > [!proof]- Proof. ([[every compact subspace of a Hausdorff space is closed]]) Let $Y$ be a [[compact]] [[subspace topology|subspace]] of a [[Hausdorff space|Hausdorff]] [[topological space]] $X$. We shall prove that $X-Y$ is open in $X$ by showing that every element of $X-Y$ is an [[topological interior|interior point]] of $X-Y$. > Let $x_{0} \in X-Y$. For each $y \in Y$, obtain open [[neighborhood]]s $V_{y}$ of $y$ and $U_{y}$ of $x$ with $V_{y} \cap U_{y} = \emptyset$. The collection $\{ V_{y} \}_{y \in Y}$ is an [[cover|open cover]] of $Y$ by sets open in $X$; using [[compact]]ness fix a finite subcover $\{ V_{i} \}_{i=1}^{n} \supset Y$ corresponding to neighborhoods $\{ U_{i} \}_{i=1}^{n}$. Note that for $i \in [n]$ with have $U_{i} \cap V_{i} =\emptyset$. From this it follows that the open set $U:=\bigcap_{i=1}^{n}U_{i}$ intersects the set $\bigcup_{i=1}^{n} V_{i}$ trivially. But $\bigcup_{i=1}^{n}V_{i}$ contains $Y$, and so we have constructed an open neighborhood $U$ of $x_{0}$ contained entirely in $X-Y$. Since $x_{0}$ was arbitrary we conclude $X-Y$ is open in $X$. ----- #### ---- #### 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 > ```