----- > [!proposition] Proposition. ([[arbitrary union of nontrivially-intersecting connected subspaces is connected]]) > The union of a collection of [[connected]] [[subspace|subspaces]] of a [[topological space]] $X$ that have a point in common is a [[connected]] subspace of $X$. > [!warning] Warning. > The intersection of connected subspaces with nontrivial intersection does not need to be connected at all. Take $\mathbb{R}^{2}$ with the standard topology. The [[unit circle]] is [[connected]] (it is [[homeomorphism|homeomorphic]] to $[0, 2\pi] /0 \sim 2\pi$), as is the $x$-axis $\mathbb{R} \times \{ 0 \}$. Their intersection is $\{ (-1,0),(1,0) \}$ which is not [[connected]] in $\mathbb{R}^{2}$. > (In $\mathbb{R}$, though, this *is* true.) ^6d042e > [!proof]- Proof. ([[arbitrary union of nontrivially-intersecting connected subspaces is connected]]) > ~ > Let $A=\{ A_{\alpha} \}$ be the collection of [[connected]] [[subspace|subspaces]] of $X$ in question; let $p$ be a point in $\bigcap_{\alpha}^{}A_{\alpha}$. Because each $A_{\alpha}$ is [[connected]], a prospective [[separation of a topological space|separation]] $A=U \sqcup V$ must entail for each $\alpha$ that either $A_{\alpha} \subset U$ or $A_{\alpha} \subset V$. Also, $U$ and $V$ are nonempty so we indeed must have 'an $A_{\alpha}$ on both sides'. But since each $A_{\alpha}$ contains $p$, this would entail $p \in U$ and $p \in V$, contradicting the fact that $U \cap V = \emptyset$. ----- #### ---- #### 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 > ```