---- > [!definition] Definition. ([[quasi-connected points]]) > Let $X$ be a [[topological space]]. We say $x$ is **quasi-connected in $X$ to $y$** if there is no [[separation of a topological space|separation]] $X=A \sqcup B$ of $X$ into disjoint open subsets such that $x \in A$ and $y \in B$. > Define an [[equivalence relation]] on $X$ by setting $x \sim y$ if $x$ is quasi-connected in $X$ to $y$. The [[equivalence class|equivalence classes]] of $\sim$ [[partition]] $X$ [[partitions are always determined uniquely by equivalence relations|into]] disjoint sets, called the **quasi-connected components** of $X$. > [!justification] > We need to show $\sim$ is an [[equivalence relation]]. Let $x,y,z \in X$. > >- $x \sim x$, for if $x \in A$ and $x \in B$ then $A$ and $B$ must clearly intersect nontrivially. >- Suppose $y \not \sim x$. Then there exist $B,A$ open in $X$ with $X=B \sqcup A$, such that $y \in B$ and $x \in A$. But if $X=B \sqcup A$ then also $X=A \sqcup B$, implying $x\not \sim y$. Hence, contrapositively, $x \sim y$ implies $y \sim x$. >- Suppose $x \not \sim z$. Then we can find open sets $A, B \in X$ such that $X=A \sqcup B$, with $x \in A$ and $z \in B$. If $y \sim z$, then $y \in B$, for otherwise $y \in A$, witnessing a separation of $X$ with $y$ on one side and $z$ on the other. But then $A \sqcup B$ witnesses a separation of $X$ with $x$ on one side and $y$ on the other. So, $x \not \sim y$. Analogous reasoning shows that if $x \sim y$ than $y \not \sim z$. Thus, the contrapositive is shown: if $x \not \sim z$, then $x \not \sim y$ or $y \not \sim z$. > (Not sure how to avoid contrapositive proofs here, since the definition is phrased in terms of something *not* existing) ^78b235 > [!basicproperties] > >1. The [[connected component]] of any $x \in X$ is contained in the quasi-connected component of $x$. >2. If $X$ is [[locally connected, locally path-connected|locally connected]], then the reverse inclusion holds. Thus, the [[connected component]]s of $X$ and the quasi-connected components of $X$ are the same when $X$ is [[locally connected, locally path-connected|locally connected]]. > [!proof] Proof of Basic Properties. > > >1. Let $x \in X$ belong to [[connected component]] $C$ and quasi-connected component $Q$. If $A \sqcup B$ constitutes a [[separation of a topological space|separation]] of $X$ with $x \in A$, then by [[connected subspace of separated set lies in one constituent]] we have $C \subset A$. Thus if $y \in C$, then $y \notin B$, meaning that $A \sqcup B$ must not witness the existence of a [[separation of a topological space|separation]] of $X$ wherein $x$ is on one side and $y$ is on the other. So, $x$ and $y$ are quasi-connected: $y \in Q$. Since $y$ was arbitrary, the result follows. > >2. We'll show the contrapositive. Suppose $X$ is [[locally connected, locally path-connected|locally connected]]. Let $x \in X$ belong to the [[quasi-connected points|quasi-connected component]] $Q$, and to the [[connected component]] $C$. Suppose $y \notin C$. ~~Since $X$ is [[locally connected, locally path-connected|locally connected]], any open [[neighborhood]] $U$ of $x$ contains a [[connected]] [[neighborhood]] $V$ of $x$. The union of all such [[neighborhood]]s is exactly $C$.~~ We know that $C$ is [[closed set|closed]] in $X$, since [[adjoining limit points preserves connectedness|closure preserves connectedness]]. $C$ is also open in $X$, ~~for it is a union of open sets (containing $x$)~~, by [[locally connected iff open sets have open components]]. Therefore, $C$ and $X-C$ are each open in $X$ with $x \in C$ and $y \in X-C$, implying that $X=C \sqcup (X-C)$ witnesses that $x$ is not quasi-connected to $y$. So $y \notin Q$. ^bba492 Let $K=\left\{ \frac{1}{n}: n \geq 1 \right\}$ and $X=K \times [0,1] \cup \{ (0,0), (0,1) \}.$ Prove that the quasi-connected component of $(0,0)$ in $X$ is different from its connected component. The [[connected component]] of $(0,0)$ is just the singleton $\{ (0,0) \}$. But the quasi-connected component of $(0,0)$ must not be this singleton, for if $C$ is a [[clopen set]] containing $(0,0)$ then we know $C$ must also contain $(0,1)$. ---- #### ---- #### 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 > ```