Source of most content: Gunhee Cho, Lecture note by Matsumura and Topology by Munkres.

Connected Space

Def. Given T.S. $(X, T)$, $\emptyset \ne E \subseteq X$ is disconnected if there exist disjoint sets $A, B$ such that (1) non-empty (2) $E \subseteq A \cup B$ (3) $\bar{A} \cap B = \emptyset = A \cap \bar{B}$.

NB1; $E$ is connected if $E$ is not disconnected.
NB2; Let $E$ be a disconnected set in $X$. Let $V= \bar{A}^c, W = \bar{B}^c$ be open sets. Then, (3) implies $B \subseteq V$ and $A \subseteq W$. So, $E \subseteq V \cup W$. Let $V’ = E \cap V$ and $W’ = E \cap W$ be open in $E$ (subspace topology). Then, $E = V’ \mathbin{\unicode{x228D}} W’$ holds since $V’ \cap W’ = E \cap V \cap W = E \cap \bar{A}^c \cap \bar{B}^c$ $= E \cap (\bar{A}\cup \bar{B})^c \subseteq E \cap (X\setminus E)^c = \emptyset$. Hence, $E$ can be separated by two proper subset of $E$ which are open and closed.
NB3; $\mathbb{Q}\subseteq \mathbb{R}$ is totally disconnected, i.e., only one point subset is connected.

Propositions

\[\text{Let } X= U \mathbin{\unicode{x228D}} V, \text{ where } U,V \in T_X. \text{ If } A\subseteq X \text{ is connected, then } A \subseteq U \text{ or } A \subseteq V.\]

Proof

If $A\subseteq A_1 \cup A_2$ for non-empty sets $ A_1 = A \cap U$ and $A_2 = A \cap V$, then $A_1$ and $A_2$ are open in $A$. Therefore, $A = A_1 \mathbin{\unicode{x228D}} A_2$ holds. Since $A_1$ and $A_2$ are non-empty open sets in $A$, $A$ is disconnected.

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\[\forall \alpha \in I, A_\alpha \subseteq X \text{ (connected), } \cap_{\alpha \in I} A_\alpha \ne \emptyset. \text{ Then, } \cup_{\alpha \in I} A_\alpha \text{ is connected.} \nonumber\]

Proof

Suppose $\cup_{\alpha \in I} A_\alpha$ is disconnected, i.e., $\cup_{\alpha \in I} A_\alpha = U \mathbin{\unicode{x228D}} V $. Let $p \in \cap_{\alpha \in I} A_\alpha$ and $p\in U$. Since $A_\alpha$ is connected, it must lie entirely in either $U$ or $V$, i.e., $A_\alpha \subset U$ for all $\alpha \in I$. So that $\cup_{\alpha \in I} A_\alpha \subset U$, i.e., $V$ is empty, which is a contradiction.

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\[\text{Let } X \text{ be a connected space and } f: X \to Y \text{ be continous. Then, } f(X) \text{ is connected.} \nonumber\]

Proof

Let $Z= f(X)$ and consider a continous surjective map $g: X \to Z$. Suppose $Z$ is disconnected, i.e., $Z = A \mathbin{\unicode{x228D}} B$ for non-empty open sets $A,B$ in $Z$. Then, $X = g^{-1}(Z) = g^{-1}(A) \mathbin{\unicode{x228D}} g^{-1}(B)$ holds since $g$ is surjective. Since $g$ is continous, $g^{-1}(A)$ and $g^{-1}(B)$ are open in $X$, which concludes the proof.

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\[\text{The union of a collection of connected subspaces of } X \text{ that have a point in common is connected.}\]

Proof

Let $\{ A_\alpha \}$ be a collection of connected subspaces of $X$ and $p$ be a common point. Suppose $A = \cup A_\alpha = U \mathbin{\unicode{x228D}} V$, i.e., $A$ is diconnected. If $p \in U$, $A_\alpha \subset U$ for all $\alpha$ by (1). Thus, $A \subset U$, which contradicts the assumption. Hence, $A$ is connected.

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\[\text{Let } X_1, \ldots, X_n \text{ be a connected space. Then, finite Cartesian product } \prod_{i=1}^K X_i \text{ is connected.} \nonumber\]

Proof

It suffices to show $X_1 \times X_2$ is connected. Note that $X_1 \cong X_1 \times \{ p \}$ and $X_2 \cong \{ p \} \times X_2$ are connected from above proposition. Fix $(a,b) \in X_1 \times X_2$. For each $x \in X_1$, Let $V_x := (\{ x \} \times X_2) \cup (X_1 \times \{ x \})$. Then, $V_x$ is connected since its subspace shares a common point $(x,b)$ by (2). Since $X_1 \times X_2 = \cup_x V_x$ and $(a,b) \in C_x$ holds, by (2), $X_1 \times X_2$ is connected.

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\[\text{If } A \subseteq X \text{ is connected, then } \forall B \subseteq X \text{ s.t. } A \subseteq B \subseteq \bar{A}, \, B\text{ is connected.} \nonumber\]

Proof

If $B = U \mathbin{\unicode{x228D}} V$ for non-empty open sets $U,V$. By (1), $A \subseteq U$ or $A\subseteq V$ and assume $A\subseteq U$ w.l.o.g. Therefore, $\bar{A} \subseteq \bar{U} = U $ holds. Note that $U \mathbin{\unicode{x228D}} V \subseteq \bar{A}$ holds by assumption. This implies that $U\mathbin{\unicode{x228D}} V \subseteq U$, i.e., $V$ is empty set, which is a contradiction.

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NB3; $(\mathbb{R}^{w}, T_{box})$ is disconnected.
NB4; $(\mathbb{R}^{w}, T_{prod})$ is connected.

Linear Continum

Def. Let $L$ be simply ordered (totally ordrered). $L$ is called linear continum if it satisfies (1) $L$ has a least upper bound property, i.e., $\exists \sup E \in L$ and (2) $L$ has a density, i.e., $\forall x, y \in L$ with $x< y$, $\exists z \in L$ s.t. $x < z < y$.

NB5; $L$ is connected.
NB6; intervals, rays are connected.

Proposition

\[\text{For linear continum with ordered topology } (L, T_{order}) \text{, let } C \subseteq L \text{ be a convex subset. Then, } C \text{ is connected.}\]

Proof

Suppose $C$ is disconnected, i.e., $C = U\mathbin{\unicode{x228D}} V$ for non-empty open sets $U,V$. Take $a \in U, \, b \in V$ and assume $a< b$ for the simplicity, i.e., $[a, b] \subset C$ since $C$ is convex. Let $[ a, b] \cap U = U’$ and $[ a, b] \cap V = V’$. Then, $\exists \sup U’ \in L$ and $a \leq \sup U’ \leq b$, i.e., $\sup U’ \in [a, b]$.

• Claim: $\sup U’ \not\in [a, b]$.
  Note that $[ a, b ] = U’ \cup V’$. Suppose $\sup U’ \in U’$, i.e., $\sup U’ \ne b$. Then, $\exists d \in U’$ s.t. $[ \sup U’ , d ) \subseteq U’$, i.e., $\sup U’ < d \in U’$, which is a contradiction. Hence, $\sup U’ \in V’$, i.e., $\sup U’ \ne a$. Then, $\exists d’$ s.t. $(d’, \sup U’) \subseteq V’$. By definition of linear continum, $\exists z \in V’$ s.t. $z < \sup U’$. Since $\sup U’$ is least upper bound and $z$ is upper bound of $U’$, it is contradiction. Hence, $\sup U’ \not\in U’ \mathbin{\unicode{x228D}} V’ = [a, b]$. Therefore, $C$ is connected.

Intermediate Value Thm

\[\text{Let } X \text{ be connected space, } Y \text{ be simply ordered space and } f:X \to Y \text{ be cts.} \nonumber\] \[\text{ Then, } \forall a,b \in X \text{ with } f(a) \ne f(b), \, \forall r \in (f(a), f(b)), \exists c \in (a,b) \text{ s.t. } f(c) =r. \nonumber\]

Proof

Suppose not, i.e., $\exists r \in (f(a), f(b))$ s.t. $\forall c \in (a,b)$, $f(c)\ne r$. Then, $[ f((a,b)) \cap (-\infty, r) ] \mathbin{\unicode{x228D}} [ f((a,b)) \cap (r, \infty) ] = f((a,b))$. Since $(a,b)$ is connected by (3), image is also connected, which is a contradiction.