Connectedness

Definition: disconnected

A topological $X$ is disconnected if and only if it can be expressed as the disjoint union of two non-empty open subsets $U\bigsqcup V=X$ ($U$ and $V$ are subsets not spaces)

Definition: connected

A space that is not disconnected is connected

Whether a space is connected or not depends on its topology

Example

• disconnected
  • $\mathbb{R}\setminus \{0\}$: can be expressed as $(-\infty ,0)\cup (0,\infty )$
  • The disjoint union of two closed disks in ${\mathbb{R}}^2$ is disconnected. (we are talking about the subset topology of those disks (as topology spaces))
  • ${\mathbb{Q}}^2$ is disconnected in ${\mathbb{R}}^2$
• connected
  • Intervals are connected
  • Open (closed) disks are connected
  • ${\mathbb{R}}^n$ is connected

Lemma

NOTE

$X$ is connected $⟺$ $\varnothing$ and $X$ are the only subsets in $X$ that are both open and closed.

Proof: $⟹$ Suppose $U\subseteq X$ is clopen (closed and open), then $U^c$ is open. Then $U\bigsqcup U^c=X$, and that would mean $X$ is disconnected.

NOTE

$X$ is connected $⟺$ $X$ is not homeomorphic to a disjoint union of some non-empty spaces

Properties: subspaces

Let $\{B_{\alpha }{\}}_{\alpha \in A}$ be a collection of connected subspaces of $X$ with a point in common. Then ${\bigcup }_{\alpha \in A}{B}_{\alpha }$ is connected.

Let $p\in {\bigcap }_{\alpha }{B}_{\alpha }$ Suppose ${\bigcup }_{\alpha }{B}_{\alpha }=U\bigsqcup V$ with $U,V$ open Then $p\in U$ or $p\in V$. Assume wlog. that $p\in U$ For each $\alpha$, $B_{\alpha }\subseteq {\bigcup }_{\alpha }=U\bigsqcup V$ $⟹B_{\alpha }\subseteq U$ or $B_{\alpha }\subseteq V$ Since $p\in B_{\alpha }$ and $p\in U$ $⟹B_{\alpha }\subseteq U$ $⟹$ ${\bigcup }_{\alpha }{B}_{\alpha }\subseteq U$ $⟹V=\varnothing$ Contradiction

Properties: real number

NOTE

The connected subsets of $\mathbb{R}$ are $\varnothing$,points, and intervals ($J\subseteq \mathbb{R}$ interval $⟺$ $a,b\in J$ and $a<c<b⟹c\in J$)

Intermediate value Theorem

NOTE

Let $X$ be a connected space and $f:X\to \mathbb{R}$ continuous. If $p,q\in X$, then $f$ attains every value between $f(p)$ and $f(q)$

Proof: $X$ connected $⟹$ $f(X)\subseteq \mathbb{R}$ connected $⟹f(X)$ is a singleton or an interval(Properties real number) Let $p,q\in X$

• If $f(p)=f(q)$, then no points are in between.
• If $f(p)<f(q)$, then $f(p),f(q)\in f(X)$ ($f(X)$ is an interval)
  • $⟹$ $\forall cf(p)<c<f(q):c\in f(X)$
  • $⟹\exists z\in X:f(z)=c$