Connectedness

In metric spaces §

Definition: Connected sets in metric spaces

Given a metric space $(X,d)$, a subset $A\subseteq X$ is said to be connected if for all other (nonempty) subsets $U,V\in X$ where

$$\overline{U}\cap V=U\cap \overline{V}=\varnothing ,$$

we have $U\cup V\ne A$. The sets $U,V$ which satisfy the equality above are said to be separated; a subset is connected if it is not the union of two (nonempty) separated sets.

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In topological spaces §

Definition: Separation, connected set in topological space

Given a topological space $(X,\mathcal{T})$, a separation of $X$ is a pair of open subsets $U,V\subset X$ which are non-empty, disjoint, and have $U\cup V=X$.

We say a set $X$ is connected if there does not exist any separation of $X$. Equivalently, $X$ is connected if and only if the only sets that are both closed and open are $\varnothing$ and $X$.

Connected subspaces §

Lemma: Facts about connectedness

• (i) If $Y\subseteq X$ is a connected subspace and $U,V$ form a separation of $X$, then $Y\subseteq U$ or $Y\subseteq V$.
• (ii) If $A\subseteq X$ is connected and $A\subseteq B\subseteq \overline{B}$, then $B$ is also connected.
• (iii) Conditions for the union of connected subspaces to be connected. Suppose $\{A_{\alpha }{\}}_{\alpha \in J}$ is a collection of connected subspaces of $X$ such that there exists a “special” index $\beta \in J$ such that for all other $\alpha \in J$, we have $A_{\beta }\cap A_{\alpha }\ne \varnothing$. Then ${\bigcup }_{\alpha \in J}{A}_{\alpha }$ is connected.

Path-connectedness and continuity §

Definition: Path, path-connected

Let $(X,\mathcal{T})$ be a topological space. Given $x,y\in X$, a path from $x\to y$ is a continuous map $p:[0,1]\to X$ such that $p(0)=x$ and $p(1)=y$.

We say $X$ is path-connected if for all $x,y\in X$, there exists a path from $x$ to $y$.

Proposition: If X is path-connected, then X is connected.

Proposition: Continuous functions preserve (path-)connectedness

Let $f:X\to Y$ be a continuous function between topological spaces. Then:

• (i) If $X$ is connected, then so is $f(X)$.
• (ii) If $X$ is path-connected, then so is $f(X)$.

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Examples §

Intervals in $\mathbb{R}$ §

Theorem: Every interval of the real numbers is connected.

On the other hand, subsets of $\mathbb{R}$ that are not intervals cannot be connected. One example is the rationals $\mathbb{Q}\subset \mathbb{R}$; then $U=\mathbb{Q}\cap (-\infty ,\sqrt{2})$ and $V=\mathbb{Q}\cap (\sqrt{2},\infty )$ is a separation.

The straight line path §

Definition: Straight line path

The straight line path between any two points $p:[0,1]\to X$, where $p(0)=x$ and $p(1)=y$, is defined by

$$p(t)=(1-t)x+ty.$$

The following sets are all path-connected, and hence connected:

• Any interval in $\mathbb{R}$;
• ${\mathbb{R}}^n$;
• Any ball in ${\mathbb{R}}^n$;
• If $n\ge 2$, the set ${\mathbb{R}}^n\backslash \{w\}$ for some $w\in {\mathbb{R}}^n$;
• ${\mathbb{R}}^n$ with countably many points removed (but not uncountably many—imagine removing the x-axis of ${\mathbb{R}}^2$!).

Any two points of the first three sets are connected via the straight line path. In the final set, points are connected either by the straight line path or, if $w\in p([0,1])$, we can define a path by

$$p(t)=\{\begin{array}{cc}f(2t)& \text{if }0\le t\le 1/2\\ g(2t-1)& \text{if }1/2\le t\le 1\end{array}$$

where $f$ is a path from $x$ to $z$ and $g$ is a path from $z$ to $y$ for some $z\in {\mathbb{R}}^n\backslash \{w\}$; note that $p$ is continuous by the pasting lemma.

🔺Exercise. Finish showing that these sets are path-connected by showing that the straight line path is indeed contained in the set.

Connected but not path-connected §

#wip Topologist’s sine curve

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Review §

• Show that any two intervals of the reals are homeomorphic, so the domain of a path can be defined with an arbitrary interval. ⭐