Connected Open Sets in Euclidean Space

From Local Paths to Paths in the Whole Set

In Euclidean space, a sufficiently small ball around any point stays inside a given open set. Balls also have a useful geometric property: the straight segment between any two of their points remains in the ball. These facts show that Euclidean space has path-connected neighborhoods at every point. The connected locally path-connected theorem then turns this local supply of paths into a path joining any two points of a connected open set.

We use the Euclidean topology on $\mathbb{R}^n$, where $n\geq 1$, and its metric $d(x,y)=\|x-y\|$. A path in a space $X$ from $p$ to $q$ is a continuous map $\gamma:I\to X$, where $I=[0,1]$, such that $\gamma(0)=p$ and $\gamma(1)=q$. The space is path-connected if every pair of its points can be joined by a path. A subspace is connected if it cannot be written as the union of two disjoint, nonempty sets that are open in that subspace.

Definition: A space $X$ is locally path-connected if, for every $x\in X$ and every open neighborhood $U$ of $x$, there is a path-connected open set $V$ such that $x\in V\subseteq U$. Equivalently, the path-connected open neighborhoods form a neighborhood base at each point.

Convex Balls Give Local Paths

A subset $C\subseteq\mathbb{R}^n$ is convex if $(1-t)p+tq\in C$ whenever $p,q\in C$ and $t\in I$. For a ball, the path given by this formula is the straight segment from $p$ to $q$.

Proposition: Every open ball in $\mathbb{R}^n$ is path-connected. Consequently, $\mathbb{R}^n$ is locally path-connected.

Proof. Let $B(c,r)=\{x\in\mathbb{R}^n:\|x-c\|<r\}$, where $r>0$, and take any $p,q\in B(c,r)$. Define $\gamma(t)=(1-t)p+tq$ for $t\in I$. Each coordinate of $\gamma$ is a linear function of $t$, so $\gamma$ is continuous, and $\gamma(0)=p$, $\gamma(1)=q$. For every $t\in I$, the triangle inequality and homogeneity of the Euclidean norm give $$ \|\gamma(t)-c\| =\|(1-t)(p-c)+t(q-c)\| \leq (1-t)\|p-c\|+t\|q-c\| < (1-t)r+tr=r. $$ The strict inequality holds because both $\|p-c\|<r$ and $\|q-c\|<r$, and the coefficients $1-t$ and $t$ are nonnegative and sum to $1$. Thus $\gamma(t)\in B(c,r)$ for every $t$; the ball is path-connected.

Now take $x\in\mathbb{R}^n$ and an open neighborhood $U$ of $x$. Since Euclidean open balls form a neighborhood base, some $r>0$ satisfies $B(x,r)\subseteq U$. The ball $B(x,r)$ is open and path-connected by the first part of the proof, so it is the required neighborhood. Hence $\mathbb{R}^n$ is locally path-connected. $\square$

Worked Example: A Path Inside a Ball

In $\mathbb{R}^3$, consider the open ball $B(c,5)$ with $c=(1,-2,0)$, and the points $p=(2,-2,1)$ and $q=(1,0,2)$. We verify that the straight-line path from $p$ to $q$ stays in the ball.

First calculate the distances from the center: $$ p-c=(1,0,1),\qquad \|p-c\|=\sqrt{1^2+0^2+1^2}=\sqrt{2}<5, $$ $$ q-c=(0,2,2),\qquad \|q-c\|=\sqrt{0^2+2^2+2^2}=\sqrt{8}=2\sqrt{2}<5. $$ Thus both endpoints belong to $B(c,5)$. The segment is $$ \gamma(t)=(1-t)p+tq=(2-t,-2+2t,1+t),\qquad 0\leq t\leq1. $$ Its displacement from the center is $\gamma(t)-c=(1-t,2t,1+t)$. The convexity estimate gives $$ \|\gamma(t)-c\| \leq (1-t)\sqrt{2}+t(2\sqrt{2}) =(1+t)\sqrt{2} \leq 2\sqrt{2}<5. $$ Therefore $\gamma(t)\in B(c,5)$ for all $t\in I$. At $t=0$, the formula gives $(2,-2,1)=p$; at $t=1$, it gives $(1,0,2)=q$, checking both endpoints.

The Connected Locally Path-Connected Theorem

Local path-connectedness by itself does not say that one path can reach every point in the entire space. Connectedness supplies the global condition. We use the following standard result, established in the earlier material on locally path-connected spaces.

Theorem (Connected locally path-connected theorem): Every connected, locally path-connected space is path-connected. The assertion includes the empty space under the convention that path-connectedness, which quantifies over pairs of points, holds vacuously when there are no points.

Here is the consequence for Euclidean open sets. The topology on a subset $G\subseteq\mathbb{R}^n$ is the subspace topology: its open sets are intersections of $G$ with open subsets of $\mathbb{R}^n$.

Theorem: If $G$ is a connected open subset of $\mathbb{R}^n$, then $G$ is path-connected.

Proof. If $G=\varnothing$, it is path-connected by the stated convention. Suppose $G\ne\varnothing$. We first verify local path-connectedness in the subspace topology. Let $x\in G$ and let $W$ be an open neighborhood of $x$ in $G$. By the definition of the subspace topology, $W=G\cap O$ for some open $O\subseteq\mathbb{R}^n$. Since $x\in W$ and both $G$ and $O$ are open in $\mathbb{R}^n$, their intersection $W$ is open in $\mathbb{R}^n$. Hence there is $r>0$ such that $B(x,r)\subseteq W$. This ball is open in $\mathbb{R}^n$, and it is also open in $G$, since $B(x,r)\subseteq G$ and $B(x,r)=G\cap B(x,r)$. The proposition shows that it is path-connected. Thus every open neighborhood of every point in $G$ contains a path-connected open neighborhood in $G$, so $G$ is locally path-connected. By hypothesis $G$ is connected. Applying the connected locally path-connected theorem shows that $G$ is path-connected. $\square$

Worked Example: An Open Rectangle

Let $G=(0,4)\times(-1,2)\subseteq\mathbb{R}^2$. It is open because it is the product of two open intervals. To see directly that it is connected, take any $p=(x_1,y_1)$ and $q=(x_2,y_2)$ in $G$. The segment $\gamma(t)=(1-t)p+tq$ has coordinates $(1-t)x_1+tx_2$ and $(1-t)y_1+ty_2$. Because $0<x_1,x_2<4$, each first coordinate remains strictly between $0$ and $4$; because $-1<y_1,y_2<2$, each second coordinate remains strictly between $-1$ and $2$. Thus the segment lies in $G$. This verifies path-connectedness directly, and path-connected spaces are connected.

For a numerical check, take $p=(1,0)$ and $q=(3,1)$. Then $\gamma(t)=(1+2t,t)$. For $0\leq t\leq1$, $1\leq1+2t\leq3$, so $0<1+2t<4$; also $0\leq t\leq1$, so $-1<t<2$. The path stays in $G$, with $\gamma(0)=(1,0)$ and $\gamma(1)=(3,1)$.

Worked Example: A Connected Open Set Built from Balls

In $\mathbb{R}^2$, let $$ G=B((0,0),2)\cup B((3,0),2)\cup B((6,0),2). $$ This is open as a union of open balls. The first and second balls intersect: $(1.5,0)$ is at distance $1.5<2$ from each center. The second and third intersect: $(4.5,0)$ is at distance $1.5<2$ from each of their centers. Each ball is path-connected by the straight-segment argument. Paths within the first and second balls can therefore be joined at their shared point, and paths within the second and third can be joined at theirs. It follows that any points in the union can be joined by paths through this chain of balls; hence $G$ is path-connected and therefore connected.

The theorem applies because $G$ is both open and connected. For a concrete path, choose $p=(0,1)\in B((0,0),2)$ and $q=(6,-1)\in B((6,0),2)$. Use the intermediate points $a=(1.5,0)$ and $b=(4.5,0)$. The segment from $p$ to $a$ lies in the first ball, since both endpoints have distance less than $2$ from its center and the ball is convex. The segment from $a$ to $b$ lies in the second ball: each endpoint is at distance $1.5<2$ from $(3,0)$. The segment from $b$ to $q$ lies in the third ball, since the distances of its endpoints from $(6,0)$ are $1.5$ and $\sqrt{2}$, respectively, and both are less than $2$.

These segments can be parametrized on successive thirds of $I$: use $p+3t(a-p)$ for $0\leq t\leq1/3$, $a+(3t-1)(b-a)$ for $1/3\leq t\leq2/3$, and $b+(3t-2)(q-b)$ for $2/3\leq t\leq1$. The formulas agree at the joining parameters: both adjacent formulas give $a$ at $t=1/3$, and both give $b$ at $t=2/3$. They define a continuous path from $p$ to $q$ in $G$, checking the theorem’s conclusion in this case.

What the Conclusion Does—and Does Not—Say

The theorem guarantees a path, not necessarily a straight segment. In particular, connected open subsets of Euclidean space need not be convex. Convexity would guarantee that every segment between two points stays in the set; connectedness and openness do not.

Worked Example: An Open Connected Set That Is Not Convex

Consider the open set $$ U=\{(x,y)\in\mathbb{R}^2:|x|<1\}\cup \{(x,y)\in\mathbb{R}^2:|y|<1\}. $$ Each strip is open, so their union is open. Each strip is convex: a convex combination of two first coordinates strictly between $-1$ and $1$ remains strictly between those bounds, and the same reasoning applies to the second coordinates. Both strips contain $(0,0)$. Given any two points of $U$, each lies in at least one strip; a segment within that strip joins it to $(0,0)$. Reversing one such segment and concatenating it with the other gives a path between the two points. Thus $U$ is path-connected and hence connected.

Nevertheless, take $p=(2,0)$ and $q=(0,2)$. Both belong to $U$: $p$ satisfies $|0|<1$, and $q$ satisfies $|0|<1$. Their midpoint is $(1,1)$, which satisfies neither $|1|<1$ nor $|1|<1$, so it is not in $U$. The segment from $p$ to $q$ therefore leaves $U$. A path that does stay in $U$ is the two-segment route from $(2,0)$ to $(0,0)$ and then to $(0,2)$: the first segment lies in the horizontal strip and the second in the vertical strip.

Common pitfall. The incorrect argument “$G$ is connected and open, so the segment between any two points lies in $G$” confuses path-connectedness with convexity. In the example, the midpoint calculation gives $(1,1)\notin U$, even though $U$ is open and connected. The correct conclusion is that some continuous path lies in $G$; straight-line paths are guaranteed when the set is convex, not merely when it is connected and open.

Openness matters because it supplies small Euclidean balls contained in the set, and those balls provide path-connected neighborhoods. Connectedness matters because it is the global hypothesis in the connected locally path-connected theorem. Together they give path-connectedness, without asserting that the entire set is convex or that every pair can be joined by a segment.

Check Your Understanding

For each question, distinguish the local role of openness from the global role of connectedness.

Why does every open neighborhood $W$ of $x$ in an open subset $G\subseteq\mathbb{R}^n$ contain a ball centered at $x$ that is also contained in $G$?
Write the straight-line path from $p$ to $q$ and explain why it stays inside $B(c,r)$ when both endpoints belong to that ball.
Which hypothesis in the connected locally path-connected theorem is verified by proving that $G$ is open in $\mathbb{R}^n$, and which is verified by assuming $G$ is connected?
In the cross-shaped set $U$, verify that $(1,1)\notin U$, while $(2,0)$ and $(0,2)$ both belong to $U$.
Does the theorem say that every connected open subset of $\mathbb{R}^n$ is convex? Explain the difference between the conclusion and convexity.