Connectedness and path-connectedness

In one sentence

Path-connected implies connected and not the other way round, the connected subsets of $\mathbb{R}$ are exactly the intervals, and one curve answers most of the questions in this topic.

Why the exam asks it

Because connectedness is defined negatively — by the absence of a separation — which makes it awkward to verify directly, and because the one implication that fails has a single famous counterexample. A candidate who knows the topologist's sine curve properly can answer most of what is asked here; one who knows only the phrase cannot.

The idea

$X$ is connected if it cannot be written as $A \cup B$ with $A, B$ non-empty, disjoint and both open in $X$. Such a pair is a separation.

$X$ is path-connected if any two points are joined by a continuous $\gamma : [0,1] \to X$.

Path-connected $\Rightarrow$ connected. The converse is false.

The connected subsets of $\mathbb{R}$ are the intervals

Exactly the intervals — including degenerate ones. If $A \subseteq \mathbb{R}$ is not an interval there are $a < c < b$ with $a, b \in A$ and $c \notin A$, and then $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ separate it. That is the whole proof, and it is worth being able to reproduce, because it is the reason the intermediate value theorem is a statement about connectedness.

What passes through continuous maps

• The continuous image of a connected set is connected. This is the intermediate value theorem: $f([a,b])$ is connected in $\mathbb{R}$, hence an interval, hence contains everything between $f(a)$ and $f(b)$.
• The continuous image of a path-connected set is path-connected.

Neither passes backwards to preimages.

Unions and closures

• If $A_i$ are connected and share a common point, $\bigcup A_i$ is connected.
• If $A$ is connected and $A \subseteq B \subseteq \overline{A}$, then $B$ is connected. In particular $\overline{A}$ is connected.
• The interior of a connected set need not be connected.

Two closed discs in the plane joined by a segment form a connected set whose interior is two disjoint open discs. The bar connects, and contributes nothing to the interior.

The topologist's sine curve

$$S = \left\{\left(x, \sin\tfrac1x\right) : 0 < x \le 1\right\} \cup \left(\{0\}\times[-1,1]\right).$$

It is connected. The graph part $G$ is a continuous image of the interval $(0,1]$, so it is connected. And $S \subseteq \overline{G}$: as $x \to 0^+$, $\sin(1/x)$ runs through every value in $[-1,1]$ infinitely often, so every point of the segment is a limit of points of the graph. A set squeezed between a connected set and its closure is connected.

It is not path-connected. No continuous path can run from a point of the segment to a point of the graph. Informally: to arrive at the segment the path must cross every oscillation, and there are infinitely many of them in any neighbourhood of the axis, so the path's second coordinate cannot settle down — it would have to oscillate between $-1$ and $1$ on every interval right up to the arrival time, contradicting continuity at that time.

This is the standard example, and it is the answer whenever an option asserts that connected implies path-connected.

Components

A component is a maximal connected subset. Components are always closed. They need not be open: in $\mathbb{Q}$ every component is a single point, and singletons are not open in $\mathbb{Q}$. A space whose components are all singletons is totally disconnected — $\mathbb{Q}$, the irrationals, and the Cantor set are the standard examples.

Path components can be smaller than components, and the topologist's sine curve shows it: one component, two path components.

The one place the converse does hold

An open connected subset of $\mathbb{R}^n$ is path-connected — in fact polygonally connected.

This is a true and useful statement, and it is exactly why the counterexample has to be an odd set: any open connected subset of the plane is fine. It also explains where the sine curve gets its strangeness, since $S$ is not open.

$\mathbb{R}$ is not $\mathbb{R}^{2}$

Removing a point from $\mathbb{R}$ disconnects it; removing a point from $\mathbb{R}^n$ for $n \ge 2$ does not. Since a homeomorphism would carry one situation to the other, $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^n$ for $n \ge 2$. This "remove a point and count the pieces" argument is the standard way these questions are answered.

Where intuition breaks

"Connected, so path-connected." The topologist's sine curve. Tempting because every connected set anyone draws is path-connected — and it is a theorem for open subsets of $\mathbb{R}^n$, which is what one usually draws.

"Components are open." In $\mathbb{Q}$ they are single points, which are closed and not open. Components are always closed; openness needs a hypothesis such as local connectedness.

"$A \cup B$ is connected and $A \cap B \ne \emptyset$, so $A$ and $B$ are connected." Backwards. Take $A = [0,1] \cup [2,3]$ and $B = [0,3]$: the union is $[0,3]$, connected, and $A$ is not. The true statement runs the other way — connected pieces with a common point have a connected union.

"The interior of a connected set is connected." The dumbbell. The closure of a connected set is connected; the interior is the one that fails, and the two get swapped in options.

"A connected subset of $\mathbb{R}$ containing $a$ and $b$ contains $[a,b]$." True, and worth recognising among false options — it is exactly the characterisation of intervals.

"The continuous preimage of a connected set is connected." $f(x) = x^2$ and the connected set $\{1\}$: the preimage is $\{-1, 1\}$. Forward only.

"$\mathbb{R}^2$ minus a countable set is disconnected." It is connected, and in fact path-connected: between any two points there are uncountably many distinct polygonal paths through a third point, and only countably many can be blocked. A pleasing argument and a genuine exam question.

The exam's angle

1. If an option says connected implies path-connected, it is false, unless the set is open in $\mathbb{R}^n$ — in which case it is true. Check openness before answering.
2. In $\mathbb{R}$, connected means interval. Most questions restricted to the line collapse to that.
3. Use "remove a point" to separate spaces. It distinguishes $\mathbb{R}$ from $\mathbb{R}^n$, an interval from a circle, and a letter from another letter in those homeomorphism questions.
4. Watch the direction of every statement about continuous maps. Images inherit connectedness and path-connectedness; preimages inherit nothing.
5. Closure preserves connectedness; interior does not. Keep the dumbbell in mind for the second.

The night before

• Path-connected $\Rightarrow$ connected. Converse false: topologist's sine curve.
• $S$ = graph of $\sin(1/x)$ on $(0,1]$ plus $\{0\}\times[-1,1]$: connected, not path-connected. One component, two path components.
• Connected subsets of $\mathbb{R}$ = intervals, exactly.
• Continuous image of connected is connected — that is the IVT. Preimages inherit nothing.
• Connected sets with a common point: union connected. $A \subseteq B \subseteq \overline A$ with $A$ connected $\Rightarrow$ $B$ connected.
• Closure of connected is connected; interior need not be — two discs joined by a bar.
• Components are closed, not always open. $\mathbb{Q}$: components are points, totally disconnected.
• Open and connected in $\mathbb{R}^n$ $\Rightarrow$ path-connected, even polygonally.
• Remove a point: $\mathbb{R}$ splits, $\mathbb{R}^n$ ($n \ge 2$) does not. So they are not homeomorphic.