Connectedness and components

In one sentence

Connectedness in a general space is still the absence of a separation, components are always closed and need not be open, and the gap between components and path components is where the questions live.

Why the exam asks it

Because $\mathbb{Q}$ and the topologist's sine curve between them refute most plausible statements, and because "locally connected" is the hypothesis that makes components behave — and it is the one options omit.

The idea

$X$ is connected if it is not the union of two disjoint non-empty open sets. Equivalently, the only clopen subsets are $\emptyset$ and $X$; equivalently, every continuous map to the two-point discrete space is constant. That last formulation is often the quickest to apply.

What passes through continuous maps: images. Connectedness and path-connectedness are both preserved forwards and neither backwards.

Unions and closures: connected sets sharing a point have connected union; if $A$ is connected and $A \subseteq B \subseteq \overline A$ then $B$ is connected. Interiors are not preserved.

Products: an arbitrary product of connected spaces is connected in the product topology. Not so for the box: $\mathbb{R}^\omega$ with the box topology is disconnected, separated into the bounded and unbounded sequences.

Components

The connected component of $x$ is the largest connected set containing it. Components:

• partition $X$;
• are always closed, being closures of themselves;
• need not be open.

$\mathbb{Q}$ is the standard witness: every component is a single point, and points are not open in $\mathbb{Q}$. A space whose components are all singletons is totally disconnected — $\mathbb{Q}$, the irrationals, the Cantor set.

Components are open when $X$ is locally connected: every point has a neighbourhood basis of connected sets. $\mathbb{R}^n$ is locally connected, $\mathbb{Q}$ is not, and the topologist's sine curve is connected without being locally connected.

Path components

Path-connected $\Rightarrow$ connected, never the reverse. The topologist's sine curve — the graph of $\sin(1/x)$ on $(0,1]$ together with $\{0\}\times[-1,1]$ — is connected because it lies between a connected set and its closure, and not path-connected because no path reaches the segment.

So it has one component and two path components, which is exactly the distinction being tested. Path components refine components, always.

The two notions coincide for open subsets of $\mathbb{R}^n$, and more generally for locally path-connected spaces. That is why every set anyone draws is path-connected and the intuition needs correcting.

Quasi-components

The quasi-component of $x$ is the intersection of all clopen sets containing $x$. Components sit inside quasi-components, and they can differ — in a compact Hausdorff space they agree. This appears occasionally and is worth recognising rather than deriving.

Where intuition breaks

"Connected implies path-connected." The topologist's sine curve. True for open subsets of $\mathbb{R}^n$, which is what everyone pictures.

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

"A connected space is locally connected." The topologist's sine curve again — connected, not locally connected. The prefix is independent, in both directions: $\mathbb{Q}$ is neither, a discrete space is locally connected and not connected.

"The continuous image of a disconnected space is disconnected." Take any constant map. Only connectedness passes forward, and preimages inherit nothing.

"The intersection of connected sets is connected." Two arcs of a circle meeting in two points intersect in a disconnected set. Unions with a common point behave; intersections do not.

"The closure of a connected set may be disconnected." It is connected, always. The interior is the one that fails — two discs joined by a segment.

"An infinite product of connected spaces is disconnected." Connected in the product topology, disconnected in the box. Another instance of the same distinction.

The exam's angle

1. To prove disconnected, exhibit a clopen set or a non-constant map to $\{0,1\}$.
2. To prove connected, build from known pieces — connected sets sharing a point, or something squeezed between a connected set and its closure.
3. For components, ask whether the space is locally connected. If not, expect components that are not open and expect the answer to be $\mathbb{Q}$-like.
4. Distinguish components from path components whenever the sine curve could be in play.
5. Check the topology on a product — the box breaks connectedness where the product preserves it.

The night before

• Connected $=$ no separation $=$ only trivial clopen sets $=$ every map to $\{0,1\}$ is constant.
• Continuous images preserve connectedness and path-connectedness; preimages preserve neither.
• Connected sets with a common point: union connected. $A\subseteq B\subseteq\overline A$ keeps $B$ connected. Interiors do not.
• Components are closed, not always open. Open $\iff$ locally connected. $\mathbb{Q}$: components are points.
• Totally disconnected: $\mathbb{Q}$, irrationals, Cantor set.
• Path-connected $\Rightarrow$ connected. Topologist's sine curve: one component, two path components.
• The two agree on open subsets of $\mathbb{R}^n$ and on locally path-connected spaces.
• Products of connected spaces are connected in the product topology; the box version is disconnected.