Connected Spaces

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We now examine connectedness, one of the fundamental concepts in topology.

Definition 1 A topological space \(X\) is called a connected space if it cannot be written as the union of two disjoint nonempty open sets. More generally, a subset \(A\) of \(X\) is connected if \(A\) with the subspace topology is connected.

In other words, a topological space is disconnected if and only if

Proposition 2 For a connected set \(A\subseteq X\), any set \(B\) satisfying \(A\subseteq B \subseteq \cl(A)\) is connected.

Proof

In the given situation,

\[\cl_B(A)=B\cap \cl_X(A)=B\]

so \(A\) is a dense subset of \(B\). (§Subspaces, ⁋Proposition 4) Now, toward a contradiction, suppose there exist two disjoint open sets \(U,V\) in \(B\) with \(U\cup V=B\). Since \(A\) is dense in \(B\), both \(U\cap A\) and \(V\cap A\) are nonempty and \(U\cap V\cap A=\emptyset\). This contradicts the assumption that \(A\) is connected.

The following proposition is also intuitively plausible.

Proposition 3 For a family \((A_i)\) of connected sets, if \(A_i\cap A_j\neq\emptyset\) for all \(i,j\), then \(A=\bigcup A_i\) is also connected.

Proof

Toward a contradiction, assume there exist two open sets \(U,V\) such that

\[A=(U\cap A)\cup (V\cap A),\qquad U\cap V\cap A=\emptyset\]

First, for each \(i\), since \(A_i\) is connected, exactly one of the two inclusions \(A_i\subseteq U\) or \(A_i\subseteq V\) must hold. On the other hand, if \(A_i\subseteq U\) and \(A_j\subseteq V\), then

\[A_i\cap A_j\subseteq (U\cap A)\cap (V\cap A)=U\cap V\cap A=\emptyset\]

which is a contradiction; thus all the \(A_i\) must be contained in \(U\), or all must be contained in \(V\). Hence either \(U\cap A=\emptyset\) or \(V\cap A=\emptyset\).

Properties of Connected Sets

Connectedness is preserved by continuous functions.

Proposition 4 For any continuous function \(f:X \rightarrow Y\) and any connected subset \(A\subseteq X\), the image \(f(A)\) is also connected.

Proof

Suppose toward a contradiction that \(f(A)\) is not connected, and choose open sets \(V_1,V_2\) in \(Y\) such that

\[f(A)=(V_1\cap f(A))\cup (V_2\cap f(A)), \qquad V_1\cap V_2\cap f(A)=\emptyset\]

Then \(f^{-1}(V_1)\) and \(f^{-1}(V_2)\) are open in \(X\), and

\[A=(A\cap f^{-1}(V_1))\cup (A\cap f^{-1}(V_2)),\qquad f^{-1}(V_1)\cap f^{-1}(V_2)\cap A=\emptyset\]

From the assumption that \(A\) is connected, we know that either \(V_1\cap f(A)=\emptyset\) or \(V_2\cap f(A)=\emptyset\).

From this the following corollary is immediate.

Corollary 5 The quotient space of a connected space is connected.

Also, the following holds.

Proposition 6 The product of connected spaces is connected. Conversely, if a product is connected then each factor is connected.

Proof

The reverse direction follows immediately from Proposition 4 applied to the projections \(\pr_i\), so there is nothing to prove.

Thus, assuming each \(X_i\) is connected, suppose toward a contradiction that \(X=\prod X_i\) is not connected. If \(X=U\cup V\) with \(U\cap V=\emptyset\) and \(U,V\neq\emptyset\), then the function \(f:X \rightarrow \{0,1\}\) defined by

\[f(x)=\begin{cases}1&\text{if $x\in U$}\\0&\text{if $x\in V$}\end{cases}\]

is continuous. (Here \(\{0,1\}\) is given the discrete topology.)

Now fix an element \(a=(a_i)\in X\), and define \(\iota_i: X_i \rightarrow X\) to be the map whose \(i\)th coordinate is \(x\) and whose remaining coordinates are taken from \(a\). Then \(f\circ\iota_i\) is a continuous function from \(X_i\) to \(\{0,1\}\), and since \(X_i\) is connected we know that \(f\circ\iota_i\) must be constant. Thus, by induction, points \(x\) in \(X\) all of whose coordinates except finitely many are equal to \(a\) must satisfy \(f(x)=f(a)\). Since such points form a dense subset of \(X\), \(f\) must be constant on all of \(X\), which is a contradiction.

Connected Components

Meanwhile, for a fixed \(x\in X\), the collection of connected sets containing \(x\) satisfies the hypothesis of Proposition 3, so the largest connected set containing \(x\) is well defined.

Definition 7 The connected component containing a point \(x\in X\) is the largest connected subset of \(X\) containing \(x\). If the connected component containing every point \(x\in X\) is always \(\{x\}\) itself, then \(X\) is called totally disconnected.

By definition, if \(X\) is connected then \(X\) has a unique connected component. More generally, any \(X\) can be written as a union of connected components

\[X=\bigcup_{i\in I} U_i\]

Meanwhile, by Proposition 2 each \(U_i\) must be closed. If \(I\) is finite, we know that the \(U_i\) must be simultaneously open and closed. Of course this does not apply to infinitely many connected components, but any clopen set in an arbitrary topological space must be a union of connected components. For if not, and some connected component \(C\) met a clopen set \(A\) while also meeting its complement, then \(C\cap A\) and \(C\setminus A\) would be two open sets partitioning \(C\).

Moreover, the following holds.

Proposition 8 Define an equivalence relation \(\sim\) on a topological space \(X\) by

\[x\sim y\iff \text{$x$ and $y$ lie in the same component}\]

Then \(X/{\sim}\) is totally disconnected.

The proof of this is immediate.

Locally Connected Spaces

Definition 9 A topological space \(X\) is locally connected at a point \(x\in X\) if, whenever a neighborhood \(U\) of \(x\) is given, there exists a connected neighborhood of \(x\) contained in \(U\). A space that is locally connected at every point is simply called a locally connected space.

Then the following holds.

Proposition 10 \(X\) is locally connected if and only if every component of every open set in \(X\) is open.

Path-Connected Spaces

Meanwhile, we