Connectedness in Topological Spaces

Connectedness Beyond Metric Spaces

Connectedness asks whether a space can be divided into two nonempty pieces that are topologically separated. In metric spaces, this idea was developed using separated sets and their consequences. The same central question applies to every topological space, whether or not it comes from a metric. The definitions below use only open sets, so they apply directly to the spaces introduced in the topology tutorials.

The topology matters: the same underlying set can be connected under one topology and disconnected under another. Once connectedness is defined in terms of open sets, it also becomes possible to prove preservation results without referring to distances. In particular, continuous maps preserve connectedness, and even products with arbitrarily many connected factors are connected.

The open sets in a separation are open in \(X\). For a subset \(A\subseteq X\), connectedness of \(A\) means connectedness in the subspace topology: the sets in a separation of \(A\) must be open relative to \(A\). They need not be open in the whole space \(X\).

The Clopen Test

A subset is called clopen if it is both open and closed. This gives a useful test that does not require us to search for two open sets independently. The clopen characterization was established for metric spaces in Connectedness Proof Workshop; the same argument works for arbitrary topological spaces.

Proof. Suppose \(A\subseteq X\) is clopen and \(A\ne\varnothing,X\). Then \(A\) and \(X\setminus A\) are nonempty, disjoint, and open, and their union is \(X\). They form a separation, so \(X\) is disconnected.

Conversely, suppose \(X\) is disconnected, with a separation \(U,V\). Since \(X=U\cup V\) and the sets are disjoint, \(X\setminus U=V\). Both \(U\) and \(V\) are open, so \(U\) is also closed. It is nonempty and is not all of \(X\). Thus \(X\) has a clopen subset other than \(\varnothing\) and \(X\). These two implications prove the characterization. \(\square\)

Continuous Images Preserve Connectedness

Continuous maps carry open-set information back to their domains by taking preimages. This is exactly what is needed to show that a separation of an image would produce a separation of the original space. The metric-space version of this result appeared in Connectedness in Metric Spaces; the proof here uses only the topological definition of continuity.

Proof. Suppose, to the contrary, that \(f[X]\) has a separation \(P,Q\). Thus \(P\) and \(Q\) are disjoint, nonempty, relatively open subsets of \(f[X]\), and \(f[X]=P\cup Q\). The map \(f\), regarded as a map from \(X\) to \(f[X]\), is continuous: for any relatively open \(W\subseteq f[X]\), there is an open set \(O\subseteq Y\) with \(W=f[X]\cap O\), and hence \(f^{-1}(W)=f^{-1}(O)\), which is open in \(X\).

It follows that \(f^{-1}(P)\) and \(f^{-1}(Q)\) are open in \(X\). They are disjoint and cover \(X\). They are both nonempty because \(P,Q\subseteq f[X]\): each contains some value of \(f\), and that value has a preimage in \(X\). Thus they form a separation of \(X\), contradicting connectedness. Therefore \(f[X]\) is connected. \(\square\)

Connectedness of Products

For two connected spaces, the product is connected. The proof illustrates how slices of a product can transfer connectedness from the factors to the whole space. It also provides the finite-product step needed to treat products with any index set.

Proof. If \(I=\varnothing\), the product consists of one point and is connected. Assume \(I\ne\varnothing\). Choose a base point \(x^0=(x_i^0)_{i\in I}\in X\); the choice of one point from each nonempty factor is understood. We first establish the connectedness of finite products.

Let \(X_1,X_2\) be connected and nonempty. Suppose \(C\) is a clopen subset of \(X_1\times X_2\). Fix \(y_0\in X_2\). The map \(x\mapsto(x,y_0)\) is continuous, so the inverse image of \(C\) is clopen in \(X_1\). By connectedness of \(X_1\), this inverse image is either empty or all of \(X_1\).

If it is all of \(X_1\), then \(C\) contains \(X_1\times\{y_0\}\). For each \(x\in X_1\), the vertical slice \(\{x\}\times X_2\) is connected, since it is homeomorphic to \(X_2\). That slice meets \(C\), and its intersections with \(C\) and with the complement of \(C\) are clopen relative to the slice. Connectedness forces the entire slice to lie in \(C\). This holds for every \(x\), so \(C=X_1\times X_2\).

If the inverse image is empty, each vertical slice meets the complement of \(C\) at \((x,y_0)\). The same connectedness argument shows that every vertical slice lies entirely in the complement, so \(C=\varnothing\). The clopen characterization now proves that \(X_1\times X_2\) is connected. Induction gives connectedness of every nonempty finite product of the factors.

Consider the subset \(D\subseteq X\) consisting of points that differ from \(x^0\) in only finitely many coordinates. For each finite \(F\subseteq I\), let \(X_F\) be the set of points whose coordinates outside \(F\) equal those of \(x^0\). The space \(X_F\) is homeomorphic to the finite product \(\prod_{i\in F}X_i\), so it is connected; when \(F=\varnothing\), it is the singleton \(\{x^0\}\). Also, every \(X_F\) contains \(x^0\), and \(D=\bigcup_{F\subseteq I,\ F\text{ finite}}X_F\).

To verify that \(D\) is connected, let \(A\) be clopen in \(D\). Its intersection with each \(X_F\) is clopen relative to \(X_F\), so it is either empty or all of \(X_F\). Since every \(X_F\) contains \(x^0\), if \(x^0\in A\), then every \(X_F\) lies in \(A\), giving \(A=D\). If \(x^0\notin A\), every \(X_F\) is disjoint from \(A\), giving \(A=\varnothing\). The clopen characterization shows that \(D\) is connected.

The set \(D\) is dense in \(X\). Indeed, a nonempty basic open set in the product topology restricts coordinates in only a finite set \(F\). Choose a point of that basic open set; the point that agrees with it on \(F\) and agrees with \(x^0\) outside \(F\) belongs both to the basic open set and to \(D\). Thus every nonempty open subset of \(X\) meets \(D\).

Finally, if \(X\) were disconnected, the clopen characterization would give a nonempty proper clopen subset \(U\subset X\). Since \(D\) is dense and \(U\) is nonempty and open, \(U\cap D\ne\varnothing\). This intersection is clopen in \(D\), so connectedness of \(D\) implies \(D\subseteq U\). But \(X\setminus U\) is also nonempty and open, so density of \(D\) requires it to meet \(D\), a contradiction. Hence \(X\) is connected. \(\square\)

How to Use Connectedness Carefully

Connectedness is determined by the open sets of the space in question. When considering a subset \(A\subseteq X\), test for a separation using sets open in \(A\), not necessarily sets open in \(X\). Likewise, the continuous-image theorem concludes that \(f[X]\) is connected with its subspace topology; it does not claim that the image is open or closed in the codomain.

The theorem does not reverse: a continuous image being connected does not imply that its domain is connected. For example, the constant map from a two-point discrete space to a one-point space is continuous, and its image is connected, although its domain is disconnected. Nor does connectedness of the underlying set mean the topology is irrelevant: the two-point examples above have different connectedness properties under different topologies.

The clopen test is particularly useful when a space has an algebraic or product description. A proposed separation immediately gives a nontrivial clopen set, while a proof that all clopen sets are trivial rules out every possible separation at once. The continuous-image theorem and the product theorem then let connectedness be carried to new spaces through maps and constructions.