Fundamental Groups of Points and Discrete Spaces

Why Discrete Spaces Have No Nonconstant Loops

A loop begins and ends at the same point, but it could in principle travel through other points in between. In a discrete space, however, each point is an open set. Continuity then prevents a path from moving between distinct points: the interval cannot be divided into two nonempty open pieces that record different values of the path. We will prove this fact and use it to compute the fundamental group at any base point of a discrete space.

Throughout, \(I=[0,1]\) has the subspace topology inherited from \(\mathbb{R}\) with its standard topology. A space \(X\) is discrete if every subset of \(X\) is open; its topology is then \(\mathcal{P}(X)\). A map \(f:X\to Y\) is continuous if \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\).

These definitions require \(x_0\in X\). In particular, the empty space has no fundamental group at a base point because it has no base point. For a nonempty discrete space, we will show that every based loop is exactly the constant loop; there will therefore be only one equivalence class.

The Connectedness Argument

We first verify the connectedness fact needed for the loop argument. The proof uses the least-upper-bound property of \(\mathbb{R}\).

Proof. Suppose, to obtain a contradiction, that \(I=U\cup V\), where \(U\) and \(V\) are disjoint, nonempty, relatively open subsets of \(I\). Choose one point from each set. By exchanging the names \(U\) and \(V\) if necessary, there are \(u\in U\) and \(v\in V\) with \(u<v\). Let \(S=U\cap[u,v]\). This set is nonempty because \(u\in S\), and it is bounded above by \(v\). Let \(c=\sup S\), so \(u\le c\le v\). Since \(c\in I=U\cup V\), either \(c\in U\) or \(c\in V\). If \(c\in U\), then \(c\ne v\), because \(v\in V\) and \(U\cap V=\varnothing\). Relative openness of \(U\) gives an \(\varepsilon>0\) such that \((c-\varepsilon,c+\varepsilon)\cap I\subseteq U\). Choose \(d\) with \(c<d\le v\) and \(d<c+\varepsilon\); such a \(d\) exists because \(c<v\). Then \(d\in U\cap[u,v]=S\), contradicting that \(c\) is an upper bound of \(S\). If \(c\in V\), then \(c\ne u\), because \(u\in U\). Relative openness of \(V\) gives an \(\varepsilon>0\) such that \((c-\varepsilon,c+\varepsilon)\cap I\subseteq V\). Since \(c=\sup S\) and \(c>u\), there is \(s\in S\) with \(c-\varepsilon<s\le c\). The inequality \(s<c\) follows from \(s\in U\), \(c\in V\), and disjointness. Thus \(s\in V\) by the chosen neighborhood of \(c\), contradicting \(s\in S\subseteq U\). Both cases are impossible, so no such \(U,V\) exist. \(\square\)

Proof. Suppose \(f(C)\) is not connected. Then there are disjoint, nonempty sets \(P,Q\), open in the subspace \(f(C)\), with \(f(C)=P\cup Q\). By the definition of subspace topology, \(P=O\cap f(C)\) and \(Q=R\cap f(C)\) for open sets \(O,R\) of \(Y\). Continuity gives open sets \(f^{-1}(O)\) and \(f^{-1}(R)\) in \(C\). Their intersections with \(C\) are disjoint, and they cover \(C\): each point of \(C\) maps into exactly one of \(P,Q\). Both are nonempty because \(P,Q\) are nonempty subsets of \(f(C)\). This separates \(C\), contradicting its connectedness. Hence \(f(C)\) is connected. \(\square\)

Proof. Theorems 1 and 2 show that \(f(I)\) is connected. If \(f(I)\) contained two distinct points \(a\) and \(b\), then \(\{a\}\) and \(f(I)\setminus\{a\}\) would be disjoint, nonempty subsets covering \(f(I)\). Both are open in the subspace \(f(I)\), because every subset of the discrete space \(X\) is open. This would contradict connectedness of \(f(I)\). Thus \(f(I)\) has at most one point. Since \(I\) is nonempty, its image has at least one point, so \(f(I)\) consists of exactly one point and \(f\) is constant. \(\square\)

Computing the Fundamental Group

Proof. Let \(\gamma:I\to X\) be any based loop at \(x_0\). By the corollary, \(\gamma\) is constant. Its value must be \(x_0\), since \(\gamma(0)=x_0\). Thus the only based loop as a map is \(c_{x_0}\), and the set of based-loop equivalence classes contains exactly \([c_{x_0}]\). Concatenating this constant loop with itself gives the same constant loop: at every parameter value, both pieces have value \(x_0\). Therefore \([c_{x_0}][c_{x_0}]=[c_{x_0}]\), and this sole element is the identity. A group with one element is the trivial group, so \(\pi_1(X,x_0)=\{[c_{x_0}]\}\). \(\square\)

The result is independent of which point is chosen as base point: the proof applies to every \(x_0\in X\). This is a statement about the group at each base point separately; it does not require a map or a homotopy between different base points.

A Common Pitfall: A Jump Is Not a Continuous Path

It may seem that a path can move between two discrete points by staying at one point for a while and then switching to another. The switch itself prevents continuity. For a concrete test, give \(X=\{a,b\}\) the discrete topology and define \(\gamma:I\to X\) by \(\gamma(t)=a\) when \(0\le t<1/2\), and \(\gamma(t)=b\) when \(1/2\le t\le1\). This map starts at \(a\) and ends at \(b\), but is not continuous: the open set \(\{b\}\subseteq X\) has preimage \(\gamma^{-1}(\{b\})=[1/2,1]\), which is not open in \(I\). In particular, this piecewise-constant construction cannot be used to produce a nonconstant loop.

The relevant condition is continuity on the entire interval, not just the values at the endpoints or the fact that the formula is constant on each side of a switch. In a discrete target, a continuous map from \(I\) cannot have even one switch between different values. The connectedness argument rules out every possible nonconstant construction, including ones with many or infinitely many changes.