Components as Topological Structure
Connectedness can distinguish spaces even when they have the same number of points or look similar as subsets of another space. A sharper version of this idea counts the maximal connected pieces of a space. A homeomorphism cannot merge two such pieces or split one into smaller pieces: it transports the pieces bijectively. We will prove this correspondence and use it to test whether concrete spaces can be homeomorphic.
For a subset \(A\subseteq X\), use the subspace topology: its open sets are the sets \(A\cap U\), where \(U\) is open in \(X\). A space is connected if it cannot be written as the union of two disjoint, nonempty open subsets. A subset is connected when it is connected with its subspace topology. A homeomorphism is a bijection that is continuous and whose inverse is continuous.
“Maximal” means that the component is not properly contained in any connected subset of \(X\); it does not mean that it is the only connected subset containing its points. The next result makes the component through a point concrete.
Proof. First note that if a family of connected subsets all contains \(x\), then its union is connected. To see this, suppose the union had a separation into disjoint, nonempty, relatively open sets \(P\) and \(Q\). The point \(x\) belongs to one of them; suppose \(x\in P\). Each connected set in the family meets \(P\), because it contains \(x\). If it also met \(Q\), its intersections with \(P\) and \(Q\) would separate that connected set. Thus every set in the family is contained in \(P\), contradicting that \(Q\) is nonempty in the union.
Now take the union \(C_x\) of all connected subsets containing \(x\). The family is nonempty because \(\{x\}\) is connected: a one-point space has no two disjoint nonempty subsets. The union argument shows that \(C_x\) is connected. If a connected set contains \(C_x\), it contains \(x\), so it was already among the sets used in forming \(C_x\); hence it is contained in \(C_x\). Therefore \(C_x\) is a component. Any component containing \(x\) is one of the connected sets in the union and is contained in \(C_x\); maximality then makes it equal to \(C_x\). This proves existence and uniqueness. \(\square\)
In particular, components partition \(X\). Indeed, if two components meet at a point, their union is connected by the same union argument, so maximality forces the two components to be equal. Every point belongs to a component by the proposition.
Why Homeomorphisms Preserve Components
Proof. Suppose \(f(A)\) had a separation \(P,Q\). Then \(f^{-1}(P)\cap A\) and \(f^{-1}(Q)\cap A\) are disjoint, relatively open subsets of \(A\), since \(f\) is continuous. They are nonempty because \(P,Q\subseteq f(A)\). They cover \(A\), since every \(a\in A\) has \(f(a)\in P\) or \(f(a)\in Q\). This would separate \(A\), contrary to its connectedness. Thus \(f(A)\) is connected. \(\square\)
Proof. Let \(C\) be a component of \(X\). By the lemma, \(h(C)\) is connected. Suppose it were properly contained in a connected subset \(D\subseteq Y\). The inverse \(h^{-1}\) is continuous, so the lemma applied to \(D\) shows that \(h^{-1}(D)\) is connected in \(X\). It contains \(C\), since \(h(C)\subseteq D\), and it properly contains \(C\), since \(D\) properly contains \(h(C)\) and \(h\) is a bijection. This contradicts maximality of \(C\). Therefore \(h(C)\) is a component of \(Y\).
Applying the same argument to \(h^{-1}\) shows that every component of \(Y\) is the image of a component of \(X\). If \(h(C_1)=h(C_2)\), injectivity of \(h\) gives \(C_1=C_2\), so the assignment is a bijection. A restriction of a continuous map is continuous, and the inverse of \(h|_C\) is \(h^{-1}|_{h(C)}\), which is continuous. Thus the restriction is a homeomorphism. Taking cardinalities of the component collections gives \(c(X)=c(Y)\), including the empty-space case. \(\square\)
The theorem preserves more than a number: it gives a particular correspondence between the components, and each corresponding pair is homeomorphic. The count is nevertheless a quick obstruction. If two spaces have different numbers of components, no homeomorphism between them can exist. Equal counts do not, by themselves, prove that spaces are homeomorphic.
Worked Component Counts
Worked Example: One Interval Versus Two Separated Intervals
Give each subset below the subspace topology inherited from \(\mathbb{R}\) with its standard topology. Compare \(X=[0,1]\) and \(Y=[0,1]\cup[2,3]\).
We use the standard interval connectedness fact, with its short justification. Suppose \([a,b]\), where \(a<b\), had a separation into nonempty relatively open sets \(E,F\). Choose points from different sets and, if necessary, exchange their names so that \(x\in E\), \(y\in F\), and \(x<y\). The set \(E\cap[x,y]\) is nonempty and bounded above, so let \(s=\sup(E\cap[x,y])\). Since \(E\) is also relatively closed in \([a,b]\), \(s\in E\): points of \(E\cap[x,y]\) can be chosen arbitrarily close to \(s\), and a relatively closed set contains their limit. Since \(y\in F\), \(s\ne y\), so \(s<y\). Relative openness of \(E\) gives points of \(E\cap[x,y]\) strictly greater than \(s\), contradicting that \(s\) is an upper bound. Thus \([a,b]\) is connected; a one-point interval is connected directly.
It follows that \(X\) is connected, so its only component is \(X\), and \(c(X)=1\). For \(Y\), put \(A=[0,1]\) and \(B=[2,3]\). These are disjoint, nonempty, and cover \(Y\). They are open in \(Y\), since \(A=Y\cap(-1,1.5)\) and \(B=Y\cap(1.5,4)\). Each is also closed in \(Y\), since its complement in \(Y\) is the other open set. This separation shows that \(Y\) is disconnected. More precisely, any connected subset of \(Y\) meeting both \(A\) and \(B\) would be separated by its intersections with \(A\) and \(B\). Since \(A\) and \(B\) are themselves connected intervals, they are exactly the components. Hence \(c(Y)=2\).
Conclusion: \(X\) and \(Y\) are not homeomorphic, because their component counts are \(1\) and \(2\), respectively. This conclusion does not depend on their lengths or endpoints; it follows from the component correspondence theorem.
Worked Example: The Same Set with Two Different Topologies
Let \(S=\{a,b,c\}\). Give \(S\) first the discrete topology, in which every subset is open, and then the indiscrete topology, whose only open sets are \(\varnothing\) and \(S\). Find the components in each case.
In the discrete topology, each singleton is open. If a subset contains at least two points, choose one point and separate its singleton from the rest of the subset; both parts are nonempty and relatively open. So no such subset is connected. Each singleton is connected, and the components are \(\{a\},\{b\},\{c\}\). Thus the discrete space has \(3\) components.
In the indiscrete topology, a subspace has only the open sets \(\varnothing\) and itself: intersecting an open set of \(S\) with a subset \(A\) gives only \(\varnothing\) or \(A\). A nonempty indiscrete space cannot be separated into two nonempty open sets. Therefore \(S\) is connected and is its only component, so it has \(1\) component.
Check: the underlying set is \(S\) in both cases, but the component counts differ. The two topological spaces cannot be homeomorphic.
Worked Example: A Homeomorphism Matching Two Components
Let \(X=[0,1]\cup[2,3]\) and \(Y=[0,2]\cup[4,6]\), both with the subspace topology from the standard real line. Define \(h:X\to Y\) by \(h(x)=2x\) for \(x\in[0,1]\) and \(h(x)=2x\) for \(x\in[2,3]\).
On the first interval, the values run from \(2\cdot0=0\) to \(2\cdot1=2\), so this piece maps onto \([0,2]\). On the second, they run from \(2\cdot2=4\) to \(2\cdot3=6\), so this piece maps onto \([4,6]\). The two image intervals are disjoint, and multiplication by \(2\) is one-to-one on each; hence \(h\) is a bijection from \(X\) to \(Y\). Its inverse is \(h^{-1}(y)=y/2\) on each of \([0,2]\) and \([4,6]\).
To verify continuity on the whole space, note that the two component intervals of \(X\) are open in \(X\), as shown in the previous example, and their union is \(X\). For any open \(O\subseteq Y\), the preimage on each interval is relatively open there because the formula \(x\mapsto2x\) is continuous. Each interval is open in \(X\), so these two relatively open preimages are open in \(X\); their union is \(h^{-1}(O)\). Thus \(h\) is continuous. The same argument, using the two open component intervals of \(Y\) and the continuous formula \(y\mapsto y/2\), proves that \(h^{-1}\) is continuous. Therefore \(h\) is a homeomorphism.
Component check: \(X\) has the two components \([0,1]\) and \([2,3]\); \(Y\) has \([0,2]\) and \([4,6]\). The formulas give \(h([0,1])=[0,2]\) and \(h([2,3])=[4,6]\), exactly the component correspondence required by the theorem.
A Pitfall: A Bijection Is Not Enough
A common mistake is to treat any bijection between spaces as if it preserved their components. The theorem requires a homeomorphism, not merely a bijection, and even continuity alone does not suffice.
Wrong claim: “The identity map between the discrete and indiscrete topologies on \(\{a,b\}\) preserves components because it is a bijection.” In fact, let the domain have the discrete topology and the codomain the indiscrete topology. The identity map is continuous: the only open sets in the codomain are \(\varnothing\) and \(\{a,b\}\), whose preimages are open. It is bijective, but the domain has two singleton components while the codomain is connected and has one component. Its inverse is not continuous, because the preimage of the open set \(\{a\}\) in the discrete domain is \(\{a\}\), which is not open in the indiscrete codomain. Thus this bijection is not a homeomorphism, and the component theorem does not apply.
Component counts are a useful first test, not a complete classification. Different counts prove that no homeomorphism exists; matching counts leave open whether the components themselves have the required topological structure and whether the spaces are homeomorphic.
Check Your Understanding
For each question, use the definition of a component and distinguish a homeomorphism from a continuous bijection.
- How many components does \([0,2]\cup[4,5]\subseteq\mathbb{R}\) have with the subspace topology? Give a separation that verifies the count.
- Explain why a homeomorphism must send a component onto a component, rather than merely into one.
- Can a space with exactly two components be homeomorphic to a connected space? State the invariant that decides.
- In the discrete topology on a finite set \(S\), identify all the components. What changes if \(S\) has the indiscrete topology and is nonempty?
- Why does the identity map from a discrete two-point space to an indiscrete two-point space not contradict component invariance under homeomorphisms?