Connectedness Is Preserved by Homeomorphisms
Connectedness describes whether a space can be split into two separated open pieces. A homeomorphism preserves the open sets in both directions, so it must preserve this property in both directions too. This makes connectedness a useful test: if one space is connected and another is not, there can be no homeomorphism between them.
We use the definition of connectedness in terms of a separation. For a space \(X\), all subsets below have the subspace topology unless another topology is specified. A set \(U\subseteq X\) is relatively open if \(U=X\cap W\) for some open set \(W\) in a space containing \(X\).
The topology on a space matters in this definition: the same underlying set can be connected with one topology and disconnected with another. A homeomorphism, unlike a bijection of underlying sets alone, matches the topologies precisely.
Proof. By definition, \(f\) is a homeomorphism if it is a bijection, is continuous, and has a continuous inverse \(f^{-1}:Y\to X\). First suppose \(X\) is connected. If \(Y\) were disconnected, it would have a separation \(U,V\). Continuity of \(f\) implies that \(f^{-1}(U)\) and \(f^{-1}(V)\) are open in \(X\). Since \(f\) is onto and \(U,V\) are nonempty, both preimages are nonempty. Since \(f\) is a function, the preimages are disjoint, and since \(U\cup V=Y\), they cover \(X\). They would therefore be a separation of \(X\), a contradiction. Thus \(Y\) is connected.
For the converse, suppose \(Y\) is connected. If \(X\) were disconnected, let \(A,B\) be a separation of \(X\). The sets \(f(A)\) and \(f(B)\) are open in \(Y\): for example, \(f(A)=(f^{-1})^{-1}(A)\), which is open because \(f^{-1}\) is continuous and \(A\) is open. Bijectivity gives that \(f(A)\) and \(f(B)\) are nonempty, disjoint, and have union \(Y\). They would separate \(Y\), a contradiction. Therefore \(X\) is connected as well, proving the equivalence. \(\square\)
The proof uses both directions of continuity. Continuity of \(f\) transfers a separation of \(Y\) back to \(X\); continuity of \(f^{-1}\) transfers a separation of \(X\) forward to \(Y\). Equivalently, a homeomorphism sends separations to separations and pulls separations back to separations.
Using the Invariant on Intervals and the Real Line
Worked Example: An Interval and a Punctured Interval
Compare \(X=[0,1]\) and \(Y=[0,1]\setminus\{1/2\}\), both with the subspace topology inherited from the standard topology on \(\mathbb{R}\). The interval theorem established earlier in the connectedness material says that every interval in \(\mathbb{R}\), including \([0,1]\), is connected.
To test \(Y\), take \(U=[0,1/2)\) and \(V=(1/2,1]\). These sets are disjoint and nonempty: \(0\in U\) and \(1\in V\). They cover \(Y\), because every point of \([0,1]\) other than \(1/2\) is either less than \(1/2\) or greater than \(1/2\). They are relatively open in \(Y\), since \(U=Y\cap(-1,1/2)\) and \(V=Y\cap(1/2,2)\), with both intervals on the right open in \(\mathbb{R}\). Thus \(U,V\) separate \(Y\), so \(Y\) is disconnected.
Conclusion and check: \(X\) is connected and \(Y\) is disconnected. The theorem rules out every homeomorphism from \(X\) to \(Y\). In the separation, \(1/2\) is excluded from \(Y\); including it in either part would not give this partition of \(Y\).
Worked Example: The Real Line and the Rationals
Give \(\mathbb{R}\) its standard topology and \(\mathbb{Q}\) the subspace topology inherited from \(\mathbb{R}\). The real line is connected, since it is an interval. We show directly that \(\mathbb{Q}\) is disconnected by cutting at the irrational number \(\sqrt{2}\).
First, \(1<\sqrt{2}<2\): squaring these positive numbers gives \(1<2<4\). Also, \(\sqrt{2}\) is irrational. Indeed, if \(\sqrt{2}=m/n\) for relatively prime positive integers \(m,n\), then \(m^2=2n^2\). Thus \(m^2\) is even, so \(m\) is even; writing \(m=2k\) gives \(n^2=2k^2\), so \(n\) is even as well. This contradicts the assumption that \(m,n\) are relatively prime.
Set \(A=\{q\in\mathbb{Q}:q<\sqrt{2}\}\) and \(B=\{q\in\mathbb{Q}:q>\sqrt{2}\}\). Both are nonempty: \(1\in A\) and \(2\in B\). They are disjoint. Since \(\sqrt{2}\notin\mathbb{Q}\), every rational belongs to exactly one of \(A,B\), so \(A\cup B=\mathbb{Q}\). They are relatively open: \(A=\mathbb{Q}\cap(-\infty,\sqrt{2})\) and \(B=\mathbb{Q}\cap(\sqrt{2},\infty)\), and the two rays are open in \(\mathbb{R}\). Hence \(A,B\) form a separation of \(\mathbb{Q}\).
Conclusion and check: \(\mathbb{Q}\) is disconnected while \(\mathbb{R}\) is connected. Therefore no homeomorphism \(\mathbb{Q}\to\mathbb{R}\) exists. The irrational cut is important: if the cut point were rational, that rational would belong to neither strict inequality set, and the two sets would fail to cover \(\mathbb{Q}\).
A Circle and an Explicit Separation
Worked Example: The Circle and a Circle with Two Points Removed
Let \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\), with the subspace topology from the standard Euclidean topology on \(\mathbb{R}^2\). Compare \(S^1\) with \(Y=S^1\setminus\{(1,0),(-1,0)\}\), also given the subspace topology.
The circle \(S^1\) is connected. To see this, the map \(f:[0,1]\to S^1\) defined by \(f(t)=(\cos(2\pi t),\sin(2\pi t))\) is continuous and onto. It is onto because every point of the unit circle has an angle \(\theta\in[0,2\pi]\), and is \(f(\theta/(2\pi))\). The interval \([0,1]\) is connected, and the earlier result that continuous images of connected spaces are connected therefore gives that \(S^1\) is connected.
For \(Y\), define \(U=\{(x,y)\in Y:y>0\}\) and \(V=\{(x,y)\in Y:y<0\}\). These sets are disjoint and relatively open: they are intersections of \(Y\) with the open upper and lower half-planes in \(\mathbb{R}^2\). Each is nonempty, since \((0,1)\in U\) and \((0,-1)\in V\). To check that they cover \(Y\), a point of \(S^1\) with \(y=0\) must satisfy \(x^2=1\), so it is either \((1,0)\) or \((-1,0)\); both points have been removed. Every point of \(Y\) therefore has either \(y>0\) or \(y<0\), and belongs to exactly one of \(U,V\).
Conclusion and check: \(U,V\) separate \(Y\), so \(Y\) is disconnected, whereas \(S^1\) is connected. No homeomorphism between them exists. The removed points are precisely the points on \(S^1\) with \(y=0\), which is why the strict upper and lower half-planes cover \(Y\).
A Common Pitfall: A Continuous Bijection Need Not Be a Homeomorphism
Connectedness is invariant under homeomorphisms, not under arbitrary bijections, and not even under all continuous bijections. The inverse-continuity hypothesis is essential. The following finite example checks each property directly.
Worked Example: The Same Two Points with Different Topologies
Let \(X=\{a,b\}\) have the discrete topology \(\mathcal{P}(X)=\{\varnothing,\{a\},\{b\},X\}\), and let \(Y=\{a,b\}\) have the indiscrete topology \(\{\varnothing,Y\}\). Consider the identity bijection \(f:X\to Y\).
The space \(X\) is disconnected: \(\{a\}\) and \(\{b\}\) are disjoint, nonempty, open, and have union \(X\). The space \(Y\) is connected: its only open sets are \(\varnothing\) and \(Y\), so there cannot be two disjoint, nonempty open sets whose union is \(Y\).
The map \(f\) is continuous because the preimages of the only open sets of \(Y\) are \(f^{-1}(\varnothing)=\varnothing\) and \(f^{-1}(Y)=X\), both open in \(X\). But the inverse is not continuous: the open set \(\{a\}\) in \(X\) has inverse image \(\{a\}\) under \(f^{-1}:Y\to X\), and \(\{a\}\) is not open in \(Y\).
Common wrong conclusion: “There is a continuous bijection between \(X\) and \(Y\), so they must have the same connectedness.” The verified separation of \(X\) and the absence of any separation of \(Y\) show that this conclusion is false. The correction is to check that the bijection has a continuous inverse before applying the homeomorphism-invariance theorem.
What the Invariant Can and Cannot Decide
Connectedness is a topological invariant: it has the same truth value on any two homeomorphic spaces. In practice, the method is to determine connectedness on each side, using a known connected interval or an explicit separation, and then compare the results. If the truth values differ, the proposed homeomorphism is impossible. This conclusion rules out every possible homeomorphism; it does not depend on guessing a particular formula for one.
The comparison is a necessary test, not a complete classification. If two spaces are both connected, this invariant alone does not decide whether they are homeomorphic. Likewise, if both are disconnected, connectedness alone does not distinguish them. The test is decisive precisely when the connectedness properties disagree.
| Space or comparison | Connectedness information | Conclusion from the invariant |
|---|---|---|
| \([0,1]\) and \([0,1]\setminus\{1/2\}\) | Connected versus disconnected | No homeomorphism exists |
| \(\mathbb{R}\) and \(\mathbb{Q}\) | Connected versus disconnected | No homeomorphism exists |
| \(S^1\) and \(S^1\setminus\{(1,0),(-1,0)\}\) | Connected versus disconnected | No homeomorphism exists |
| A continuous bijection alone | May relate spaces with different connectedness | The homeomorphism theorem does not apply |
Check Your Understanding
For each question, identify the topology being used and justify any claim of connectedness or disconnection.
- Explain why a separation of \(Y\) pulls back under a homeomorphism \(f:X\to Y\) to a separation of \(X\).
- In the subspace topology on \(\mathbb{R}\setminus\{0\}\), do the sets \((-\infty,0)\) and \((0,\infty)\) form a separation? What does this imply when comparing this space with \(\mathbb{R}\)?
- Why does cutting \(\mathbb{Q}\) at an irrational number ensure that the two strict sides cover \(\mathbb{Q}\)?
- In the circle example, why would the upper and lower half-circles fail to cover \(S^1\) if neither point with \(y=0\) were removed?
- For the finite identity map, which continuity condition fails, and why is that condition needed for connectedness to be invariant in both directions?