What Connectedness Means
Compactness asks whether open covers have finite subcovers. Connectedness asks a different question: can a space be split into two nonempty pieces that are open within the space and do not overlap? If it cannot, the space is connected. This idea applies both to an entire metric space and to a subset equipped with the restricted metric.
The phrase “open within the space” matters. A subset can be open relative to a metric space even when it is not open in a larger ambient space. We will use relative openness to define connectedness, prove that continuous maps preserve it, and show that the connected subsets of the real line are exactly its intervals.
In a separation, each side is also relatively closed, since it is the complement of the other side. Thus connectedness can equivalently be expressed by saying that the only subsets that are both open and closed in \(X\) are \(\varnothing\) and \(X\). A singleton is connected, and the empty space is connected under this definition: neither can be partitioned into two nonempty sets.
Worked Example: A Discrete Metric Space with More Than One Point
Let \(X\) contain at least two points and give it the discrete metric \(d(x,y)=1\) when \(x\ne y\), and \(d(x,x)=0\). For any \(x\in X\), the open ball \(B_{1/2}(x)\) is \(\{x\}\): the distance from \(x\) to itself is \(0<1/2\), while the distance from \(x\) to every other point is \(1\), which is not less than \(1/2\). Therefore \(\{x\}\) is open. Its complement is open as well, because it is a union of open singletons.
Choose \(x\in X\). Both \(\{x\}\) and \(X\setminus\{x\}\) are nonempty, disjoint, open, and have union \(X\). Hence \(X\) is disconnected. This example shows that having a metric does not by itself prevent a space from being split into open pieces.
Continuous Maps Preserve Connectedness
A continuous map can change distances and shape, but it cannot turn a connected domain into an image that splits into two relatively open pieces. The reason is that a proposed split of the image would pull back to a split of the domain.
Proof. Suppose, to the contrary, that \(f[X]\) is disconnected. Then there are disjoint, nonempty sets \(A,B\) that are open relative to \(f[X]\), with \(f[X]=A\cup B\). Continuity of \(f\), viewed as a map from \(X\) into the subspace \(f[X]\), implies that \(f^{-1}(A)\) and \(f^{-1}(B)\) are open in \(X\). They are disjoint, and
Both preimages are nonempty: since \(A\) and \(B\) are subsets of \(f[X]\), each contains a value \(f(x)\) for some \(x\in X\). Thus these preimages form a separation of \(X\), contradicting its connectedness. Therefore \(f[X]\) is connected. \(\square\)
The theorem concerns the image as a subspace of \(Y\); it does not claim that \(f[X]\) is all of \(Y\). The map also need not be one-to-one. What matters is continuity and connectedness of the domain.
Worked Example: The Image of a Square Function
Define \(f:[0,1]\to\mathbb{R}\) by \(f(x)=x^2\). For \(x,y\in[0,1]\),
because \(0\leq x+y\leq2\). This inequality proves that \(f\) is Lipschitz, hence continuous. The interval theorem proved below shows that \([0,1]\) is connected, so the Continuous Images Theorem implies that \(f[[0,1]]\) is connected.
We can also identify the image exactly. If \(x\in[0,1]\), then \(0\leq x^2\leq1\), so \(f[[0,1]]\subseteq[0,1]\). Conversely, for each \(y\in[0,1]\), the number \(x=\sqrt y\) belongs to \([0,1]\) and satisfies \(f(x)=(\sqrt y)^2=y\). Thus \(f[[0,1]]=[0,1]\), a connected image.
Intervals Are Connected
The real line provides a fundamental illustration of connectedness. An interval contains every real number between any two of its points. That order property prevents a partition into two nonempty relatively open pieces. The proof uses the least-upper-bound property of the real numbers.
Proof. The empty interval and a one-point interval are connected by the definition, so suppose that \(I\) contains at least two points. Assume for contradiction that \(I=A\cup B\) is a separation: \(A\) and \(B\) are disjoint, nonempty, and relatively open in \(I\). Choose one point from each side. By ordering them and, if necessary, interchanging the names of the sides, we can choose \(a\in A\) and \(b\in B\) with \(a<b\). Since \(I\) is an interval, \([a,b]\subseteq I\).
Consider \(S=A\cap[a,b]\). It is nonempty because \(a\in S\), and it is bounded above by \(b\). Let \(c=\sup S\). Then \(a\leq c\leq b\), so \(c\in I\). For every positive integer \(n\), the definition of supremum gives a point \(a_n\in S\) such that \(c-1/n<a_n\leq c\). Hence \(a_n\to c\). Since \(B\) is relatively open in \(I\), its complement \(A\) is relatively closed in \(I\). Therefore \(c\in A\).
If \(c=b\), this contradicts \(b\in B\), because \(A\) and \(B\) are disjoint. If \(c<b\), relative openness of \(A\) gives some \(r>0\) such that \(I\cap(c-r,c+r)\subseteq A\). Choose \(t\) with \(c<t<b\) and \(t-c<r\); for instance, \(t=c+\min\{r/2,(b-c)/2\}\). Then \(t\in I\), and the relative-open-neighborhood inclusion gives \(t\in A\). Thus \(t\in S\) and \(t>c\), contradicting that \(c\) is an upper bound for \(S\). Both cases lead to a contradiction. Therefore \(I\) is connected. \(\square\)
The proof uses the supremum only after restricting attention to the bounded segment \([a,b]\). This is why it works for intervals that are unbounded as well as for bounded ones: the two points witnessing a proposed separation always determine a bounded segment inside the interval.
Worked Example: A Closed Interval Is Connected
Consider \(I=[-2,4]\). It is an interval: whenever \(x,z\in[-2,4]\) and \(x\leq y\leq z\), we have \(-2\leq x\leq y\leq z\leq4\), so \(y\in[-2,4]\). The Intervals Theorem therefore gives that \([-2,4]\) is connected.
The proof is not based on the endpoints being included. For example, \((-2,4)\) and \([-2,4)\) are also intervals and hence connected. The relevant property is containing every point between any two of its points, not whether the endpoints belong to the set.
Connected Subsets of the Real Line
The interval theorem has a converse for subsets of the real line: connectedness forces a set to contain every point between any two of its points. Together these statements give a complete description of connected subsets of \(\mathbb{R}\).
Proof. If \(E\) is an interval, it is connected by the Intervals Theorem. For the converse, suppose \(E\) is connected. If \(E\) is empty or has only one point, it is an interval. Otherwise, take any \(x,z\in E\) with \(x<z\), and let \(y\) satisfy \(x<y<z\). Suppose \(y\notin E\). Then
These two sets are disjoint and nonempty, since the first contains \(x\) and the second contains \(z\). Each is open relative to \(E\): both \((-\infty,y)\) and \((y,\infty)\) are open in \(\mathbb{R}\), so their intersections with \(E\) are relatively open. They would therefore form a separation of \(E\), contradicting connectedness. It follows that every \(y\) strictly between any two points of \(E\) also belongs to \(E\). This is exactly the defining interval property, so \(E\) is an interval. \(\square\)
Worked Example: A Finite Set with Two Distinct Points Is Disconnected
Let \(E=\{-3,2,7\}\subseteq\mathbb{R}\). The point \(0\) lies between \(-3\) and \(2\), but \(0\notin E\). Thus \(E\) is not an interval, so the Connected Subsets Theorem shows that it is not connected.
The partition can also be exhibited directly: take \(U=E\cap(-\infty,0)=\{-3\}\) and \(V=E\cap(0,\infty)=\{2,7\}\). The sets are disjoint, nonempty, relatively open in \(E\), and their union is \(E\). This explicit separation makes the failure of connectedness visible.
Why Connectedness Is Useful
Connectedness is a global property: it rules out a split of the entire space, not merely a gap detectable near one point. In real-valued analysis, the interval characterization explains why continuous functions on intervals cannot jump from one value to another while omitting all intermediate values. More generally, the continuous-image theorem lets us establish connectedness of complicated subsets by representing them as continuous images of simpler connected spaces.
A common pitfall is to confuse connectedness with path connectedness. A path-connected space is connected, but the definition of connectedness itself asks only whether a separation exists; it does not require that points can be joined by continuous paths. Another pitfall is to use openness in the ambient space when the definition calls for relative openness in the subset. For a subset \(E\subseteq X\), the sets in a proposed partition must be open in \(E\), not necessarily open in \(X\).
Check Your Understanding
Use the definition and the results proved above to answer the following questions.
- Why is a subset in a separation both relatively open and relatively closed?
- In the proof that continuous images preserve connectedness, why must both preimages of the proposed image partition be nonempty?
- Where does the least-upper-bound property enter the proof that intervals are connected?
- Why does a missing point strictly between two points of a subset of \(\mathbb{R}\) produce a separation?
- Give an example of a connected interval that does not contain either of its endpoints as boundary points of a larger closed interval.