Why Connectedness Constrains a Map’s Range
A continuous map may change distances, shape, and even the topology of the space it maps into. But if its domain is connected, its image cannot split into two separated pieces. The key is that a separation of the image would pull back to a separation of the domain.
For a subset \(A\subseteq X\), the subspace topology on \(A\) consists of 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 sets. For a subset, connectedness always refers to its subspace topology. A map \(f:X\to Y\) is continuous if \(f^{-1}(V)\) is open in \(X\) for every open set \(V\subseteq Y\).
Thus, when the image \(f(X)\) is discussed as a space, it is given the subspace topology inherited from \(Y\). A set open in \(f(X)\) need not be open in all of \(Y\); it is the intersection of \(f(X)\) with some open subset of \(Y\).
The Continuous Image Theorem
Proof. If \(X=\varnothing\), then \(f(X)=\varnothing\), which has no separation and is connected. Now suppose \(X\ne\varnothing\). Assume, for contradiction, that \(f(X)\) has a separation \(U,V\). Then \(U,V\) are disjoint, nonempty, open in \(f(X)\), and \(f(X)=U\cup V\). Since they are open in the subspace topology, there are open subsets \(G,H\subseteq Y\) such that \(U=f(X)\cap G\) and \(V=f(X)\cap H\). Continuity makes \(f^{-1}(G)\) and \(f^{-1}(H)\) open in \(X\). They are disjoint, and their union is \(X\), because every \(f(x)\) belongs to exactly one of \(U,V\). They are both nonempty: a point of \(U\) has a preimage under \(f\), and that preimage belongs to \(f^{-1}(G)\); likewise a point of \(V\) has a preimage in \(f^{-1}(H)\). Hence \(f^{-1}(G),f^{-1}(H)\) form a separation of \(X\), contradicting connectedness. Therefore \(f(X)\) is connected. \(\square\)
The argument depends on preimages of open sets being open. It does not require \(f\) to be one-to-one or onto all of \(Y\): it is onto its image by the definition of \(f(X)\), and that is all the proof needs.
Real-Valued Maps Have Interval Ranges
We use the standard result on connected subsets of the real line: a subset \(J\subseteq\mathbb{R}\), with the subspace topology from the standard topology on \(\mathbb{R}\), is connected if and only if it is an interval. Here “interval” means that whenever \(a,b\in J\) and \(a<r<b\), then \(r\in J\); intervals may be open, closed, half-open, unbounded, or singletons. The classification is established in the earlier material on connected subsets of the real line.
Proof. The continuous image theorem says that \(f(X)\) is connected in its subspace topology. By the stated classification of connected subsets of \(\mathbb{R}\), \(f(X)\) is an interval. If \(X=\varnothing\), then \(f(X)=\varnothing\), which also satisfies the interval condition stated above. \(\square\)
In particular, if \(f(x)=a\) and \(f(y)=b\) for points \(x,y\in X\), with \(a<b\), then every real number strictly between \(a\) and \(b\) belongs to \(f(X)\). This conclusion concerns values between two values already in the range; it does not say that the range has a minimum or maximum. For example, an interval may omit either endpoint.
Worked Example: The Square Function
Let \(X=\mathbb{R}\) with its standard topology and define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x^2\). Find its range and check how connectedness applies.
The real line is connected, and the polynomial map \(f\) is continuous. For every \(x\in\mathbb{R}\), \(x^2\ge 0\), so \(f(\mathbb{R})\subseteq[0,\infty)\). Conversely, if \(y\in[0,\infty)\), then \(x=\sqrt{y}\) is real and \(f(x)=(\sqrt{y})^2=y\). Thus \([0,\infty)\subseteq f(\mathbb{R})\), and \(f(\mathbb{R})=[0,\infty)\).
Check: \(f(0)=0^2=0\), and \(f(3)=3^2=9\). The value \(4\) lies between \(0\) and \(9\), and indeed \(f(2)=2^2=4\). This is consistent with the range-interval conclusion. The range is unbounded, so connectedness alone has not supplied a largest value.
Worked Example: A Range with One Included Endpoint
Let \(X=(0,1)\) with the subspace topology from the standard real line, and define \(f:X\to\mathbb{R}\) by \(f(t)=t(1-t)\). Determine the range, taking care with strict and non-strict inequalities.
The interval \(X\) is connected, and \(f\) is continuous because it is the restriction of a polynomial. For every \(t\in(0,1)\), both \(t\) and \(1-t\) are strictly positive, so \(f(t)>0\). Also, completing the square gives $$ f(t)=t(1-t)=\frac14-\left(t-\frac12\right)^2\le\frac14. $$ Hence \(f(X)\subseteq(0,\frac14]\).
To verify that every value in this proposed range occurs, let \(y\in(0,\frac14]\) and set \(t=\frac12-\sqrt{\frac14-y}\). Since \(y>0\), we have \(0\le\sqrt{\frac14-y}<\frac12\), so \(0<t\le\frac12<1\). Substitution gives $$ t(1-t)=\frac14-\left(t-\frac12\right)^2 =\frac14-\left(\sqrt{\frac14-y}\right)^2=y. $$ Therefore \(f(X)=(0,\frac14]\).
Check: At \(t=\frac12\), \(f(t)=\frac12(1-\frac12)=\frac14\), so the upper endpoint is attained. The value \(0\) is not attained: \(t(1-t)>0\) for every \(t\in(0,1)\). The computed range is an interval, as the corollary requires.
Worked Example: A Connected Domain with a Non-Hausdorff Topology
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Show that every continuous map \(g:X\to\mathbb{R}\) is constant.
The only subsets of \(X\) that are both open and closed are \(\varnothing\) and \(X\): the open set \(\{a\}\) has complement \(\{b\}\), which is not open. If \(X\) had a separation, each member of the separation would be both open and closed, since it is the complement of the other member. Neither member could then be a nonempty proper subset, which is impossible for a separation. Thus \(X\) is connected.
Now let \(g:X\to\mathbb{R}\) be continuous. The continuous image theorem gives that \(g(X)\) is connected. Since \(X\) has two points, \(g(X)\) has at most two points. A connected subset of \(\mathbb{R}\) is an interval, and an interval containing two distinct real numbers must also contain every real number strictly between them. A set with at most two points cannot do that. Therefore \(g(X)\) has only one point, so \(g(a)=g(b)\).
Check: Constant maps really are continuous here: the preimage of any open subset of \(\mathbb{R}\) is either \(\varnothing\) or \(X\), both open in \(\mathcal{T}\). The conclusion does not depend on the domain being a familiar interval or Hausdorff space.
What the Hypothesis Rules Out
Connectedness of the domain is essential. Continuity by itself does not prevent an image from being disconnected. For instance, let \(X=\{0,1\}\) have the discrete topology \(\mathcal{P}(X)\), and let \(f:X\to\mathbb{R}\) be \(f(x)=x\). Every subset of \(X\) is open, so the preimage of every open subset of \(\mathbb{R}\) is open; thus \(f\) is continuous. But \(f(X)=\{0,1\}\) is disconnected: its two singleton subsets are disjoint, nonempty, and open in the subspace topology. The domain is also disconnected, since \(\{0\}\) and \(\{1\}\) separate it.
Common pitfall. A mistaken argument might say, “\(f\) is continuous, so it sends the open set \(\{0\}\) to an open set; therefore the image is separated.” Continuity guarantees that preimages of open sets are open, not that images of open sets are open. In the example, \(f(\{0\})=\{0\}\) is not open in \(\mathbb{R}\), although \(f\) is continuous. The correct proof starts with a separation in the image and pulls its open sets back to the domain.
The real-valued conclusion is useful even when the range has not been computed exactly. It tells us that a continuous real-valued function on a connected domain cannot have a range with a gap: if two values occur, every value between them occurs too. It does not claim that an endpoint occurs, that the range is bounded, or that a continuous map sends open sets to open sets. Those are separate questions requiring other hypotheses or arguments.
Check Your Understanding
For each question, distinguish the topology on the domain from the subspace topology on the image.
- In the proof of the continuous image theorem, why are the two preimages nonempty when the two pieces of the image separation are nonempty?
- Give a continuous map from a connected space into \(\mathbb{R}\) whose range is not closed.
- For \(f(t)=t(1-t)\) on \((0,1)\), explain why \(0\) is not in the range but \(\frac14\) is.
- Why does the identity map from the discrete space \(\{0,1\}\) not contradict the continuous image theorem?
- If \(X\) is connected and a continuous \(f:X\to\mathbb{R}\) takes the values \(-2\) and \(5\), what can be concluded about the numbers strictly between them?