The Open-Cover Strategy
Compactness is defined by a property of open covers: every open cover has a finite subcover. A continuous map lets us transfer a cover of its image to the domain, where compactness supplies that finite selection. Mapping the selected sets back then covers the image. The argument requires neither injectivity nor surjectivity onto the entire codomain.
Throughout, a topological space is written \((X,\mathcal{T})\). A family \(\mathcal{U}\) of open subsets of \(X\) is an open cover of \(X\) if \(X=\bigcup_{U\in\mathcal{U}}U\). A finite subcover is a finite subfamily that still covers \(X\). A space is compact if every open cover has a finite subcover. If \(A\subseteq X\), its subspace topology consists of the sets \(A\cap U\), where \(U\in\mathcal{T}\).
For a set \(V\) open in the subspace \(f(X)\), there is an open set \(W\) in \(Y\) such that \(V=f(X)\cap W\). Since every value \(f(x)\) belongs to \(f(X)\), we have \(f^{-1}(V)=f^{-1}(W)\). Thus continuity on \(Y\) ensures that the preimage of a relatively open set of the image is open in the domain.
Proof. Each \(V_\alpha\) is open in \(f(X)\), so it has the form \(f(X)\cap W_\alpha\) for some open \(W_\alpha\subseteq Y\). Because \(f(x)\in f(X)\) for every \(x\in X\), \(f^{-1}(V_\alpha)=f^{-1}(W_\alpha)\), which is open by continuity. To check coverage, let \(x\in X\). The point \(f(x)\in f(X)\) belongs to some \(V_\alpha\), since the sets \(V_\alpha\) cover \(f(X)\). Therefore \(x\in f^{-1}(V_\alpha)\). Every point of \(X\) lies in one of the preimages, as required. If \(X=\varnothing\), the preimages cover the empty set as well. \(\square\)
Continuous Images Preserve Compactness
Proof. Let \(\{V_\alpha:\alpha\in A\}\) be any open cover of \(f(X)\). By the pullback-cover lemma, the sets \(f^{-1}(V_\alpha)\) form an open cover of \(X\). Compactness of \(X\) gives a finite set of indices \(\alpha_1,\ldots,\alpha_n\) for which \(\{f^{-1}(V_{\alpha_1}),\ldots,f^{-1}(V_{\alpha_n})\}\) covers \(X\). We show that the corresponding \(V_{\alpha_i}\) cover \(f(X)\). If \(y\in f(X)\), choose \(x\in X\) with \(f(x)=y\). Some \(i\) has \(x\in f^{-1}(V_{\alpha_i})\), so \(y=f(x)\in V_{\alpha_i}\). Hence the selected sets cover \(f(X)\), and they are a finite subcover of the original cover. If \(X=\varnothing\), then \(f(X)=\varnothing\), which is compact because the empty subfamily covers it. Thus the result holds in every case. \(\square\)
The proof uses compactness only on the domain and continuity only to make the pulled-back sets open. It does not assume that the map is one-to-one, that it is onto \(Y\), or that it sends open sets to open sets.
Take an arbitrary open cover of \(f(X)\), using the subspace topology on the image.
Continuity makes each preimage open, and the preimages cover \(X\).
Choose finitely many of those preimages that cover \(X\).
Every image point comes from a domain point in one of the chosen preimages, so the corresponding original sets cover \(f(X)\).
A useful immediate consequence is the surjective case.
Proof. Surjectivity means \(f(X)=Y\). The theorem says that \(f(X)\) is compact with the subspace topology from \(Y\). The subspace topology on \(Y\subseteq Y\) is exactly its original topology, since \(Y\cap W=W\) for every open \(W\subseteq Y\). Therefore \(Y\) is compact. \(\square\)
Worked Applications
Worked Example: A Nonconstant Map Between Finite Spaces
Let \(X=\{a,b,c\}\) have the discrete topology, so every subset is open, and let \(Y=\{r,s\}\) have the indiscrete topology \(\{\varnothing,Y\}\). Define \(f(a)=r\), \(f(b)=s\), and \(f(c)=s\). Show that \(f(X)\) is compact.
The domain is compact: given any open cover of \(X\), choose one cover member containing \(a\), one containing \(b\), and one containing \(c\). These at most three selected sets cover \(X\). The map is continuous because the only open sets of \(Y\) are \(\varnothing\) and \(Y\), whose preimages are \(\varnothing\) and \(X\), both open in \(X\). Its image is \(\{r,s\}=Y\).
For the cover argument, the only open cover of \(Y\) that needs consideration must include \(Y\): neither point belongs to the open set \(\varnothing\). Thus the single set \(Y\) is a finite subcover. This verifies compactness of the image directly and agrees with the theorem. The map is not injective, since \(f(b)=f(c)=s\); injectivity is not needed.
Worked Example: A Polynomial Image of the Unit Interval
Give \([0,1]\) its standard subspace topology from \(\mathbb{R}\), and define \(f:[0,1]\to[0,1]\) by \(f(x)=x(1-x)\). Find its image and deduce that the image is compact.
The function is continuous because it is the restriction of the polynomial \(x-x^2\), which is continuous on \(\mathbb{R}\). Completing the square gives \(f(x)=\frac14-(x-\frac12)^2\). Since \(0\leq x\leq1\), we have \(-\frac12\leq x-\frac12\leq\frac12\), so \(0\leq(x-\frac12)^2\leq\frac14\). It follows that \(0\leq f(x)\leq\frac14\).
To verify that every value in this interval is attained, let \(t\in[0,\frac14]\). Solving \(x(1-x)=t\) gives \(x=\frac{1\pm\sqrt{1-4t}}2\). Here \(0\leq1-4t\leq1\), so \(0\leq\sqrt{1-4t}\leq1\). In particular, \(x=\frac{1-\sqrt{1-4t}}2\) belongs to \([0,\frac12]\subseteq[0,1]\) and satisfies \(f(x)=t\). Thus \(f([0,1])=[0,\frac14]\). At the endpoints, \(f(0)=0\), and \(f(\frac12)=\frac12(1-\frac12)=\frac14\), confirming the extreme values by substitution.
The unit interval is compact in its standard topology by the open-cover theorem for closed intervals. The continuous-image theorem therefore shows that \([0,\frac14]\), with its subspace topology from \([0,1]\), is compact. The result concerns the image, not a claim that every subset of the codomain is compact.
Worked Example: A Compact Cofinite Domain
Let \(X=\mathbb{R}\) with the finite-complement topology: its open sets are \(\varnothing\) and all subsets whose complement in \(\mathbb{R}\) is finite. Let \(Y=\{0,1\}\) have the indiscrete topology, and define \(f:\mathbb{R}\to Y\) by \(f(x)=0\) for \(x\in\mathbb{Q}\) and \(f(x)=1\) for \(x\notin\mathbb{Q}\). Show that the image is compact.
The map is continuous: the only open sets in \(Y\) are \(\varnothing\) and \(Y\), and their preimages are \(\varnothing\) and \(\mathbb{R}\), both open in \(X\). Both values occur: \(0=f(0)\), while \(1=f(\sqrt{2})\), since \(\sqrt{2}\) is irrational. Thus \(f(X)=Y\).
The domain \(X\) is compact. Indeed, take an open cover and choose one member \(U\) containing \(0\). This member is nonempty, so \(\mathbb{R}\setminus U\) is finite. For each point of that finite complement, choose one member of the cover containing it. Together with \(U\), these finitely many sets cover \(\mathbb{R}\). The theorem now gives compactness of \(Y\); directly, every open cover of the nonempty indiscrete space \(Y\) must contain \(Y\), which alone is a finite subcover.
A Common Mistake: Mapping the Cover Forward
Continuity controls preimages of open sets, not images of open sets. So it is generally invalid to take an open cover of the domain and assume its images form an open cover of the image. For a concrete check, use the continuous function \(f:[0,1]\to[0,1]\) given by \(f(x)=x(1-x)\). The set \(U=(0,1)\) is open in the subspace \([0,1]\). For \(x\in U\), the completed-square formula gives \(f(x)=\frac14-(x-\frac12)^2\). The square term is zero at \(x=\frac12\), and is positive at every other \(x\in U\); it is less than \(\frac14\) throughout \(U\). Hence \(f(U)=(0,\frac14]\). This is not open in \([0,1]\), because any relatively open neighborhood of \(\frac14\) contains points of \([0,1]\) greater than \(\frac14\), which are not in \(f(U)\).
The correction is the direction used in the theorem: start with an open cover of the image and take its preimages. These are open by continuity. Compactness makes finitely many sufficient on the domain, and the proof then checks that the same finite selection covers the image.
Check Your Understanding
Use the open-cover definitions and identify precisely where continuity or compactness is needed.
- In the theorem, why is \(f^{-1}(V)\) open when \(V\) is open in the subspace \(f(X)\), rather than necessarily open in all of \(Y\)?
- Does the continuous-image theorem require \(f\) to be injective? Explain which step of the proof answers this.
- For a continuous surjection from a compact space \(X\) onto \(Y\), identify the image \(f(X)\) and explain why its subspace topology is the topology of \(Y\).
- In the polynomial example, solve \(x(1-x)=t\) and verify that one solution lies in \([0,1]\) for every \(t\in[0,\frac14]\).
- Why does the set \(f((0,1))\) in the polynomial example fail to be open in \([0,1]\), despite continuity of \(f\)?