Compactness Passes Through Continuous Maps
Compactness is defined by a condition on open covers: every open cover of the set has a finite subcover. Continuity gives a way to carry an open cover of an image back to the domain. If the domain set is compact, that pulled-back cover has a finite subcover, and the corresponding finitely many sets cover the image. This simple transfer is one of the central uses of compactness.
We work with topological spaces, which includes metric spaces with their usual open sets. A subset \(K\) of a space \(X\) is compact if every cover of \(K\) by open sets in \(X\) has a finite subcover, as in the previous tutorial. A function \(f:X\to Y\) is continuous if the inverse image of every open subset of \(Y\) is open in \(X\). For \(K\subseteq X\), write \(f[K]=\{f(x):x\in K\}\) for its image.
Proof. Let \(\mathcal{V}\) be any cover of \(f[K]\) by open sets in \(Y\). For each \(V\in\mathcal{V}\), continuity implies that \(f^{-1}(V)\) is open in \(X\). These inverse images cover \(K\): if \(x\in K\), then \(f(x)\in f[K]\), so some \(V\in\mathcal{V}\) contains \(f(x)\), and therefore \(x\in f^{-1}(V)\).
Since \(K\) is compact, finitely many members \(V_1,\ldots,V_m\) of \(\mathcal{V}\) have inverse images that cover \(K\). For every \(y\in f[K]\), choose \(x\in K\) with \(f(x)=y\). The finite cover of \(K\) puts \(x\) in some \(f^{-1}(V_j)\), so \(y=f(x)\in V_j\). Thus \(V_1,\ldots,V_m\) cover \(f[K]\). Since the original open cover was arbitrary, \(f[K]\) is compact. \(\square\)
The proof does not need \(f\) to be one-to-one or onto all of \(Y\). The only image under consideration is \(f[K]\), and every point in that image has at least one preimage in \(K\). Nor does the proof require a metric: it uses only the open-set definition of continuity and compactness.
Worked Examples
Worked Example: The Image of an Interval Under a Polynomial
Let \(K=[-1,2]\) and define \(f:K\to\mathbb{R}\) by \(f(x)=x^2\). The interval \(K\) is compact, and the polynomial \(f\) is continuous. The theorem therefore guarantees that \(f[K]\) is compact.
We can also identify the image exactly. For every \(x\in[-1,2]\), \(x^2\geq0\), and \(x^2\leq4\), so \(f[K]\subseteq[0,4]\). Conversely, let \(y\in[0,4]\). Then \(\sqrt{y}\in[0,2]\subseteq[-1,2]\), and \(f(\sqrt{y})=(\sqrt{y})^2=y\). Hence every \(y\in[0,4]\) is in the image, and
The minimum value is attained at \(x=0\), and the maximum value is attained at \(x=2\). In this example, the explicit calculation identifies the compact image as a familiar interval; the theorem itself did not require finding its endpoints.
Worked Example: A Compact Parabola Arc
Define \(f:[0,1]\to\mathbb{R}^2\) by \(f(t)=(t,t^2)\), where \(\mathbb{R}^2\) has its Euclidean metric. The domain is compact, and both coordinate functions \(t\) and \(t^2\) are continuous. Therefore \(f\) is continuous and its image is compact.
To describe the image, write \(x=t\). As \(t\) ranges over \([0,1]\), \(x\) ranges over \([0,1]\), and the second coordinate is \(x^2\). Thus
This is the indicated arc of the parabola. The conclusion is about compactness as a subset of the plane, not about whether the arc is an interval: its shape and dimension can change under a continuous map, while compactness is preserved.
Worked Example: A Parametrization of the Unit Circle
Define \(g:[0,2\pi]\to\mathbb{R}^2\) by \(g(t)=(\cos t,\sin t)\). The interval \([0,2\pi]\) is compact, and \(g\) is continuous because its coordinate functions are continuous. The image is compact by the continuous-image theorem.
For every \(t\), the identity \(\cos^2t+\sin^2t=1\) shows that \(g(t)\) lies on the unit circle. Conversely, each point on the unit circle has the form \((\cos t,\sin t)\) for some \(t\in[0,2\pi]\), by the standard parametrization of the circle. Consequently,
The map is not one-to-one, since \(g(0)=g(2\pi)=(1,0)\), but injectivity is not part of the theorem's hypotheses. This is a useful example of compactness passing to an image even when different domain points produce the same output.
Compact Images in Hausdorff Spaces
The previous tutorial established that compact subsets of Hausdorff spaces are closed. Combining that result with the theorem above gives an immediate way to prove that an image is closed: first show the image is compact, then use Hausdorffness of the target.
Proof. By the theorem on continuous images of compact sets, \(f[K]\) is compact in \(Y\). By the theorem Compact Sets Are Closed in Hausdorff Spaces, every compact subset of a Hausdorff space is closed. Hence \(f[K]\) is closed in \(Y\). \(\square\)
This conclusion depends on the target being Hausdorff, not merely on the domain being compact. The previous tutorial's indiscrete-space example shows why the separation hypothesis cannot simply be omitted from the compact-implies-closed step.
A Consequence for Continuous Bijections
A particularly useful application is that a continuous bijection from a compact space to a Hausdorff space has a continuous inverse. Such a map is a homeomorphism. The key point is that the image theorem makes images of closed subsets compact, and Hausdorffness then makes those images closed.
Proof. It remains to prove that the inverse \(f^{-1}:Y\to K\) is continuous. Let \(C\) be any closed subset of \(K\). We first show that \(C\) is compact. Given an open cover of \(C\) by sets open in \(K\), adjoin the open set \(K\setminus C\). The resulting family covers \(K\), so compactness of \(K\) gives a finite subcover. After removing \(K\setminus C\), the remaining finitely many sets still cover \(C\). Thus \(C\) is compact.
The restriction of \(f\) to \(C\) is continuous, so the continuous-image theorem implies that \(f[C]\) is compact in \(Y\). Since \(Y\) is Hausdorff, the previous corollary implies that \(f[C]\) is closed in \(Y\). Because \(f\) is bijective, the inverse image of \(C\) under \(f^{-1}\) is
We have shown that this set is closed in \(Y\) for every closed \(C\subseteq K\). Taking complements, the inverse image under \(f^{-1}\) of every open subset of \(K\) is open in \(Y\). Therefore \(f^{-1}\) is continuous, and \(f\) is a homeomorphism. \(\square\)
The Hausdorff condition rules out a possible failure: a continuous bijection need not have a continuous inverse if the target is not Hausdorff. Compactness of the domain alone is not enough to guarantee that images of closed subsets are closed in an arbitrary target.
What the Theorem Does Not Say
Continuity is essential. For example, define \(h:[0,1]\to\mathbb{R}\) by \(h(0)=2\) and \(h(x)=x\) for \(0<x\leq1\). This function is discontinuous at \(0\), and its image is \((0,1]\cup\{2\}\). To see that this image is not compact, consider the open cover consisting of \(\{(1/n,3/2):n\geq2\}\) together with \((1.9,2.1)\). The intervals \((1/n,3/2)\) cover \((0,1]\), and the last interval covers \(2\). Any finite selection of the first intervals has a largest index \(N\), so their union is \((1/N,3/2)\); it misses, for instance, \(1/(2N)\). The additional interval around \(2\) does not cover that point. Thus no finite subcollection covers the image.
The converse also fails: a compact image does not imply a compact domain. The constant function \(c:\mathbb{R}\to\mathbb{R}\), given by \(c(x)=0\), is continuous and has image \(\{0\}\), which is compact, although its domain \(\mathbb{R}\) is not compact. Compactness is guaranteed to pass from a compact domain to its continuous image; it cannot generally be inferred in the reverse direction.
Check Your Understanding
Use the open-cover proof and its consequences to answer the following questions.
- Given an open cover of \(f[K]\), why do the inverse images of its members cover \(K\)?
- Where in the proof of compactness of \(f[K]\) is continuity used, and where is compactness used?
- Why does the image theorem alone not imply that \(f[K]\) is closed in every topological space?
- In the compact-to-Hausdorff bijection theorem, why is it useful to first show that each closed subset of \(K\) is compact?
- Which hypothesis fails in the example of a continuous function on \([0,1]\) whose image is not compact?