Proof Continuous Images of Compact Sets Are Compact

A Second Route to the Compact-Image Theorem

The previous tutorial stated that a continuous image of a compact set is compact and used it in examples, and proved that the image is totally bounded. Here we develop a different proof, directly from the open-cover definition of compactness. The central technique is to pull an open cover of the image back to the domain: continuity ensures that the resulting sets are open relative to the domain, and compactness supplies a finite subcover.

The theorem itself is not new here; the goal is to understand its open-cover proof and the role played by continuity. This proof is useful beyond the real line because the same argument applies whenever compactness is defined using open covers and continuity is expressed through open preimages.

Let \(K\subseteq\mathbb{R}\). A set \(V\subseteq K\) is open relative to \(K\) if there is an open set \(G\subseteq\mathbb{R}\) such that \(V=K\cap G\). Such a set may fail to be open in all of \(\mathbb{R}\). For example, \([0,1)\) is open relative to \([0,1]\), since \([0,1)=[0,1]\cap(-1,1)\).

Proof. First suppose \(f\) is continuous at every point of \(K\), and let \(U\subseteq\mathbb{R}\) be open. If \(f^{-1}(U)\) is empty, it is relatively open. Otherwise, take \(x\in f^{-1}(U)\). Since \(U\) is open and \(f(x)\in U\), there is an \(\varepsilon>0\) such that \((f(x)-\varepsilon,f(x)+\varepsilon)\subseteq U\). By continuity at \(x\), there is a \(\delta>0\) such that for \(y\in K\), \(|y-x|<\delta\) implies \(|f(y)-f(x)|<\varepsilon\). Therefore \[ K\cap(x-\delta,x+\delta)\subseteq f^{-1}(U). \] Every \(x\in f^{-1}(U)\) thus has a relative neighborhood contained in \(f^{-1}(U)\), so this preimage is open relative to \(K\).

Conversely, suppose every open subset of \(\mathbb{R}\) has a relatively open preimage under \(f\). Fix \(x\in K\) and \(\varepsilon>0\). The interval \(U=(f(x)-\varepsilon,f(x)+\varepsilon)\) is open, and \(x\in f^{-1}(U)\). Relative openness means there is an open set \(G\subseteq\mathbb{R}\) with \(f^{-1}(U)=K\cap G\). Because \(x\in G\), some \(\delta>0\) satisfies \((x-\delta,x+\delta)\subseteq G\). For any \(y\in K\) with \(|y-x|<\delta\), we have \(y\in K\cap G=f^{-1}(U)\), hence \(|f(y)-f(x)|<\varepsilon\). This proves continuity at \(x\), and \(x\) was arbitrary. \(\square\)

Pulling Back an Open Cover

Suppose \(\{U_\alpha:\alpha\in A\}\) is an open cover of \(f(K)\). This means each \(U_\alpha\) is open in \(\mathbb{R}\), and every value \(f(x)\), \(x\in K\), belongs to at least one \(U_\alpha\). For each index \(\alpha\), consider the inverse image \[ f^{-1}(U_\alpha)=\{x\in K:f(x)\in U_\alpha\}. \] By the lemma, each inverse image is open relative to \(K\). These sets cover \(K\): for any \(x\in K\), the value \(f(x)\) lies in some \(U_\alpha\), so \(x\in f^{-1}(U_\alpha)\).

Compactness is often stated for covers by open subsets of \(\mathbb{R}\), while these inverse images are only relatively open. This difference causes no problem. For each \(\alpha\), choose an open set \(G_\alpha\subseteq\mathbb{R}\) such that \[ f^{-1}(U_\alpha)=K\cap G_\alpha. \] Since the inverse images cover \(K\), the sets \(G_\alpha\) also cover \(K\). Compactness gives finitely many indices \(\alpha_1,\ldots,\alpha_m\) such that \(G_{\alpha_1},\ldots,G_{\alpha_m}\) cover \(K\). For \(x\in K\), membership in one of these ambient open sets means \(x\in K\cap G_{\alpha_i}=f^{-1}(U_{\alpha_i})\). Thus the corresponding inverse images still cover \(K\). This is the finite-subcover step that transfers the argument back to the image.

Proof. If \(K=\varnothing\), then \(f(K)=\varnothing\), which is compact. Now suppose \(K\neq\varnothing\), and let \(\{U_\alpha:\alpha\in A\}\) be any open cover of \(f(K)\). The inverse images \(f^{-1}(U_\alpha)\) are relatively open in \(K\) by the lemma, and they cover \(K\). Write \(f^{-1}(U_\alpha)=K\cap G_\alpha\) with \(G_\alpha\) open in \(\mathbb{R}\). The sets \(G_\alpha\) cover \(K\), so compactness gives indices \(\alpha_1,\ldots,\alpha_m\) for which \(G_{\alpha_1},\ldots,G_{\alpha_m}\) cover \(K\). Consequently, \(f^{-1}(U_{\alpha_1}),\ldots,f^{-1}(U_{\alpha_m})\) cover \(K\). If \(z\in f(K)\), write \(z=f(x)\) for some \(x\in K\). That \(x\) belongs to \(f^{-1}(U_{\alpha_i})\) for some \(i\), so \(z=f(x)\in U_{\alpha_i}\). The selected \(U_{\alpha_i}\) therefore cover \(f(K)\). Every open cover has a finite subcover, which proves compactness. \(\square\)

Worked Examples

Why This Proof Is Useful

The sequential proof and the open-cover proof emphasize different features of the same theorem. The sequential argument follows a sequence of image values back to a sequence in the domain. The open-cover argument starts with a family of sets covering the image and takes inverse images. In both cases continuity connects the domain and image, while compactness supplies the finite or convergent structure needed to finish.

A common pitfall is to treat a relatively open subset of \(K\) as though it must be open in \(\mathbb{R}\). It need not be: \([0,1)\) is relatively open in \([0,1]\) but is not open in the real line. The proof handles this by writing each relative open set as \(K\cap G_\alpha\), where \(G_\alpha\) is ambient open, and applying compactness to the \(G_\alpha\). Another pitfall is to forget that the finite selection must still cover the image. The key check is that a point \(z\in f(K)\) has the form \(z=f(x)\); a selected inverse image containing \(x\) corresponds to a selected open set containing \(z\).

The theorem does not say that the image has to be an interval, nor does continuity alone guarantee a compact image. What it guarantees is preservation of compactness when both hypotheses hold. Earlier results, including the Heine–Borel Theorem, then let us recognize compact images in \(\mathbb{R}\) as closed and bounded.