From an Open Cover to Finite Data
Compactness is often used by beginning with information that varies from point to point. Each point may have its own open neighborhood, its own bound, or its own relevant member of a cover. Compactness lets us select finitely many of those neighborhoods that still cover the space. The useful next step is to work with just those finitely many choices: finitely many bounds have a maximum, finitely many lists have a finite union, and finitely many selected sets can cover the whole space.
Throughout, a compact space means a topological space in which every open cover has a finite subcover. A family \(\mathcal U\) is an open cover of \(K\) if each member is open in \(K\) and \(\bigcup_{U\in\mathcal U}U=K\). A finite subcover is a finite subfamily whose union is still \(K\). If \(K\) is a subset of a space \(X\), its topology here is the subspace topology unless stated otherwise.
The distinction between these two steps matters. Compactness does not itself calculate a bound or finish a contradiction. It supplies finitely many sets. The proof must then explain how the information attached to those sets combines.
Worked Example: Extracting a Finite Cover of the Unit Interval
Worked Example: Extracting a Finite Cover of the Unit Interval
Let \(I=[0,1]\) have the subspace topology inherited from \(\mathbb R\), and let \(\mathcal U\) be any open cover of \(I\). We show directly how to obtain a finite subcover.
Define \(S\) to be the set of \(t\in I\) such that \([0,t]\) is covered by finitely many members of \(\mathcal U\). Since \(\mathcal U\) covers \(0\), some \(O_0\in\mathcal U\) contains \(0\). Because \(O_0\) is open in \(I\), there is a \(\delta>0\) with \([0,\delta)\cap I\subseteq O_0\). Taking \(t=\min(\delta/2,1/2)>0\) gives \([0,t]\subseteq O_0\), so \(t\in S\). Thus \(S\ne\varnothing\), and \(c=\sup S\) exists with \(0<c\le 1\).
Choose \(O\in\mathcal U\) containing \(c\). Relative openness gives a \(\delta>0\) such that \((c-\delta,c+\delta)\cap I\subseteq O\). By the definition of supremum, there is \(s\in S\) with \(c-\delta<s\le c\). A finite subfamily covers \([0,s]\), and \(O\) covers every point of \([s,c]\), since that interval lies in \((c-\delta,c+\delta)\cap I\). Therefore a finite subfamily covers \([0,c]\).
If \(c<1\), choose \(\eta=\min(\delta/2,(1-c)/2)>0\). The set \(O\) covers \([c,c+\eta]\), while the finite subfamily just obtained covers \([0,c]\). Together they cover \([0,c+\eta]\), so \(c+\eta\in S\), contradicting \(c=\sup S\). Hence \(c=1\). The finite subfamily covering \([0,c]\) is the required finite subcover of \(I\).
Check: the argument uses openness at the points \(0\) and \(c\) to obtain intervals inside cover members, and uses the supremum property to extend a finite cover up to \(c\). It therefore works for an arbitrary open cover, not only for covers by intervals of a particular form.
Finite Selections Turn Local Bounds into a Global Bound
Here is a general way finite extraction is used. Suppose that for each \(x\in K\) there is an open neighborhood \(W_x\) in \(K\) and a finite number \(M_x\) such that a function \(g:K\to\mathbb R\) satisfies \(|g(y)|\le M_x\) whenever \(y\in W_x\). The neighborhoods \(W_x\) cover \(K\). If \(K\) is compact, select \(W_{x_1},\ldots,W_{x_m}\) that still cover it. The finitely many local bounds can now be combined.
Proof. If \(K=\varnothing\), the assertion that \(|g(y)|\) is bounded for all \(y\in K\) holds vacuously. Otherwise, the family \(\{W_x:x\in K\}\) is an open cover of \(K\). Compactness gives a finite subcover \(W_{x_1},\ldots,W_{x_m}\). Let \(M=\max\{M_{x_1},\ldots,M_{x_m}\}\), which exists because this is a nonempty finite set of real numbers. For any \(y\in K\), some \(W_{x_i}\) contains \(y\), so \(|g(y)|\le M_{x_i}\le M\). Thus \(g\) is bounded on \(K\). \(\square\)
Worked Example: Bounding a Continuous Function by Local Data
Let \(K\) be compact and \(g:K\to\mathbb R\) continuous. For each \(x\in K\), the interval \((g(x)-1,g(x)+1)\) is an open neighborhood of \(g(x)\). By continuity, its preimage \(W_x=g^{-1}((g(x)-1,g(x)+1))\) is open in \(K\) and contains \(x\). For every \(y\in W_x\), the triangle inequality gives \[ |g(y)|\le |g(y)-g(x)|+|g(x)|<1+|g(x)|. \]
The sets \(W_x\) cover \(K\). Select a finite subcover \(W_{x_1},\ldots,W_{x_m}\). For each selected set the displayed inequality gives a local bound \(1+|g(x_i)|\). Set \(M=\max_{1\le i\le m}(1+|g(x_i)|)\). For any \(y\in K\), choose \(i\) with \(y\in W_{x_i}\); then \(|g(y)|<1+|g(x_i)|\le M\). This proves boundedness using only finitely many of the local bounds. If \(K=\varnothing\), boundedness is vacuous and no subcover choices are needed.
Check: the final bound is a maximum over the selected finite subcover, not over all \(x\in K\). That distinction is important: the method needs only finitely many values \(g(x_i)\) to produce the bound.
Pulling a Cover Back Through a Map
Finite extraction also transfers compactness through continuous maps. The key move is to turn a cover of the image into a cover of the domain by taking preimages.
Proof. If \(K=\varnothing\), then \(f(K)=\varnothing\), which is compact because the empty cover has a finite subcover. Now suppose \(K\ne\varnothing\), and let \(\mathcal V\) be an open cover of \(f(K)\). For each \(V\in\mathcal V\), continuity implies that \(f^{-1}(V)\) is open in \(K\). These preimages cover \(K\): if \(x\in K\), then \(f(x)\in f(K)\), so some \(V\in\mathcal V\) contains \(f(x)\), and hence \(x\in f^{-1}(V)\). By compactness, finitely many preimages \(f^{-1}(V_1),\ldots,f^{-1}(V_m)\) cover \(K\). For any \(z\in f(K)\), write \(z=f(x)\) for some \(x\in K\). Since the preimages cover \(K\), some \(i\) has \(x\in f^{-1}(V_i)\), so \(z=f(x)\in V_i\). Therefore \(V_1,\ldots,V_m\) cover \(f(K)\), proving it compact. \(\square\)
Worked Example: A Continuous Image of the Unit Interval
Give \(I=[0,1]\) its usual topology and define \(f:I\to\mathbb R\) by \(f(x)=x^2\). The map is continuous: for \(x,y\in I\), \[ |f(x)-f(y)|=|x-y||x+y|\le 2|x-y|, \] because \(0\le x+y\le 2\). Thus if \(|x-y|<\varepsilon/2\), then \(|f(x)-f(y)|<\varepsilon\).
Every value \(x^2\) with \(x\in[0,1]\) lies in \([0,1]\). Conversely, if \(z\in[0,1]\), then \(\sqrt z\in[0,1]\) and \(f(\sqrt z)=z\). Hence \(f(I)=[0,1]\). For any open cover of \(f(I)\), take its preimages under \(f\). They form an open cover of \(I\); finite extraction gives finitely many preimages covering \(I\), and the corresponding original cover members cover \(f(I)\). Thus the image is compact, with the finite selection taking place on the domain.
Check: the preimages cover the domain because every \(x\in I\) maps to a point of the covered image. The selected image sets cover the image because every image point has the form \(f(x)\) for some \(x\in I\).
Finite Extraction and Locally Finite Families
A family \(\mathcal A\) of subsets of \(K\) is locally finite if every point of \(K\) has an open neighborhood that intersects only finitely many members of \(\mathcal A\). The condition concerns which sets a neighborhood meets, not just which sets contain its center.
Proof. If \(K=\varnothing\), there are no nonempty subsets of \(K\), so the family is empty and finite. Suppose \(K\ne\varnothing\). For each \(x\in K\), local finiteness supplies an open neighborhood \(W_x\) that meets only finitely many members of \(\mathcal A\). The sets \(W_x\) cover \(K\), so compactness gives a finite subcover \(W_{x_1},\ldots,W_{x_m}\). Let \(\mathcal A_i\) be the subfamily of members that meet \(W_{x_i}\); each \(\mathcal A_i\) is finite. If \(A\in\mathcal A\), then \(A\ne\varnothing\), so choose \(a\in A\). Some selected \(W_{x_i}\) contains \(a\), and therefore \(A\) meets \(W_{x_i}\). Hence \(A\in\mathcal A_i\). This proves \(\mathcal A\subseteq\bigcup_{i=1}^m\mathcal A_i\), a finite union of finite families. Thus \(\mathcal A\) is finite. \(\square\)
Worked Example: Checking Local Finiteness Near an Endpoint
For each \(n\in\mathbb N\), let \(A_n=(1/(n+1),1/n)\), viewed as a subset of \(I=[0,1]\) with its usual subspace topology. Each \(A_n\) is nonempty: \(1/(n+1)<1/n\), so the interval between its endpoints contains points. The family \(\{A_n:n\in\mathbb N\}\) is not locally finite at \(0\).
Indeed, every open neighborhood \(W\) of \(0\) in \(I\) contains \([0,\delta)\cap I\) for some \(\delta>0\). Choose \(N\in\mathbb N\) with \(1/N<\delta\). For every \(n\ge N\), each point of \(A_n\) is positive and less than \(1/n\le 1/N<\delta\), so \(A_n\subseteq W\). Thus \(W\) meets every \(A_n\) for \(n\ge N\), which is infinitely many members. No neighborhood of \(0\) meets only finitely many members.
Check: although \(0\notin A_n\) for every \(n\), neighborhoods of \(0\) meet infinitely many \(A_n\). Membership of the center alone does not test local finiteness.
A Common Pitfall: Pointwise Finiteness Is Not Local Finiteness
A tempting but incorrect argument is to check that each point belongs to only finitely many members of a family and then conclude that the family is locally finite. For the intervals \(A_n\) above, any fixed \(x>0\) belongs to at most one \(A_n\): the intervals are disjoint. The point \(0\) belongs to none. One might therefore write, “Each point belongs to finitely many \(A_n\), so the family is locally finite.” The correction is that local finiteness asks whether some neighborhood of each point meets only finitely many family members. At \(0\), every neighborhood meets infinitely many \(A_n\), as calculated above. A finite-subcover argument can be used only after the required local neighborhoods have actually been established.
More generally, finite extraction produces finitely many chosen neighborhoods, not a single neighborhood with every desired property and not automatically a common numerical scale. State what each selected set controls, then show explicitly how the finite collection supplies the conclusion. That final finite step is where maxima, finite unions, and contradiction arguments enter.
Check Your Understanding
For each question, identify the open cover and the finite data that compactness supplies.
- In the unit-interval proof, why must a cover member containing \(0\) contain \([0,t]\) for some \(t>0\)?
- In the boundedness proposition, how does the finite subcover produce one bound for all of \(K\)?
- Why are the preimages of a cover of \(f(K)\) an open cover of \(K\) when \(f\) is continuous?
- In the locally finite family proposition, where is the assumption that every family member is nonempty used?
- For \(A_n=(1/(n+1),1/n)\subseteq[0,1]\), explain why every neighborhood of \(0\) meets infinitely many \(A_n\), even though \(0\) belongs to none of them.