An Elementary Counterexample in the Compact-Open Topology

We give a short proof that the space of continuous functions from [0, 1] to [0, 1] is not compact in the compact-open topology. Suppose X and Y are compact topological spaces. Let C(X, Y ) be the space of continuous functions from X to Y , and give this space the compact-open topology. An interesting problem from topology is to prove or disprove that C(X, Y ) is compact. http://dx.doi.org/10.4169/amer.math.monthly.119.08.693 MSC: Primary 54C35 October 2012] NOTES 693 What is the compact-open topology on C(X, Y )? Let C be a compact subset of X and U an open subset of Y . Let S(C,U ) be the set of all functions f ∈ C(X, Y ) such that f (C) ⊂ U . Then the sets S(C,U ) form a subbasis of the compact-open topology on C(X, Y ). It turns out that C(X, Y ) need not be compact even if X and Y are. This is known to experts, but not found in elementary texts such as [1], [2], and [3]. The purpose of this note is to provide an elementary counterexample; all we need is the intermediate value theorem. In our counterexample, we let X = Y = I , the closed unit interval [0, 1]with the usual subspace topology inherited from R. A common proof that C(I, I ) is not compact notes that the compact-open topology agrees with the uniform topology on C(I, I ) and that the sequence ( fn) defined by fn(x) = xn has no uniformly convergent subsequence since the limiting function is not continuous. For our proof, pick < 1/2. For x ∈ I , let Ux = S({x}, (x − , x + ) ∩ I ). These sets form an open cover of C(I, I ) because, by the intermediate value theorem, every continuous function from I to I has a fixed point. We now prove that this open cover has no finite subcover. Let Ux1,Ux2, . . . ,Uxn be a finite subcollection of this open cover and, without loss of generality, assume x1 < x2 < · · · < xn . Since < 1/2, no set Uxi covers C(I, I ). Choose yi ∈ I \ (xi − , xi + ) for all i = 1, 2, . . . , n. Let f be the piecewise linear function connecting (0, f (0)), (x1, y1), (x2, y2), . . . , (xn, yn), and (1, f (1)), where f (0) is taken to be 0 if x1 6= 0 and f (1) is taken to be 1 if xn 6= 1. Then it is clear that f 6∈ Uxi for all i , but f ∈ C(I, I ), which proves that this finite subcollection does not cover C(I, I ). Thus, C(I, I ) is not compact in the compact-open topology. I like this proof because it is a good illustration of the definitions of compactness and the compact open topology, and is a good application of the intermediate value theorem. A comparison of both this proof and the more common proof should be valuable to students.