Show that in the compact-open topology, \(\mathcal{C}(X, Y)\) is Hausdorff if \(Y\) is Hausdorff, and regular if \(Y\) is regular. [Hint: If \(\overline{U} \subset V\), then \(\overline{S(C, U)} \subset S(C, V)\).]
Solution
Assume \(Y\) Hausdorff. If \(f\ne g\) in \(C(X,Y)\), choose \(x\) with \(f(x)\ne g(x)\) and disjoint open neighborhoods \(U,V\) of these values. Since \(\{x\}\) is compact, the subbasic sets \[ S(\{x\},U),\qquad S(\{x\},V) \] are disjoint neighborhoods of \(f,g\). Hence \(C(X,Y)\) is Hausdorff.
Now assume \(Y\) regular. It suffices to shrink a basic neighborhood \[ N=\bigcap_{i=1}^n S(C_i,V_i) \] of \(f\). For each \(i\), compactness of \(f(C_i)\) and regularity of \(Y\) give an open \(U_i\) with \[ f(C_i)\subset U_i,\qquad \overline{U_i}\subset V_i. \] Put \(M=\bigcap_iS(C_i,U_i)\). Then \(f\in M\). The hinted inclusion \[ \overline{S(C,U)}\subset S(C,V)\quad(\overline U\subset V) \] follows because if \(h(C)\) is not contained in \(V\), choose \(x\in C\) with \(h(x)\notin V\); the evaluation neighborhood at \(x\) avoiding \(\overline U\) separates \(h\) from \(S(C,U)\). Therefore \[ \overline M\subset\bigcap_i\overline{S(C_i,U_i)}\subset N. \] So the function space is regular.