Show that the sets \(B_C(f, \epsilon)\) form a basis for a topology on \(Y^X\).
Solution
For compact \(C\subset X\), \(f\in Y^X\), and \(\varepsilon>0\), let \[ B_C(f,\varepsilon)=\{g:\sup_{x\in C}d(f(x),g(x))<\varepsilon\}. \] Every \(f\) belongs to such a set (take any compact \(C\), including a singleton when \(X\ne\varnothing\)). If \[ h\in B_C(f,\varepsilon)\cap B_D(g,\delta), \] set \[ r=\min\{\varepsilon-d_C(f,h),\delta-d_D(g,h)\}>0, \] where \(d_C(u,v)=\sup_C d(u,v)\). Since \(C\cup D\) is compact, \(B_{C\cup D}(h,r)\) is defined, and the triangle inequality gives \[ B_{C\cup D}(h,r)\subset B_C(f,\varepsilon)\cap B_D(g,\delta). \] Thus these sets satisfy the basis intersection axiom and form a basis for the topology of compact convergence.