Talk:Combination Theorem for Continuous Functions/Real

Why need the subset be open? None of the proofs use this. The functions are ideally defined on an open neighbourhood of the subset, but that's something different. --Lord_Farin 12:40, 9 March 2012 (EST)


 * I'm racking my brains ... nope, haven't a clue. Must be a mistake, like a copypasta from something else. --prime mover 13:13, 9 March 2012 (EST)