So right now I am reading on continuity in topology, which is stated as a function is continuous if the open subset of the image has an open subset in the inverse image..... that is not the issue. I just read the pasting lemma which states: Let X = A[itex]\cup[/itex]B, where A and B are closed in X. Let f: A-->Y and g: B-->Y. If f(x) = g(x) for every x [itex]\in[/itex] A[itex]\cap[/itex]B, then f and g combine to give a continuous function h: X-->Y, defined by setting h(x)= f(x) if x[itex]\in[/itex]A and h(x) = g(x) if x[itex]\in[/itex]B. Now they showed a couple of examples which I got except for this one: l(x) = x-2, for x < 0 AND x+2 for x [itex]\geq[/itex]0 they go on to say that l(x) is not continuous because the inverse image of the open set (1,3) for instance is the nonopen set [0,1) It's this last statment I am confused on how are they taking the inverse image of (1,3) and mapping it to the nonopen set? Seems like a simple enough concept but it is driving me bezerk. Thanks in advance.