- #1
gonzo
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I'm having trouble with the third part of a three part problem (part of the problem is that I don't even see how what I'm trying to prove can be true).
The problem is:
Let X and Y be topological spaces with X=E u F. We have two functions: f: from E to Y, and g: from F to Y, with f=g on the intersection of E and F. f and g are continuous with respect to the relative topologies. We are interested in the function h=f u g, from X into Y.
There were three questions and the one I can't get is to prove that if E and F are both closed then h is continuous.
The problem is this doesn't seem to be true to me. Nothing is said about which topologies are used, so we can assume pretty much anything for Y. Now the intersection of E and F is closed, since they are both closed. We can then assume Y takes on a single value in this intersection (for example, if graphing on the real line we could have E=[0,2] and F=[1,3] then the intersection = [1,2], and then we could have h be a trapezoidal function ... f(X)=x if x is in [0,1] f(x)=g(x)=1 if x is in [1,2], and g(x)=3-x if x is in [2,3] ... if we assume that {1} is an open set in Y, then the inverse image is the closed set [1,2] in X, so this isn't continuous under these topologies).
Clearly I am missing something and my reasoning must be off. Any help would be appeciated.
The problem is:
Let X and Y be topological spaces with X=E u F. We have two functions: f: from E to Y, and g: from F to Y, with f=g on the intersection of E and F. f and g are continuous with respect to the relative topologies. We are interested in the function h=f u g, from X into Y.
There were three questions and the one I can't get is to prove that if E and F are both closed then h is continuous.
The problem is this doesn't seem to be true to me. Nothing is said about which topologies are used, so we can assume pretty much anything for Y. Now the intersection of E and F is closed, since they are both closed. We can then assume Y takes on a single value in this intersection (for example, if graphing on the real line we could have E=[0,2] and F=[1,3] then the intersection = [1,2], and then we could have h be a trapezoidal function ... f(X)=x if x is in [0,1] f(x)=g(x)=1 if x is in [1,2], and g(x)=3-x if x is in [2,3] ... if we assume that {1} is an open set in Y, then the inverse image is the closed set [1,2] in X, so this isn't continuous under these topologies).
Clearly I am missing something and my reasoning must be off. Any help would be appeciated.