Proving a function f is continuous given A U B = X

  • #1
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Summary:
Let X & Y be topological spaces and let f: X --> Y. Suppose A U B = X. And that f limited to A gives: A ---> Y and f limited to B gives: B ---> Y and these are continuous.
Basically with this problem, I need to show that f is continuous if A and B are open and if A and B are closed. My initial thoughts are that in the first case X must be open since unions of open sets are open. My question is that am I allowed to assume open sets exist in Y? Because then I can feasibly apply the topological continuity theorem (it's called theorem 2.6 in Basic Topology by M.A. Armstrong) which essentially says that the preimage of open sets are open. And I think that would prove it is continuous. The second case troubles me because I don't think I can say conclusively that the preimage of closed sets are closed. I know that if X is closed then X complement must be open. But I'm pretty sure X complement is not in the domain of f. Apparently it's also possible to find a function such that given these constraints, the function is not continuous. Any help on this is greatly appreciated. Thanks!
 

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  • #2
Office_Shredder
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I think one of the axioms of a space being a topological space is that it's an open set, so you know X is open before you're told anything about A and B.

Y is also a topological space, so it has at least two open sets, Y and the empty set. Beyond that you haven't been told anything.

I think you should just start with the definition of what it means for f to be continuous on A and B, and then write down what you need to prove for f to be continuous on X. It seems like your thoughts are a bit too scattered right now.
 
  • #3
pasmith
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Summary:: Let X & Y be topological spaces and let f: X --> Y. Suppose A U B = X. And that f limited to A gives: A ---> Y and f limited to B gives: B ---> Y and these are continuous.

Basically with this problem, I need to show that f is continuous if A and B are open and if A and B are closed. My initial thoughts are that in the first case X must be open since unions of open sets are open. My question is that am I allowed to assume open sets exist in Y?

[itex]X[/itex] is open by definition of being a topological space. Again by definition, [itex]Y[/itex] has at least two open subsets: [itex]Y[/itex] and [itex]\emptyset[/itex].

Because then I can feasibly apply the topological continuity theorem (it's called theorem 2.6 in Basic Topology by M.A. Armstrong) which essentially says that the preimage of open sets are open.

That is the idea, yes.

The second case troubles me because I don't think I can say conclusively that the preimage of closed sets are closed. I know that if X is closed then X complement must be open. But I'm pretty sure X complement is not in the domain of f.

What is the complement of [itex]X[/itex]? It must be with respect to some set, and that set must be [itex]X[/itex]. So the complement of [itex]X[/itex] is empty. As this is open, [itex]X[/itex] itself is closed. (Remember that "closed" is not the logical negation of "open" and that a set can be both.)

If [itex]V \subset Y[/itex] is closed, then [itex]Y \setminus V[/itex] is open. Now [itex]f^{-1}(Y \setminus V)[/itex] is everything in the domain of [itex]f[/itex] whose image is not in [itex]V[/itex], which is exactly the complement of [itex]f^{-1}(V)[/itex].

Apparently it's also possible to find a function such that given these constraints, the function is not continuous. Any help on this is greatly appreciated. Thanks!

I assume you mean in the case that [itex]A[/itex] and [itex]B[/itex] are not both open or both closed, since we know the result holds for those cases. Consider a non-constant function [itex]f: [0,1] \to \{0,1\}[/itex]. Then [itex]f[/itex] is necessarily continuous when restricted to [itex]A = f^{-1}(\{0\})[/itex] and [itex]B = f^{-1}(\{1\})[/itex] respectively, and clearly [itex][0,1] = A \cup B[/itex]. Can [itex]f[/itex] be continuous on [itex][0,1][/itex]?
 
  • #4
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[itex]X[/itex] is open by definition of being a topological space. Again by definition, [itex]Y[/itex] has at least two open subsets: [itex]Y[/itex] and [itex]\emptyset[/itex].



That is the idea, yes.



What is the complement of [itex]X[/itex]? It must be with respect to some set, and that set must be [itex]X[/itex]. So the complement of [itex]X[/itex] is empty. As this is open, [itex]X[/itex] itself is closed. (Remember that "closed" is not the logical negation of "open" and that a set can be both.)

If [itex]V \subset Y[/itex] is closed, then [itex]Y \setminus V[/itex] is open. Now [itex]f^{-1}(Y \setminus V)[/itex] is everything in the domain of [itex]f[/itex] whose image is not in [itex]V[/itex], which is exactly the complement of [itex]f^{-1}(V)[/itex].



I assume you mean in the case that [itex]A[/itex] and [itex]B[/itex] are not both open or both closed, since we know the result holds for those cases. Consider a non-constant function [itex]f: [0,1] \to \{0,1\}[/itex]. Then [itex]f[/itex] is necessarily continuous when restricted to [itex]A = f^{-1}(\{0\})[/itex] and [itex]B = f^{-1}(\{1\})[/itex] respectively, and clearly [itex][0,1] = A \cup B[/itex]. Can [itex]f[/itex] be continuous on [itex][0,1][/itex]?
I don't really understand how unioning two numbers 0 and 1 makes a closed interval.
 
  • #6
mathwonk
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By one definition of continuity, it suffices show that the inverse image C of an open set of Y under f, is open in X. By hypothesis its restriction, i.e. intersection, with both A and B are open. So we want to show that if X = AunionB, and both A and B are open, and if CmeetA is open in A, and CmeetB is open in B, then C is itself open in X. Fortunately, since A and B are open in X, being open in A or B is the same as being open in X. Then use the formula C = CmeetX = Cmeet(AunionB) = (CmeetA) union (CmeetB).

Similarly, continuity of f is equivalent to showing that the inverse image D of a closed set in Y, is closed in X. We are given that DmeetA and Dmeet B, are closed in A and B respectively, Again, being closed in a closed subset of X is the same as being closed in X........

If it is any consolation to you, presumably as a newbie, I admit that even as a seasoned old professional, I found this tedious and had to use pen and paper.

However, as an aid to future problems, continuity is a "local" property, which means if it is true on each set of an open cover, then it is true. So the case where A and B are open, is "obvious" to someone who knows that. It had not occurred to me that it is also true for A and B closed, and presumably this is because we need the closed cover to be finite, i.e. we need to use the fact that a finite union of closed sets is closed, whereas any union of open sets is open.

i.e. I would guess that the correct generalization of this problem is to show that if f is continuous when restricted to each set of any open cover of X, then f is continuous on X, and if f is continuous on each set of any finite closed cover of X, then f is continuous on X. You might try showing these when you finish the current problem. This would be more useful later on than just the result stated in the problem.
 
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