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tjackson3
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Homework Statement
Let [itex]E,E'[/itex] be metric spaces, [itex]f:E\rightarrow E'[/itex] a function, and suppose that [itex]S_1,S_2[/itex] are closed subsets of [itex]E[/itex] such that [itex]E = S_1 \cup S_2[/itex]. Show that if the restrictions of [itex]f[/itex] to [itex]S_1,S_2[/itex] are continuous, then [itex]f[/itex] is continuous. Also, if the restriction that [itex]S_1,S_2[/itex] are closed is dropped, does the assertion still hold? Explain.
Homework Equations
Definition of continuity of a function from [itex](E,d)[/itex] to [itex](E',d')[/itex] at a point [itex]p_0[/itex]: A function is continuous at [itex]p_0[/itex] if for any [itex]\epsilon > 0[/itex] and [itex]p \in E[/itex], there exists [itex]\delta > 0[/itex] such that if [itex]d(p,p_0) < \delta[/itex], [itex] d'(f(p),f(p_0)) < \epsilon[/itex].
The Attempt at a Solution
Let [itex]p, p_0 \in E[/itex] and let [itex]\epsilon > 0[/itex]. Then there are two cases to consider:
1.) Since in [itex]S_1,S_2[/itex], for any [itex]\epsilon > 0[/itex], there exists [itex]\delta > 0[/itex] such that if [itex]p \in S_1[/itex] (or equivalently [itex]S_2[/itex]), if [itex]d(p,p_0) < \delta, d'(f(p),f(p_0)) < \epsilon[/itex] by the definition of continuity. If, for the given choice of [itex]\epsilon[/itex], the corresponding [itex]\delta[/itex] is such that [itex]B_{\delta}^E(p_0) = B_{\delta}^{S_1}(p_0)[/itex] (i.e. if the ball of radius [itex]\delta[/itex] is entirely contained in [itex]S_1[/itex]), then by hypothesis, [itex]f[/itex] is continuous at [itex]p_0[/itex]. The same logic applies for [itex]S_2[/itex].
2.) Now suppose that the corresponding [itex]\delta[/itex] is such that [itex]B_{\delta}^E(p_0) \neq B_{\delta}^{S_1}(p_0)[/itex]. In other words, some of the elements of the ball computed in [itex]E[/itex] are in [itex]S_1[/itex], while some are in [itex]S_2[/itex].
I'm not sure where to go from here. I have an idea (just separating into two separate balls those elements of [itex]S_1,S_2[/itex] and applying the hypothesis), but this idea does not use the fact that the two sets are closed.
As for the final part of the question, if both sets are open, then f is continuous (due to a theorem involving the preimage of a function being open implying that f is continuous), but I feel like if one of the sets is open, while the other is closed, the assertion wouldn't hold. I think figuring out how to involve the fact that the sets are closed into the main part of the proof would go a long way to figuring this out, but I'm not sure.
Thanks so much for your help!