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Continuity on Restrictions Implies Continuity Everywhere

  1. Oct 23, 2008 #1
    The problem statement, all variables and given/known data
    Let (E, m) and (E', m') be metric spaces, let A and B be closed subsets of E such that their union equals E, and let f be a function from E into E'. Prove that if f is continuous on A and on B, then f is continuous on E.

    The attempt at a solution
    I have approached this problem in three ways, but in each I get stuck.

    First approach: Let S' be a closed subset of E' and let S = f-1(S'). If I can demonstrate the S is closed, then f is continuous on E. I wrote S as [tex](S \cap A) \cup (S \cap B)[/tex] and attempted to demonstrate that each intersection is closed. However, each intersection is closed if S is closed. This is a circular argument so it won't work. What else can I do here?

    Second approach: Let p belong to E and let e > 0. Then either p belongs to A or p belongs to B. If the former, then there is a d(A) > 0 such that m'(f(p), f(q)) < e for all q in A satisfying m(p, q) < d(A). Now suppose there is a s in E such that m(p, s) < d(A) but m'(f(p), f(s)) ≥ e. This s must be in B. Since f is continuous on B, there is a d(B) > 0 such that m'(f(s), f(q)) < e for all in B satisfying m(s, q) < d(B). I don't know how to proceed from here.

    Third approach: Suppose by way of contradiction that f is not continuous on E. Then for some p in E, for some e > 0, for all d > 0, there is a q in E such m(p, q) < d but m'(f(p), f(q)) ≥ e. I stopped here when I realized this is proceeding similarly as the second approach.
     
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  3. Oct 23, 2008 #2

    Dick

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    I don't see anything circular about your first approach. The inverse image of each closed subset is closed. The union is f^(-1)(S'). The union of two closed sets is closed. Hence f^(-1)(S') is closed if S' is closed. Sounds pretty airtight to me.
     
  4. Oct 23, 2008 #3
    That only works if f is continuous in E, which is what I'm trying to prove.

    The intersection of S and A is closed if both S and A are closed. I already know that A is closed. That means I must only worry about S, but that's what I was worrying about from the beginning. In other words: S is closed iff the intersection of S and A (and S and B) is closed iff S is closed. This is the circular argument I'm referring to.
     
  5. Oct 23, 2008 #4

    Dick

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    But S=f^(-1)(S')=(f|A)^(-1)(S') union (f|B)^(-1)(S') where f|A and f|B are the restrictions. (f|A)^(-1)(S') is a closed subset of A (technically, 'in the topology of E restricted to A') since it's continuous ON A. There is a little bit to worry about here. For example, if O is an open set in E, a closed subset of O is not necessarily a closed subset of E. Take the example of E=R (reals), O=(-1,1), and A=[0,1). A is closed in O. But A is not closed in R. Do you see what I mean? But a closed subset of a closed set is closed.
     
    Last edited: Oct 23, 2008
  6. Oct 23, 2008 #5
    This never occurred to me. Jeez...

    Right.

    Yes. A may be written as the intersection of O with [0,1], which is closed in R, hence A is closed in O.

    You mean: if A is closed in B and B is closed in E, then A is closed in E? In other words, "closedness" is transitive?
     
    Last edited: Oct 23, 2008
  7. Oct 23, 2008 #6

    Dick

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    Yes, exactly. "closedness" is transitive. It's easy to prove.
     
  8. Oct 23, 2008 #7
    Maybe for you but no for me: Let A', B' be the complements of A, B in E. Then B' is a subset of A' since A is a subset of B. Pick a p in A'. If p is in B', then we can find an open ball centered at p entirely contained in B' and consequently in A'. But what if p is not in B', i.e. what if p is in B?
     
  9. Oct 23, 2008 #8

    Dick

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    Ok, pick a p not in A. Either p is in B-A, in which case there is a ball around it that doesn't touch A because A IS CLOSED IN B. Or p is in E-B which means there is a ball around it that doesn't touch B because B IS CLOSED IN E. In the second case you also know the ball doesn't touch A because A is contained in B.
     
  10. Oct 23, 2008 #9
    Neat. Thank you for the perspicuous explanation.
     
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