Continuity on Restrictions Implies Continuity Everywhere

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In summary, the author has tried three different approaches to solving this problem, but none of them have worked.
  • #1
e(ho0n3
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Homework Statement
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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  • #2
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.
 
  • #3
Dick said:
The inverse image of each closed subset is closed.
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.
 
  • #4
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.
 
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  • #5
Dick said:
But S=f^(-1)(S')=(f|A)^(-1)(S') union (f|B)^(-1)(S') where f|A and f|B are the restrictions.
This never occurred to me. Jeez...

(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.
Right.

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

But a closed subset of a closed set is closed.
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?
 
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  • #6
e(ho0n3 said:
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?

Yes, exactly. "closedness" is transitive. It's easy to prove.
 
  • #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?
 
  • #8
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.
 
  • #9
Neat. Thank you for the perspicuous explanation.
 

1. What is meant by "continuity on restrictions implies continuity everywhere"?

This statement refers to the fact that if a function is continuous on a restricted domain, it will also be continuous on its entire domain. In other words, if a function has no discontinuities on a certain part of its domain, it will also have no discontinuities on its entire domain.

2. Why is this statement important in mathematics and science?

This statement is important because it allows us to analyze the continuity of a function on a smaller, more manageable domain, and then extend that analysis to the entire domain. This can simplify problem-solving and make it easier to understand the behavior of a function.

3. How is this statement related to the concept of limits?

This statement is related to limits because continuity is closely tied to the limit of a function. If a function is continuous, it means that its limit exists at every point, and this also holds true for a restricted domain. Therefore, if a function is continuous on a restricted domain, it will also have a limit at every point on its entire domain.

4. Can a function be continuous on its entire domain but not on a restricted domain?

No, this is not possible. If a function is continuous on its entire domain, it means that it has no discontinuities anywhere. Therefore, it will also be continuous on any restricted domain.

5. Are there any exceptions to this statement?

Yes, there are some cases where this statement may not hold true. For example, if a function is defined differently on different parts of its domain, it may not be continuous on a restricted domain. Additionally, in some cases, the restricted domain may not include certain points where the function is discontinuous, so it may not accurately represent the behavior of the function on its entire domain.

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