metric spaces question?

Question

Hi I've been struggling with this problem and I would really appreciate any help or hints anybody could give me.
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Let (X,d) and (Y,p) be metric spaces and
f:X->Y is a continuous function

Suppose C is a compact subset of X such that:

the restriction f|_C: C->Y is injective and for each x in C there exists a real number r(x)>0 such that, on the open ball (neighborhood) with radius r(x) around x consists of all x' in X such that d(x,x')
i.e. B(x,r(x))={x' in X|d(x,x')
and the restriction f|_B(x,r(x)):B(x,r(x))->Y is injective.
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prove that there exists a real number R>0 such that on the open subset G of X where G consists of the Union of all open balls of radius R centered at x

[i.e. B(x,R)] for all x in C
G=U_(x in C) B(x,R)

That the restriction f|_G: G->Y is injective
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Thank you so much for taking the time to help me

Answer 1

It's no wonder you're struggling with it -- this is a hard problem. I have a solution, although it's fairly complicated and inelegant, so I can't help but think that there's a simpler solution. The solution I have relies on proving the following lemma:

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Lemma 1: If X and Y are metric spaces, f:X → Y is continuous, C is a compact subset of X and A₁ and A₂ are open subsets of X, C⊆A₁∪A₂, and f|C, f|A₁, and f|A₂ are all injective, then there is an open set U containing C such that f|U is injective.

Proof: First note that C\A₁ = C∩∁A₁, thus is the intersection of a compact set and a closed set, and is thus compact (∁ is an old symbol meaning complement -- I use it because I can't find a good way to implement the superscript c notation in unicode). Similarly, C\A₂ = C∩∁A₂, so C\A₂ is compact. Since f is continuous, f(C\A₁) and f(C\A₂) are both compact, and thus closed (since Y is metric, thus Hausdorff). Further, these sets are disjoint -- suppose that they are not. Then ∃y∈f(C\A₁) ∩ f(C\A₂), which means that ∃x₁, x₂ such that x₁∈C\A₁, x₂∈C\A₂, and f(x₁) = f(x₂) = y. Now, since f|C is injective, this means x₁ = x₂, so x₁∈C\A₂ as well. Then x₁∈(C\A₁) ∩ (C\A₂). But since C⊆A₁∪A₂, (C\A₁) ∩ (C\A₂) = ∅, so this is a contradiction. It follows that there is no element in f(C\A₁) ∩ f(C\A₂), and these sets are disjoint.

Now, since Y is metric, it is normal. f(C\A₁) and f(C\A₂) are disjoint closed sets, so there exist open sets E₁ and E₂ in Y such that f(C\A₁)⊆E₁, f(C\A₂)⊆E₂, and E₁∩E₂=∅. So now let U = (f⁻¹(E₁)∩A₂) ∪ (A₁∩A₂) ∪ (f⁻¹(E₂)∩A₁). I claim that U is open, C⊆U, and also that f|U is injective.

First, to show that U is open, note that f⁻¹(E₁) and f⁻¹(E₂) are open, since they are the preimages of open sets under a continuous function. Thus (f⁻¹(E₁)∩A₂), (A₁∩A₂), and (f⁻¹(E₂)∩A₁) are open since they are the intersections of two open sets. So finally U is open since it is the union of open sets.

To see that C⊆U, let x∈C. Then there are three possibilities -- either x∈A₁∩A₂, x∈C\A₁, or x∈C\A₂. We examine them each:

1: If x∈A₁∩A₂, then clearly x∈U.
2: If x∈C\A₁, then f(x)∈f(C\A₁) ⊆ E₁, so x∈f⁻¹(E₁). Also, since C⊆A₁∪A₂, either x∈A₁ or x∈A₂. But x∉A₁, so x∈A₂. Thus x∈f⁻¹(E₁)∩A₂, so x∈U.
3: If x∈C\A₂, then f(x)∈f(C\A₂) ⊆ E₂, so x∈f⁻¹(E₂). Also, since C⊆A₁∪A₂, either x∈A₁ or x∈A₂. But x∉A₂, so x∈A₁. Thus x∈f⁻¹(E₂)∩A₁, so x∈U

Thus in any case x∈C ⇒ x∈U, so C⊆U. Finally, to show that f|U is injective, suppose that x, y∈U and f(x) = f(y). Now, suppose that x∉A₁ and y∉A₂. Then it is easy to see from the definition of U that x∈f⁻¹(E₁) and y∈f⁻¹(E₂), so f(x)∈E₁ and f(y)∈E₂. But since f(x) = f(y), that means f(x)∈E₂ as well, which means E₁∩E₂≠∅, a contradiction. By similar logic, we can deduce that it is not the case that x∉A₂ and y∉A₁. So we thus have the following possibilities:

1: x∉A₁. In this case, we must have y∈A₂. But since x∈U⊆A₁∩A₂ (that U⊆A₁∩A₂ follows easily from the definition), it follows that x∈A₂ as well, so from the injectivity of f|A₂ and the fact that f(x) = f(y), it follows that x=y.
2: y∉A₁. In this case, we must have x∈A₂. But since y∈U⊆A₁∩A₂, it follows that y∈A₂ as well, so from the injectivity of f|A₂ and the fact that f(x) = f(y), it still follows that x=y.
3: x∈A₁ and y∈A₁. In this case it follows from the injectivity of f|A₁ that x=y.

In any case, f(x) = f(y) ⇒ x=y, so f|U is injective. Thus the lemma is proved.

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That was a hard lemma. Thankfully, the rest of the proof is merely tedious, but fairly straightforward.

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Lemma 2: If X and Y are metric spaces, f:X → Y is continuous, C is a compact subset of X and A₁, A₂, A₃... A_n are finitely many open subsets of X, C⊆A₁∪A₂∪A₃... A_n, and f|C, f|A₁, f|A₂, f|A₃... f|A_n are all injective, then there is an open set U containing C such that f|U is injective.

Proof: We proceed by induction on n. Clearly, this holds when n=1, since then A₁ itself is such an open set. Suppose this holds for some n, and consider the case of n+1 sets. Now, C\A_(n+1) is compact, f|C\A_(n+1) is injective (since f|C is), and A₁, A₂, A₃... A_n cover C\A_(n+1), so by our inductive hypothesis, there is an open set T containing C\A_(n+1) such that f|T is injective. Then f|C, f|T, f|A_(n+1) are all injective, T and A_(n+1) are open, and C⊆T∪A_(n+1), so by lemma 1, there is an open set U containing C such that f|U is injective. Thus, if lemma 2 holds for n, then it also holds for n+1, so by induction, it holds for all finite n.

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Lemma 3: If X and Y are metric spaces, f:X → Y is continuous, C is a compact subset of X, f|C is injective, and for every point x∈C, ∃r s.t. f|B(x, r) is injective, then there is an open set U containing C such that f|U is injective.

Proof: Let S={A: A is open and f|A is injective}. Now, let x∈C, then ∃r such that f|B(x, r) is injective, thus x∈B(x, r) and B(x, r)∈S. Since we can find such a set for any x∈C, S covers C, so by the compactness of C, S has a finite subcover, A₁, A₂, A₃... A_n. By the definition of S, f|A₁, f|A₂, f|A₃... f|A_n are all injective, and by hypothesis f|C is injective, so per lemma 2, there is an open set U containing C such that f|U is injective.

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Lemma 4: If C is a compact subset of a metric space X and U is an open set containing C, then ∃R>0 s.t. [x∈C]⋃B(x, R) ⊆ U.

Proof: Define g(x) = inf {d(x, y): y∉U}. Obviously, since U is open, g(x) is positive for every point c in C. Further, it is easily shown that g is continuous, so it follows that g(C) is compact in ℝ. But that implies that it must have a minimum. So let R = 1/2 min {g(x): x∈C}. Then R is positive, and if y∈[x∈C]⋃B(x, R), then there exists some x∈C such that d(x, y) < R < min {g(x): x∈C} < g(x), so y∈U. Therefore, [x∈C]⋃B(x, R) ⊆ U.

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Main theorem: If X and Y are metric spaces, f:X → Y is continuous, C is a compact subset of X, f|C is injective, and for every point x∈C, ∃r s.t. f|B(x, r) is injective, then there is an R>0 such that f|[x∈C]⋃B(x, R) is injective.

Proof: By lemma 3, find an open set U containing C such that f|U is injective. Then by lemma 4, find R>0 such that [x∈C]⋃B(x, R) ⊆ U. Then f|[x∈C]⋃B(x, R) is injective. Q.E.D.