Is the proof for this claim correct? State that it is correct, or circle the first error you see if it isn't, and explain if it can be corrected to be proven true.
Claim: If f: (0,1] → R and g: [1,2) → R are uniformly continuous on their domains, and f(1) =g(1), then the function h: (0,2) → R, defined by h(x) =f(x) for x ∈ (0,1] and h(x)=g(x) for x ∈ [1,2), is uniformly continuous on (0,2).
Proof:
Let ε >0.
Since f is uniformly continuous on (0,1], there exists δ1 > 0 such that if x,y ∈ (0,1] and |x−y|< δ1, then |f(x)−f(y)|< ε/2.
Since g is uniformly continuous on [1,2), there exists δ2 > 0 such that if x,y ∈ [1,2) and |x−y| < δ2, then |g(x)−g(y)| < ε/2.
Let δ= min{δ1, δ2}.
Now suppose x,y ∈ (0,2) with x < y and |x−y|< δ.
If x,y ∈ (0,1], then |x−y| < δ ≤ δ1 and so |h(x)−h(y)| = |f(x)−f(y)| < ε/2 < ε.
If x,y ∈ [1,2), then |x−y| < δ ≤ δ2 and so |h(x)−h(y)| = |g(x)−g(y)| < ε/2 < ε.
If x ∈ (0,1) and y ∈ (1,2), then |x−1| < |x−y| < δ ≤ δ1 and |1−y| < |x−y| < δ ≤ δ2 so that |h(x)−h(y)| = |f(x)−f(1) + g(1)−g(y)| ≤ |f(x)−f(1)|+|g(1)−g(y)| < ε/2 +ε/2 =ε.
So in all cases, |h(x)−h(y)| < ε, as needed.
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