# Is the proof correct? Either state that it is, or circle the first error and explain what is incorrect about it.

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− 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)| < e/2 + e/2 = e. So in all cases, |h(x) − h(y)| < e , as needed. |

Is the proof correct? Either state that it is, or circle the first error and explain what is incorrect about it. If the proof is not correct, can it be fixed to prove the claim 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] g(x) for x ∈ [1, 2) , is uniformly continuous on (0, 2). Proof: Let e > 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)| < e/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)| < e/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)| < e/2 < e. If x, y ∈ [1, 2), then |x − y| < δ ≤ δ2 and so |h(x) − h(y)| = |g(x) − g(y)| < e/2 < e. 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)| < e/2 + e/2 = e. So in all cases, |h(x) − h(y)| < e , as needed.

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