Problem 2. Let A = {an : n ∈ N} = {a1, a2, . . . } be an infinite subset of R, and supposethat no element of A is listed twice: an 6 = am for all n 6 = m. This implies that the functionf : R → R defined byf (x) ={ 1n if x = an for some n ∈ N,0 if x /∈ A,is actually a function (but you do not need to prove this). Prove that limx→x0f (x) = 0 for anyx0 ∈ R.
Question
Problem 2. Let A = {an : n ∈ N} = {a1, a2, . . . } be an infinite subset of R, and supposethat no element of A is listed twice: an 6 = am for all n 6 = m. This implies that the functionf : R → R defined byf (x) ={ 1n if x = an for some n ∈ N,0 if x /∈ A,is actually a function (but you do not need to prove this). Prove that limx→x0f (x) = 0 for anyx0 ∈ R.
Solution
To prove that limx→x0 f(x) = 0 for any x0 ∈ R, we need to show that for any ε > 0, there exists a δ > 0 such that if 0 < |x - x0| < δ, then |f(x) - 0| < ε.
Let's consider two cases:
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If x0 is not in A, then f(x) = 0 for all x in a neighborhood of x0 (except possibly at x0 itself). So, for any ε > 0, we can choose δ = 1. Then, if 0 < |x - x0| < δ, we have |f(x) - 0| = |0 - 0| = 0 < ε.
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If x0 is in A, then x0 = an for some n. Since no element of A is listed twice, f(x) = 0 for all x in a neighborhood of x0 (except possibly at x0 itself). So, for any ε > 0, we can choose δ = 1/n. Then, if 0 < |x - x0| < δ, we have |f(x) - 0| = |0 - 0| = 0 < ε.
In both cases, we have shown that for any ε > 0, there exists a δ > 0 such that if 0 < |x - x0| < δ, then |f(x) - 0| < ε. Therefore, limx→x0 f(x) = 0 for any x0 ∈ R.
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