Problem 2. Let f : R → R be a function. Suppose that f is increasing: for any x, y ∈ R,if x ≤ y then f (x) ≤ f (y). Prove that limx→a− f (x) exists for any a ∈ R. (Hint: What shouldthe value of this limit be? Try sup{f (x) : x < a}.)
Question
Problem 2. Let f : R → R be a function. Suppose that f is increasing: for any x, y ∈ R,if x ≤ y then f (x) ≤ f (y). Prove that limx→a− f (x) exists for any a ∈ R. (Hint: What shouldthe value of this limit be? Try sup{f (x) : x < a}.)
Solution
To prove that the limit of f(x) as x approaches a from the left exists for any a in R, we can use the concept of the least upper bound (supremum).
Step 1: Define the set S = {f(x) : x < a}. This set includes all values of f(x) for x less than a.
Step 2: Since f is increasing, all elements in S are less than or equal to f(a). Therefore, f(a) is an upper bound of S.
Step 3: By the least upper bound property of real numbers, S has a supremum. Let's denote it as L = sup S.
Step 4: We need to show that for any ε > 0, there exists a δ > 0 such that if 0 < a - x < δ, then |f(x) - L| < ε. This will prove that the limit of f(x) as x approaches a from the left is L.
Step 5: Since L = sup S, for any ε > 0, there exists an element in S, say f(b), such that L - ε < f(b) ≤ L. Since f is increasing, for x in (b, a), we have f(b) ≤ f(x) ≤ f(a) ≤ L.
Step 6: Therefore, if we choose δ = a - b, then for any x in (a - δ, a), we have |f(x) - L| < ε. This completes the proof.
So, the limit of f(x) as x approaches a from the left exists and is equal to the supremum of the set of all f(x) for x less than a.
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