Let a, b ∈ R, and suppose that for every ε > 0, we have a ≤ b + ε. Show that a ≤ b.
Question
Let a, b ∈ R, and suppose that for every ε > 0, we have a ≤ b + ε. Show that a ≤ b.
Solution
Sure, here is the proof:
Assume for contradiction that a > b. Then, let ε = (a - b) / 2. This is a positive number because we assumed a > b.
By our assumption, we know that a ≤ b + ε. Substituting ε, we get a ≤ b + (a - b) / 2. Simplifying this, we get a ≤ (2b + a - b) / 2, which simplifies to a ≤ (a + b) / 2.
However, this contradicts our assumption that a > b, because (a + b) / 2 is strictly greater than b (since a > b).
Therefore, our assumption that a > b must be false. Hence, a ≤ b.
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