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Set inclusion: intervals and the empty set

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    Solution steps

      Sets · Parameters

      Question

      Let A = {x | −1 ≤ x ≤ 1} and B = {x | a − 1 ≤ x ≤ 2a − 1}. If B ⊆ A, find all real values of a.

      Set inclusion and intervals on a number lineA is the fixed interval [−1, 1]. B is empty for a < 0, and is a non-empty interval inside A for 0 ≤ a ≤ 1. The combined result is a ≤ 1.-2-1012xA = [−1, 1]① When B = ∅:a − 1 > 2a − 1 ⇒ a < 0 (valid)② When B ≠ ∅:−1 ≤ a−1 and 2a−1 ≤ 1 ⇒ 0 ≤ a ≤ 1Union of the cases: a ≤ 1, or (−∞, 1]
      B is empty when a < 0. For 0 ≤ a ≤ 1, B is an interval within [−1, 1]. The union of both cases is a ≤ 1.

      Step-by-step solution

      1. Start with the definition of a subset

      Every element of B must belong to A = [−1, 1]. The empty set is a subset of every set, so it must be considered separately.

      2. Consider B = ∅

      B is empty when its left bound exceeds its right bound: a − 1 > 2a − 1, giving a < 0. Every such value satisfies B ⊆ A.

      3. Consider a non-empty B

      Non-emptiness requires a ≥ 0. Its endpoints must lie within A: a − 1 ≥ −1 and 2a − 1 ≤ 1. Together these give 0 ≤ a ≤ 1.

      4. Take the union of the cases

      Combining a < 0 with 0 ≤ a ≤ 1 gives a ≤ 1, or (−∞, 1].

      The parameter range is a ≤ 1.

      Which method does this problem use?

      Check the empty-set case first. For a non-empty interval, convert inclusion into endpoint inequalities. Take the union of all valid cases and check boundary values.