Prove that the set (a, infinity) has no minimum, by ending up with contradictory results.
To prove that the set (a, ∞) has no minimum, we need to show that there is no element x in the set (a, ∞) such that x ≤ y for all y in the set (a, ∞).
To do this, let's assume that there exists an element x in the set (a, ∞) that is a minimum.
By definition, x is a minimum if and only if x ≤ y for all y in the set (a, ∞).
However, since the set (a, ∞) contains all real numbers greater than a, we can always find a number that is greater than x, let's call it z. Then, we have z ∈ (a, ∞) and z > x.
This contradicts the assumption that x is the minimum because we found a number z in the set (a, ∞) that is greater than x.
Therefore, we have reached a contradiction, and we can conclude that the set (a, ∞) has no minimum.
By showing the contradiction, we have proven that the set (a, ∞) has no minimum element.