Final answer:
To prove that a set {x:f(x)>α} is open, we show that around any point in the set there exists an interval entirely contained within the set. This confirms the openness as per the definition of an open set in real numbers.
Step-by-step explanation:
The question asks to prove that for any real number α, the set {x:f(x)>α} is open. We say a set S is open in the context of real numbers if, for every point x in S, there exists an ε > 0 such that the interval (x-ε, x+ε) is entirely contained within S. To prove that the given set is open, consider any point x₀ inside the set where f(x₀) > α. By the definition of inequality, there exists a positive number δ such that f(x₀) - α > δ. Now, due to the continuity of f, there will be a corresponding ε > 0 for which any x in the open interval (x₀-ε, x₀+ε) will satisfy f(x) > α-δ which is greater than α. Therefore, this interval lies completely within the set {x:f(x)>α}, confirming the set is open.