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Let (X,T) be a topological space. Show that f:X→ℜ is lower semicontinuous if {x∈ X∣ f(x) >r }∈ T for every rational number r. You can proceed with showing that a function f is lower semicontinuous under the given condition.

User Pcoronel
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A function f from a topological space X to the reals is lower semicontinuous if the set of points where f(x) is greater than any rational r is open in X. The given condition is equivalent to the definition of lower semicontinuity, satisfying the open neighborhood condition around any point in X.

A function f:X→ℝ is said to be lower semicontinuous if for every x in X and every ε > 0, there exists an open neighborhood U of x such that f(y) > f(x) - ε for all y in U. The question states that for a topological space (X,T), the function f is lower semicontinuous if x ∈ X is an open set in T for every rational number r.

This implies that for any point x∈X and any ε > 0, one can find a rational number r such that f(x) > r > f(x) - ε, and f(y) > r will be an open neighborhood of x meeting the condition for lower semicontinuity. Therefore, the function f is lower semicontinuous because it satisfies the open neighborhood condition for all points in its domain.

User Tom Studee
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