An open set in the real numbers is a set that contains an open interval around each of its points. The set of all numbers greater than 0 is an open set because for any number in the set, we can find an open interval around that number that is entirely contained in the set.

A closed set in the real numbers is a set that contains all of its limit points. The set of all numbers greater than or equal to 0 is a closed set because all of its limit points are also in the set.

The limit point of a set is a point that is arbitrarily close to infinitely many points in the set. In this case, the set {1/n : n is a natural number} has a limit point of 0 because for any ε > 0, we can find a natural number N such that 1/N < ε, which means that there is a point in the set that is less than ε away from 0.

A function is continuous at a point if the limit of the function as x approaches that point is equal to the value of the function at that point. In this case, the limit of f(x) = x^2 as x approaches 0 is 0, and f(0) = 0, so f(x) is continuous at x = 0.

A function is discontinuous at a point if the limit of the function as x approaches that point does not exist or is not equal to the value of the function at that point. In this case, the limit of f(x) = 1/x as x approaches 0 does not exist, so f(x) is discontinuous at x = 0.

The intermediate value theorem states that if a function is continuous on a closed interval, then for any value between the minimum and maximum values of the function on that interval, there exists a point in the interval where the function takes on that value.

The intermediate value theorem states that if a function is continuous on a closed interval, then it takes on every value between its minimum and maximum values on that interval. In this case, f(x) = x^2 is continuous on [0, 1] and its minimum and maximum values on that interval are 0 and 1, respectively. Therefore, f(x) takes on every value between 0 and 1 on [0, 1].

The intermediate value theorem states that if a function is continuous on a closed interval, then it takes on every value between its minimum and maximum values on that interval. In this case, f(x) = 1/x is not continuous on [0, 1] because it has an infinite discontinuity at x = 0. Therefore, f(x) does not satisfy the intermediate value theorem on [0, 1].

A limit point of a set is a point that is arbitrarily close to infinitely many points in the set. This means that for any ε > 0, there exist infinitely many points in the set that are less than ε away from the limit point.

A set has a limit point 0 if for any ε > 0, there exist infinitely many points in the set that are less than ε away from 0. In this case, the set {1/n : n is a natural number} has a limit point 0 because for any ε > 0, we can find a natural number N such that 1/N < ε, which means that there is a point in the set that is less than ε away from 0.

A set does not have a limit point 0 if there exists an ε > 0 such that there are only finitely many points in the set that are less than ε away from 0. In this case, the set {(-1)^n : n is a natural number} does not have a limit point 0 because for ε = 1/2, there are only finitely many points in the set that are less than 1/2 away from 0.

A function is continuous at a point if the limit of the function as x approaches that point is equal to the value of the function at that point. This means that for any ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - f(a)| < ε.

A function is continuous at a point if the limit of the function as x approaches that point is equal to the value of the function at that point. In this case, the limit of f(x) = x^2 as x approaches 0 is 0, and f(0) = 0, so f(x) is continuous at x = 0.

A function is discontinuous at a point if the limit of the function as x approaches that point does not exist or is not equal to the value of the function at that point. In this case, the limit of f(x) = 1/x as x approaches 0 does not exist, so f(x) is discontinuous at x = 0.