Real Analysis True/False Quiz: Open Sets, Closed Sets

The statement is true. There is no sequence in the set Q of all rational numbers such that the set of accumulation points Q' equals R, the set of all real numbers.

In other words, there is no sequence of rational numbers that, when its limit points are considered, covers all real numbers. This is because the set of rational numbers Q is countable, while the set of real numbers R is uncountable. Countable sets like Q have fewer elements than uncountable sets like R, so there is no way to construct a sequence of rational numbers that includes all real numbers as its limit points.

A nonempty set with no isolated point does not necessarily have to be a closed set. A closed set is a set that contains all of its limit points. While a nonempty set with no isolated point does not have any isolated points, it does not guarantee that it contains all of its limit points. Therefore, it is possible for a nonempty set with no isolated point to be an open set.

It is possible to construct a set A⊆R such that the set of all accumulation points A' equals the open interval (0, 1). One way to construct such a set is to consider the set A defined as follows:
A = (1/2, 1) ∪ (1/3, 1/2) ∪ (1/4, 1/3) ∪ (1/5, 1/4) ∪ ...
In other words, A is the union of open intervals, each of which is (1/(n+1), 1/n) for n = 2, 3, 4, 5, ... and so on.
The set of accumulation points A' of this set A is indeed the open interval (0, 1). This is because every real number x in the open interval (0, 1) is an accumulation point of A, as it belongs to infinitely many of the open intervals in the construction of A. Additionally, there are no other accumulation points of A outside the interval (0, 1).

The statement is false because there does exist a countable subset A of R (the set of real numbers), such that A' (the set of limit points of A) is equal to R. One example of such a subset is the set of rational numbers Q, which is countable and has all real numbers as its limit points. Therefore, the correct answer is false.

A nonempty set with no interior points and no isolated points can indeed exist. An example is the set of all rational numbers within an interval on the real number line. This set has no interior points because it is not open in the standard topology, and it has no isolated points because between any two rationals, there is always another rational. Therefore, the statement is true.

The statement is true. There is no set with an uncountable number of isolated points. In mathematical terms, a set with isolated points is a set where each point in the set is not an accumulation point of the set. In other words, each point is surrounded by some open neighborhood that contains no other points from the set.
If a set has an uncountable number of isolated points, it would mean that each of these isolated points has its own separate open neighborhood, and since there are uncountably many of them, it would imply that the set itself is uncountable. However, this contradicts the fact that sets with isolated points are countable because each isolated point can be associated with a rational number from a countable set.
Therefore, there cannot be a set with an uncountable number of isolated points.

The open interval (0, 1) can be expressed as a union of closed sets because we can define a sequence of closed intervals [1/n, 1-1/n], where n is a positive integer. As n approaches infinity, the union of these closed intervals covers the entire open interval (0, 1). Therefore, the statement is true.

The statement is false because it is possible to have a set S such that the sequence of derived sets S', S'', S''', ... has all distinct terms. For example, consider the set S = {1, 2, 3}. The first derived set S' would be {2, 3}, the second derived set S'' would be {3}, and so on. Each derived set in this sequence is distinct from the previous one.

The statement is false. The intersection of an infinite sequence of closed sets may be empty. For example, consider the sequence of closed sets in R^2 defined by Fn = {(x, y) | ||(x, y)|| >=n} for n = 1, 2, 3, ... The intersection of all these sets is the empty set, as there is no point in R^2 that satisfies the condition for all values of n. Therefore, the statement is false.