R is open because any of its points have at least one neighborhood (in fact all) included in it; R is closed because any of its points have every neighborhood having non-empty intersection with R (equivalently punctured neighborhood instead of neighborhood).

The rational numbers are “closed” under addition, subtraction, and multiplication. Irrational numbers are “not closed” under addition, subtraction, multiplication or division.

From Rational Numbers under Addition form Abelian Group, (Q,+) forms a group. Thus: ∀a,b∈Q:a+(−b)∈Q. Therefore rational number subtraction is closed.

Under Addition (Subtraction): By definition, a rational number can be expressed as a fraction with integer values in the numerator and denominator (denominator not zero). Thus, adding two rational numbers produces another rational number. Rationals are closed under addition (subtraction).

In any topological space X, the empty set is open by definition, as is X. Since the complement of an open set is closed and the empty set and X are complements of each other, the empty set is also closed, making it a clopen set. The closure of the empty set is empty.

1. An open interval (0, 1) is an open set in R with its usual metric. Proof.
2. Let X = [0, 1] with its usual metric (which it inherits from R).
3. A set like {(x, y)
4. Any metric space is an open subset of itself.
5. In a discrete metric space (in which d(x, y) = 1 for every x.

A subset S of R3 is said to be open if for every point (x,y,z) ∈ S there is an open ball B such that (x,y,z) ∈ B ⊆ S. Definition Let A be a subset of R2. Let A be a subset of R3. The complement of A, denoted Ac, is the set Ac = {(x,y,z) ∈ R3 | (x,y,z) /∈ A}.

In our class, a set is called “open” if around every point in the set, there is a small ball that is also contained entirely within the set. If we just look at the real number line, the interval (0,1)—the set of all numbers strictly greater than 0 and strictly less than 1—is an open set.

Singleton sets are open because {x} is a subset of itself. There are no points in the neighborhood of x.

Thus since every singleton is open and any subset A is the union of all the singleton sets of points in A we get the result that every subset is open. Since all the complements are open too, every set is also closed. Since all inverse images are open, every function from a discrete space is continuous.

Given a singleton {x}, consider y in its complement, i.e. y = x. Then there is a neighborhood V about y that does not contain x. Thus, y is an interior point of the complement of {x}. Since y = x was arbitrary, this shows the complement of {x} is open, hence the singleton {x} is closed.

In Shilov’s book, ” Elementary Real and Complex Analysis”, there is the theorem: If M is a complete metric space consisting of only countably many points, then M has an isolated point. In his proof, he stated: Every one-element subset of M is closed.