Is an open set finite or infinite

Closed sets

Since the open sets are not closed with the formation of a complement, we obtain a new type of set through the formation of the complement:

definition(completed)

An A ⊆ ℝ is called completedif ℝ - A is open.

Preferred letters for closed sets are A, B, C (the "C", which is reminiscent of "closed", is also used for the Cantor set, which we will define below). Next is also F for French. "Fermé" common.

The empty set and ℝ are closed because ℝ and ∅ are open. They thus form examples of sets that are both closed and open. Beyond that there are no further such sets: If P ⊆ ℝ is open and closed, then P = ∅ or P = ℝ. We shall use this fact in the next chapter for a topological proof of the intermediate value theorem.

Every closed interval [a, b] is closed, since its complement is the union of two open sets. In addition, the intervals of the form] −∞, b] and [a, ∞ [are closed. Every finite set is closed, and the sets ℕ and ℤ are closed too. While the open sets only have the powers 0 and “uncountable”, the closed sets can be finite, countable, infinite or uncountable. This is a first example of the remarkable fact that closed sets are more complicated than open sets. And nontrivial questions arise such as:

Is every uncountable closed set A ⊆ ℝ equal to ℝ?

The answer is “yes”, but in contrast to the analogous question for the open sets, the proof is no longer easy.

The dual properties of the open sets apply to the closed sets:

sentence(Closeness properties of closed sets)

Let I be a set and let A.i ⊆ ℝ closed for all i ∈ I. Then:

(a)

i ∈ I A.i is closed.

(b)

If I is finite, then ⋃i ∈ I A.i completed.

The second statement is generally no longer correct for countable unions:

example

The set P = {1 / n | n ≥ 1} is the countable union of the closed sets An = {1 / n}, n ≥ 1, but P is not closed. The set “lacks” the point 0. The set P ∪ {0}, on the other hand, is closed, because the complement of this set breaks down into the open intervals

] −∞, 0 [,  ] 1, ∞ [,  ] 1/2, 1 [,  ] 1/3, 1/2 [,  …

In this context we recall the concept of the accumulation point of a set:

definition(Accumulation point of a set)

Let P ⊆ ℝ and p ∈ ℝ. Then p is called a Cluster point of P, if for all neighborhoods U of p holds:

(U - {p}) ∩ P is not empty. (Cluster point condition)

In fact, a neighborhood U of an accumulation point p of P always contains an infinite number of points of P, so that “U ∩ P is infinite” instead of “U ∩ P contains a point different from p” can be used in the definition. But if one wants to show that a point p ∈ ℝ is an accumulation point of P, then it suffices to find a single x different from p for every (without restriction open) neighborhood U of p, which is in P ∩ U. The condition of the definition can later be adopted unchanged for topological spaces.

We emphasize again that accumulation points may or may not belong to P. The zero is an accumulation point of] 0, 1 [and of [0, 1]. In general, a point p is an accumulation point of P - {p} if and only if p is an accumulation point of P ∪ {p}.

We now show that the complement formation of the open sets can be formulated with the help of accumulation points.

sentence(Accumulation point characterization of the seclusion)

Let A ⊆ ℝ. Then the following statements are equivalent:

(b)

For all p ∈ ℝ the following applies: If p is an accumulation point of A, then p ∈ A.

proof

(a) implies (b): Let p be an accumulation point of A. For all neighborhoods U of p then U ∩ A is not empty, i.e. U is not a subset of ℝ - A. Since ℝ - A is open, we have p ∉ ℝ - A and thus p ∈ A.

(b) implies (a): Let p ∈ ℝ - A. Then there is a neighborhood U of p with U ∩ A = ∅, since otherwise p would be an accumulation point of A and p ∈ A would hold. But U ∩ A = ∅ is equivalent to U ⊆ ℝ - A. So ℝ - A is a neighborhood of p. Since this holds for all p ∈ ℝ - A, ℝ - A is open and therefore A closed.

A set is therefore closed if and only if it contains all of its accumulation points. The following variant of the characterization theorem for consequences is easy to see:

sentence(Characterization of the consequences of the closed sets)

Let A ⊆ ℝ. Then the following statements are equivalent:

(b)

Every Cauchy sequence in A converges in A, i.e. i.e., is (xn)n ∈ ℕ a Cauchy sequence with xn ∈ A for all n, then limn xn ∈ A.

Historically, the closed sets were even examined before the open sets. The term “completed”, which goes back to Georg Cantor, is motivated by the two characterization sentences. Completed quantities are completed with the formation of accumulation points and limit values. Today, the open sets (or, equivalently, the environments) are usually at the top in topology. As we have already indicated and will see again in the discussion of the Cantor set, they are simpler than the closed sets. Since the open sets are the complements of the closed sets, one can in principle also begin with the closed sets. Let's sit

𝒰 = {U ⊆ ℝ | U open}, 𝒜 = {A ⊆ ℝ | A completed},

then 𝒰 = {ℝ - A | A ∈ 𝒜}, 𝒜 = {ℝ - U | U ∈ 𝒰}. However, it does not hold that 𝒰 = ℘ (ℝ) - 𝒜 = {X ⊆ ℝ | X ∉ 𝒜}, and likewise 𝒜 = ℘ (ℝ) - 𝒰 is not correct, consider the interval] 0, 1], which belongs neither to 𝒰 nor to 𝒜. The area ℘ (ℝ) - (𝒰 ∪ 𝒜) of the subsets of ℝ that are neither open nor closed, we will examine more closely in the outlook.