Uncertainty is the only certainty there is, and knowing how to live with insecurity is the only security, John Allen Paulos
Definition. A point z0∈ S is an interior point of a subset S ⊆ ℂ if ∃R > 0 (meaning, there exist a positive radius R) such that DR(z0) ⊆ S. This means that there is an open disc around z0 that lies entirely within S (a neighborhood exists fully within the set).
Definition. A set S is an open set if all its points are interior points. Formally, ∀z0 ∈ S, ∃R > 0 : (such that) DR(z0) ⊆ S. In other words, an open set contains all their interior points, i.e., every point has a neighborhood entirely within the set.
Alternative definition. A set A ⊆ ℂ is open if for any arbitrary element of the set ∀z ∈ A, there is a radius ∃r > 0 (possibly depending on z) such that an open disk or ball of radius r centered at it is completely contained in the set, B(z; r) ⊆ A.
Examples and counterexamples:
B(a; r) is an open set. Proof (Figure 5).
∀z ∈ B(a; r), |z - a| < r ↭ 0 < r - |z -a|. Let δ be such that 0 < δ < r - |z -a|, we aim to demonstrate that B(z, δ) ⊆ B(a; r).
∀w ∈ B(z, δ) ⇒ |w -z| < δ ⇒ |w -z + z -a| ≤ |w -z| + |z -a| < δ + |z -a| < r ⇒ z ∈ B(z, r) ∎
A1 ∩ A2 is an open set.
Let A1 and A2 be open sets in ℂ. By definition, for every z∈A1, there exists a radius r1 > 0 such that B(z; r1) ⊆ A:. Similarly, for every z∈A2, there exists a radius r2 > 0 such that B(z; r2) ⊆ A2.
Let z∈A1∩A2. Since z∈A1, there exists r1 > 0 such that B(z; r1) ⊆ A1. Similarly, since z∈A2, there exists r2 > 0 such that B(z; r2) ⊆ A2
Let r be the smaller of the two radii r1 and r2. Then: B(z; r) ⊆ B(z;r1) ⊆ A1 and B(z;r) ⊆ B(z; r2) ⊆ A2. Since B(z; r) is contained in both A1 and A2, it is also contained in their intersection: B(z; r) ⊆ B(z; r1) ∩ B(z; r2) ∎
{Sα}α∈A an arbitrary collection of open sets in ℂ, then $\cup_{\alpha ∈ A} S_{\alpha}$ is also an open set.
Proof.
Suppose $z \in \cup_{\alpha ∈ A} S_{\alpha} \leadsto z \in S_{\alpha}$ for some α ∈ A. Since Sα is an open set, there is an r > 0 such that B(z; r) ⊆ Sα ⊆ $\cup_{\alpha ∈ A} S_{\alpha}$ ∎
Definition. A point z0 is an exterior point of S ⊆ ℂ if ∃R > 0 (meaning, there exist a positive radius) such that DR(z0) contains no points of S ↭ ∃ε > 0 s.t. B(z; ε) ∩ S = ∅. In other words, DR(z0) is entirely contained within the complement of S (ℂ - S), DR(z0) ⊆ ℂ - S.
Every interior point is a limit point, but every limit point need not to be an interior point.
An interior point of a set S is a point that has a neighborhood entirely contained in S. Since every neighborhood of an interior point contains infinitely many points of S, it is also a limit point of S. However, a limit point of S is a point where every neighborhood contains at least one point of S other than itself, e.g., 1 is a limit point of S = [0,1], but not an interior point.
Isolated points are never interior points and interior points are never isolated points.
An isolated point of a set S is a point x∈S such that there exists a neighborhood of x containing no other points of S. By definition, an isolated point cannot be an interior point because it lacks a neighborhood entirely contained in S. Conversely, an interior point has a neighborhood entirely contained in S, meaning it cannot be isolated.
Every boundary point of a set S (e.g., S = {3} ∪ B(0, 1)) is an isolated point (e.g., {3}) or a limit point (every point of the circle is both a limit point and a boundary point {z: |z| = 1}).
Every limit point might not be a boundary point.
A limit point of S is a point where every neighborhood contains at least one point of S other than itself. However, a limit point does not have to lie on the boundary of S, e.g., every point in S = (0, 1) is a limit point, but none of them are boundary points (the boundary points are 0 and 1).
Theorem. If A and B are open sets in the complex plane (or any metric space), then their union (A ∪ B) and intersection (A ∩ B) are also open sets.
Suppose A and B are open sets. This means for any arbitrary point z in A ∪ B, there exists an open disk centered at z that is entirely contained within A ∪ B.
By the definition of union, this means that z ∈ A or z ∈ B (or both).
Case 1: z ∈ A. Since A is open, there exists an open disk Dₐ(z) centered at z such that Dₐ ⊆ A. Because A ⊆ A ∪ B, it follows that Dₐ(z) ⊆ A ∪ B.
Case 2: z ∈ B. Since B is open, there exists an open disk DB(z) centered at z such that DB ⊆ B. Because B ⊆ A ∪ B, it follows that DB(z) ⊆ A ∪ B.
In either case, we have found an open disk centered at z that is contained within A ∪ B. Since z was an arbitrary point in A ∪ B, this proves that A ∪ B is open.
Suppose A and B are open sets. We aim to show that A ∩ B is open. This means for any point z in A ∩ B, there exists an open disk centered at z that is entirely contained within A ∩ B.
Let z be an arbitrary point in A ∩ B, this means that z ∈ A and z ∈ B. Since z ∈ A and A is open, there exists an open disk Dₐ centered at z with radius RA > 0 such that DA(z) ⊆ A. Similarly, since z ∈ B and B is open, there exists an open disk DB(z) centered at z with radius RB > 0 such that DB(z) ⊆ B.
Now, choose R = min(RA, RB). Since both εₐ and εB are positive, ε is also positive. Consider the open disk D centered at z with radius ε:
DR(z) = {w ∈ ℂ | |w - z| < R}
Because R ≤ RA, DR(z) ⊆ DA(z) ⊆ A, hence DR(z) ⊆ A. Similarly, because R ≤ RB, D ⊆ DB(z) ⊆ B, so we have DR(z) ⊆ B, and taking both facts into account DR(z) ⊆ A ∩ B.
Since we have found an open disk D centered at z that is entirely contained within A ∩ B, and z was an arbitrary point in A ∩ B, we have proven that A ∩ B is open.
We need to show that for every point z₀ in S, there exists an open disk centered at z₀ that is entirely contained within S.
Let z₀ be an arbitrary point in S. This means that z₀ can be written as z₀ = x₀ + iy₀, where x₀ > 2.
Choose a radius R. We need to find a radius R > 0 such that the open disk centered at z₀ with radius R, denoted by DR(z₀) = {z ∈ ℂ | |z - z₀| < R} ⊆ S
Let’s choose R = x₀ - 2. Since x₀ > 2, R is strictly greater than 0, satisfying the condition for a radius.
We aim to prove that DR(z₀) ⊆ S. Let z be an arbitrary point in DR(z₀). This means |z - z₀| < R. We want to show that z is also in S.
Let z = x + iy. Then: |z - z₀| = |(x + iy) - (x₀ + iy₀)| = |(x - x₀) + i(y - y₀)| = $\sqrt{(x - x₀)² + (y - y₀)²}$
Since |z - z₀| < R, we have: $\sqrt{(x - x₀)² + (y - y₀)²} < ε$
Squaring both sides (since both sides are positive): (x - x₀)² + (y - y₀)² < R²
Because squares are non-negative, we know that: (x - x₀)² < R², and taking the square root of both sides: |x - x₀| < R ↔ -R < x - x₀ < R ↔[Adding x₀ to all sides:] x₀ - R < x < x₀ + R ↔[Now, substitute our choice of R = x₀ - 2:] x₀ - (x₀ - 2) < x < x₀ + (x₀ - 2) ↔ 2 < x < 2x₀ - 2. More importantly, we have shown that x > 2, so z ∈ S
Since x > 2, z = x + iy is in S. We have shown that if z ∈ DR(z₀), then z is also in S. Therefore, DR(z₀) ⊆ S. Since z₀ was an arbitrary point in S, we have shown that for every point in S, there exists an open disk centered at that point that is entirely contained within S. This is the definition of an open set. Therefore, S is an open set.