Defn Suppose (X,d) is a metric space and A is a subset of X. 1.5 Limit Points and Closure As usual, let (X,d) be a metric space. Theorem In a any metric space arbitrary intersections and finite unions of closed sets are closed. If a subset of a metric space is not closed, this subset can not be sequentially compact: just consider a sequence converging to a point outside of the subset! Proposition A set O in a metric space is open if and only if each of its points are interior points. Let S be a closed subspace of a complete metric space X. Metric Spaces A metric space is a set X that has a notion of the distance d(x,y) between every pair of points x,y ∈ X. A metric space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X. Theorem 9.6 (Metric space is a topological space) Let (X,d)be a metric space. Definition Let E be a subset of a metric space X. A point p is a limit point of the set E if every neighbourhood of p contains a point q ≠ p such that q ∈ E. Theorem Let E be a subset of a metric space … A subset is called -net if A metric space is called totally bounded if finite -net. In most cases, the proofs First, if pis a point in a metric space Xand r2 (0;1), the set (A.2) Br(p) = fx2 X: d(x;p)
0. Set N of all natural numbers: No interior point. 1. The closure of A, denoted by A¯, is the union of Aand the set of limit points of A, A¯ = … Proof Exercise. Suppose that A⊆ X. Definition. Interior, Closure, Exterior and Boundary Let (X;d) be a metric space and A ˆX. A metric space (X,d) is said to be complete if every Cauchy sequence in X converges (to a point in X). Example: Any bounded subset of 1. The point x o ∈ Xis a limit point of Aif for every neighborhood U(x o, ) of x o, the set U(x o, ) is an infinite set. 2. Set Q of all rationals: No interior points. Definition 3. We do not develop their theory in detail, and we leave the verifications and proofs as an exercise. Theorem 4. It depends on the topology we adopt. Let be a metric space. FACTS A point is interior if and only if it has an open ball that is a subset of the set x 2intA , 9">0;B "(x) ˆA A point is in the closure if and only if any open ball around it intersects the set x 2A , 8">0;B "(x) \A 6= ? Whole of N is its boundary, Its complement is the set of its exterior points (In the metric space R). (0,1] is not sequentially compact … This distance function :×→ℝ must satisfy the following properties: Metric Spaces, Open Balls, and Limit Points DEFINITION: A set , whose elements we shall call points, is said to be a metric space if with any two points and of there is associated a real number ( , ) called the distance from to . A metric space X is compact if every open cover of X has a finite subcover. In the standard topology or $\mathbb{R}$ it is $\operatorname{int}\mathbb{Q}=\varnothing$ because there is no basic open set (open interval of the form $(a,b)$) inside $\mathbb{Q}$ and $\mathrm{cl}\mathbb{Q}=\mathbb{R}$ because every real number can be written as the limit of a sequence of rational numbers. A set Uˆ Xis called open if it contains a neighborhood of each of its Definition 1.14. Proof. 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