Advanced Calculus (II)
WEN-CHINGLIEN
Department of Mathematics National Cheng Kung University
2009
Ch10: Metric Spaces
10.1: Introduction
Definition (10.1)
A metric space is a set X together with a function ρ :X × X → R (call the metric of X ), that satisfies the following properties for all x , y , z ∈ X :
Positive Definite: ρ(x , y ) ≥ 0 with ρ(x , y ) = 0 if and only if x = y ,
Symmetric: ρ(x , y ) = ρ(y , x ),
Triangle Inequality: ρ(x , y ) ≤ ρ(x , z) + ρ(z, y ).
(Notice that by definition, ρ(x , y ) is finite-valued for all x , y ∈ X .)
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Example (10.2)
For each n ∈N, Rnis a metric space with metric ρ(x, y) = kx − yk. (We shall call this the usual metric on Rn. Unless specified otherwise, we shall always use the usual metric onRn.)
Example (10.3)
R is a metric space with metric
σ(x , y ) = 0 x = y 1 x 6= y . (This metric is called the discrete metric.)
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Example (10.6)
Let C[a, b] represent the collection of continuous f : [a, b] →R and
kf k := sup
x ∈[a,b]
|f (x)|.
Then ρ(f , g) := kf − gk is a metric on C[a, b].
Proof.
By the Extreme Value Theorem,kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f ,and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b]. Thus ρ is positive definite. Since ρ is obviously symmetric, it remains to verify the triangle inequality. But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
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Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b].By definition, kf k ≥ 0 for all f , and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b].Thus ρ is positive definite. Since ρ is obviously symmetric, it remains to verify the triangle inequality. But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f ,and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b]. Thus ρ is positive definite.Since ρ is obviously symmetric, it remains to verify the triangle inequality. But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
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Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f , and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b].Thus ρ is positive definite. Since ρ is obviously symmetric,it remains to verify the triangle inequality. But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f , and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b]. Thus ρ is positive definite.Since ρ is obviously symmetric, it remains to verify the triangle inequality.But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
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Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f , and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b]. Thus ρ is positive definite. Since ρ is obviously symmetric,it remains to verify the triangle inequality. But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f , and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b]. Thus ρ is positive definite. Since ρ is obviously symmetric, it remains to verify the triangle inequality.But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
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Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f , and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b]. Thus ρ is positive definite. Since ρ is obviously symmetric, it remains to verify the triangle inequality. But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f , and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b]. Thus ρ is positive definite. Since ρ is obviously symmetric, it remains to verify the triangle inequality. But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
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Proof.
By the Extreme Value Theorem, kf k is finite for each f ∈ C[a, b]. By definition, kf k ≥ 0 for all f , and kf k = 0 if and only if f (x ) = 0 for every x ∈ [a, b]. Thus ρ is positive definite. Since ρ is obviously symmetric, it remains to verify the triangle inequality. But
kf + gk = sup
x ∈[a,b]
|f (x) + g(x)|
≤ sup
x ∈[a,b]
|f (x)| + sup
x ∈[a,b]
|g(x)| = kf k + kgk .
Definition (10.7)
Let a ∈ X and r > 0. The open ball (in X ) with center a and radius r is the set
Br(a) := {x ∈ X : ρ(x , a) < r },
and the closed ball (in X ) with center a and radius r is the set
{x ∈ X : ρ(x, a) ≤ r }.
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Definition (10.8)
(i) A set V ⊆ X is said to be open if and only if for every x ∈ V there is an ε > 0 such that the open ball Bε(x ) is contained in V .
(ii) A set E ⊆ X is said to be closed if and only if Ec :=X \E is open.
Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a).Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r . Thus by Definition 10.7, y ∈ Br(a). In particular, Bε(x ) ⊆ Br(a). Similarly, we can show that
{x ∈ X : ρ(x, a) > r } is also open. Hence, every closed ball is closed.
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Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.)If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r .
Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a).Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ),then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r . Thus by Definition 10.7, y ∈ Br(a). In particular, Bε(x ) ⊆ Br(a). Similarly, we can show that
{x ∈ X : ρ(x, a) > r } is also open. Hence, every closed ball is closed.
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Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.)If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r .
Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ),then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x) + ρ(x, a) < ε + ρ(x, a) = r . Thus by Definition 10.7, y ∈ Br(a). In particular, Bε(x ) ⊆ Br(a). Similarly, we can show that
{x ∈ X : ρ(x, a) > r } is also open. Hence, every closed ball is closed.
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Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r .
Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x) + ρ(x, a) < ε + ρ(x, a) = r . Thus by Definition 10.7, y ∈ Br(a). In particular, Bε(x ) ⊆ Br(a). Similarly, we can show that
{x ∈ X : ρ(x, a) > r } is also open. Hence, every closed ball is closed.
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Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r .
Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r . Thus by Definition 10.7, y ∈ Br(a). In particular, Bε(x ) ⊆ Br(a).Similarly, we can show that
{x ∈ X : ρ(x, a) > r } is also open. Hence, every closed ball is closed.
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Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r .
Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r . Thus by Definition 10.7, y ∈ Br(a). In particular, Bε(x ) ⊆ Br(a).Similarly, we can show that
{x ∈ X : ρ(x, a) > r } is also open. Hence, every closed ball is closed.
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Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r .
Remark (10.9)
Every open ball is open,and every closed ball is closed.
Proof.
Let Br(a) be an open ball. By definition, we must prove that given x ∈ Br(a) there is an ε > 0 such that
Bε(x ) ⊆ Br(a). Let x ∈ Br(a) and set ε = r − ρ(x , a). (Look at Figure 8.5 to see why this choice of ε should work.) If y ∈ Bε(x ), then by the triangle inequality, assumption, and the choice of ε,
ρ(y , a) ≤ ρ(y , x ) + ρ(x , a) < ε + ρ(x , a) = r . Thus by Definition 10.7, y ∈ Br(a). In particular, Bε(x ) ⊆ Br(a). Similarly, we can show that
{x ∈ X : ρ(x, a) > r } is also open. Hence, every closed ball is closed.
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Remark (10.11)
In an arbitrary metric space, the empty set ∅ and the whole space X are both open and closed.
Definition (10.13)
Let {xn} be a sequence in a metric space X .
(i) {xn} converges (in X ) if there is a point a ∈ X (called the limit of xn) such that for every ε > 0 there is an N ∈ N such that
n ≥ N implies ρ(xn,a) < ε.
(ii) {xn} is Cauchy if for every ε > 0 there is an N ∈ N such that
n, m ≥ N implies ρ(xn,xm) < ε.
(iii) {xn} is bounded if there is an M > 0 and a, b ∈ X such that ρ(xn,b) ≤ M for all n ∈N.
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Theorem (10.14)
Let X be a metric space.
(i) A sequence in X can have at most one limit.
(ii) If xn ∈ X converges to a and {xnk} is any subsequence of {xn}, then xnk converges to a as k → ∞.
(iii) Every convergent sequence in X is bounded.
(iv) Every convergent sequence in X is Cauchy.
Theorem (10.14)
Let X be a metric space.
(i) A sequence in X can have at most one limit.
(ii) If xn ∈ X converges to a and {xnk} is any subsequence of {xn}, then xnk converges to a as k → ∞.
(iii) Every convergent sequence in X is bounded.
(iv) Every convergent sequence in X is Cauchy.
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Theorem (10.14)
Let X be a metric space.
(i) A sequence in X can have at most one limit.
(ii) If xn ∈ X converges to a and {xnk} is any subsequence of {xn}, then xnk converges to a as k → ∞.
(iii) Every convergent sequence in X is bounded.
(iv) Every convergent sequence in X is Cauchy.
Theorem (10.14)
Let X be a metric space.
(i) A sequence in X can have at most one limit.
(ii) If xn ∈ X converges to a and {xnk} is any subsequence of {xn}, then xnk converges to a as k → ∞.
(iii) Every convergent sequence in X is bounded.
(iv) Every convergent sequence in X is Cauchy.
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Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a. By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V .Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a). By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a). Then V is an
Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a.By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V . Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a).By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a). Then V is an open set that contains a; hence, by hypothesis, there is an N ∈N such that n ≥ N implies xn∈ V . In particular,
ρ(xn,a) < ε for all n ≥ N.
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Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a. By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V .Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a). By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a). Then V is an
Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a. By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V . Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a).By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a).Then V is an open set that contains a; hence, by hypothesis, there is an N ∈N such that n ≥ N implies xn∈ V . In particular,
ρ(xn,a) < ε for all n ≥ N.
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Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a. By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V . Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a). By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a). Then V is an
Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a. By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V . Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a). By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a).Then V is an open set that contains a; hence, by hypothesis, there is an N ∈N such that n ≥ N implies xn∈ V . In particular,
ρ(xn,a) < ε for all n ≥ N.
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Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a. By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V . Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a). By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a). Then V is an
Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a. By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V . Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a). By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a). Then V is an open set that contains a; hence, by hypothesis, there is an N ∈N such that n ≥ N implies xn∈ V . In particular,
ρ(xn,a) < ε for all n ≥ N.
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Remark (10.15)
Let xn ∈ X . Then xn → a as n → ∞ if and only if for every open set V that contains a, there is an N ∈N such that n ≥ N implies xn∈ V .
Proof.
Suppose that xn → a, and let V be an open set that contains a. By Definition 10.8, there is an ε > 0 such that Bε(a) ⊆ V . Given this ε, use Definition 10.13 to choose an N ∈N such that n ≥ N implies xn∈ Bε(a). By the choice of ε, xn∈ V for all n ≥ N.
Conversely, let ε > 0 and set V = Bε(a). Then V is an
Theorem (10.16)
Let E ⊆ X . Then E is closed if and only if the limit of every convergent sequence xk ∈ E satisfies
k →∞lim xk ∈ E.
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Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open.Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E
N. Now 1/k → 0 as k → ∞,
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec.Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E and ρ(xk,x ) < 1/k for all k ∈N. Now 1/k → 0 as k → ∞, so it follows from the Squeeze Theorem (these are real sequences) that ρ(xk,x ) → 0 as k → ∞; i.e., xk → x as k → ∞. Thus by hypothesis, x ∈ E , a contradiction.
WEN-CHINGLIEN Advanced Calculus (II)
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open.Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E .If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E
N. Now 1/k → 0 as k → ∞,
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed,then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E and ρ(xk,x ) < 1/k for all k ∈N. Now 1/k → 0 as k → ∞, so it follows from the Squeeze Theorem (these are real sequences) that ρ(xk,x ) → 0 as k → ∞; i.e., xk → x as k → ∞. Thus by hypothesis, x ∈ E , a contradiction.
WEN-CHINGLIEN Advanced Calculus (II)
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E .If E is not closed, then by Remark 10.11,E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E
N. Now 1/k → 0 as k → ∞,
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed,then by Remark 10.11, E 6= X ,and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E and ρ(xk,x ) < 1/k for all k ∈N. Now 1/k → 0 as k → ∞, so it follows from the Squeeze Theorem (these are real sequences) that ρ(xk,x ) → 0 as k → ∞; i.e., xk → x as k → ∞. Thus by hypothesis, x ∈ E , a contradiction.
WEN-CHINGLIEN Advanced Calculus (II)
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11,E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E
N. Now 1/k → 0 as k → ∞,
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X ,and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec.Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E and ρ(xk,x ) < 1/k for all k ∈N. Now 1/k → 0 as k → ∞, so it follows from the Squeeze Theorem (these are real sequences) that ρ(xk,x ) → 0 as k → ∞; i.e., xk → x as k → ∞. Thus by hypothesis, x ∈ E , a contradiction.
WEN-CHINGLIEN Advanced Calculus (II)
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . ..Then xk ∈ E
N. Now 1/k → 0 as k → ∞,
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec.Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E and ρ(xk,x ) < 1/k for all k ∈N.Now 1/k → 0 as k → ∞, so it follows from the Squeeze Theorem (these are real sequences) that ρ(xk,x ) → 0 as k → ∞; i.e., xk → x as k → ∞. Thus by hypothesis, x ∈ E , a contradiction.
WEN-CHINGLIEN Advanced Calculus (II)
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . ..Then xk ∈ E
N. Now 1/k → 0 as k → ∞,
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E and ρ(xk,x ) < 1/k for all k ∈N.Now 1/k → 0 as k → ∞, so it follows from the Squeeze Theorem (these are real sequences) that ρ(xk,x ) → 0 as k → ∞;i.e., xk → x as k → ∞. Thus by hypothesis, x ∈ E , a contradiction.
WEN-CHINGLIEN Advanced Calculus (II)
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E
N. Now 1/k → 0 as k → ∞,
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E and ρ(xk,x ) < 1/k for all k ∈N. Now 1/k → 0 as k → ∞, so it follows from the Squeeze Theorem (these are real sequences) that ρ(xk,x ) → 0 as k → ∞;i.e., xk → x as k → ∞. Thus by hypothesis, x ∈ E , a contradiction.
WEN-CHINGLIEN Advanced Calculus (II)
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E
N. Now 1/k → 0 as k → ∞,
Proof.
The theorem is vacuously satisfied if E is the empty set.
Suppose hat E 6= ∅ is closed but some sequence xn ∈ E converges to a point x ∈ Ec. Since E is closed, Ec is open. Thus, by Remark 10.15, there is an N ∈N such that n ≥ N implies xn∈ Ec, a contradiction.
Conversely, suppose that E is a nonempty set such that every convergent sequence in E has its limit in E . If E is not closed, then by Remark 10.11, E 6= X , and by
definition, Ec is nonempty and not open. Thus, there is at least one point x ∈ Ec such that no ball Br(x ) is contained in Ec. Let xk ∈ B1/k(x ) ∩ E for k = 1, 2, . . .. Then xk ∈ E and ρ(xk,x ) < 1/k for all k ∈N. Now 1/k → 0 as k → ∞, so it follows from the Squeeze Theorem (these are real sequences) that ρ(xk,x ) → 0 as k → ∞; i.e., xk → x as k → ∞. Thus by hypothesis, x ∈ E , a contradiction.
WEN-CHINGLIEN Advanced Calculus (II)
Definition (10.19)
A metric space X is said to be complete if and only if every Cauchy sequence xn∈ X converges to some point in X.
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv,{xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical, {xn} is Cauchy in X . Since X is complete, it follows that xn → x, as n → ∞, for some x ∈ X . But E is closed, so x must belong to E . Thus E is complete by definition.
WEN-CHINGLIEN Advanced Calculus (II)
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges.By Theorem 10.14iv, {xn} is Cauchy.Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical,
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv,{xn} is Cauchy. Since E is complete,it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical, {xn} is Cauchy in X . Since X is complete, it follows that xn → x, as n → ∞, for some x ∈ X . But E is closed, so x must belong to E . Thus E is complete by definition.
WEN-CHINGLIEN Advanced Calculus (II)
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy.Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E .Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical,
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete,it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical, {xn} is Cauchy in X . Since X is complete, it follows that xn → x, as n → ∞, for some x ∈ X . But E is closed, so x must belong to E . Thus E is complete by definition.
WEN-CHINGLIEN Advanced Calculus (II)
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E .Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E .Since the metric on X and E are identical,
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical, {xn} is Cauchy in X .Since X is complete, it follows that xn → x, as n → ∞, for some x ∈ X . But E is closed, so x must belong to E . Thus E is complete by definition.
WEN-CHINGLIEN Advanced Calculus (II)
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E .Since the metric on X and E are identical,
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical, {xn} is Cauchy in X .Since X is complete, it follows that xn → x, as n → ∞, for some x ∈ X .But E is closed, so x must belong to E . Thus E is complete by definition.
WEN-CHINGLIEN Advanced Calculus (II)
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical,
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical, {xn} is Cauchy in X . Since X is complete, it follows that xn → x, as n → ∞, for some x ∈ X .But E is closed, so x must belong to E .Thus E is complete by definition.
WEN-CHINGLIEN Advanced Calculus (II)
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical,
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical, {xn} is Cauchy in X . Since X is complete, it follows that xn → x, as n → ∞, for some x ∈ X . But E is closed, so x must belong to E .Thus E is complete by definition.
WEN-CHINGLIEN Advanced Calculus (II)
Theorem (10.21)
Let X be a complete metric space and E be a subset of X . Then E (as a subspace) is complete if and only if E (as a subset) is closed.
Proof.
Suppose that E is complete and xn ∈ E converges. By Theorem 10.14iv, {xn} is Cauchy. Since E is complete, it follows from Definition 10.19 that the limit of {xn} belongs to E . Thus, by Theorem 10.16, E is closed.
Conversely, suppose that E is closed and xn ∈ E is Cauchy in E . Since the metric on X and E are identical,
Thank you.
WEN-CHINGLIEN Advanced Calculus (II)
