0590313

UNIVERSITY OF YORK

BA, BSc and MMath Examinations 2006 MATHEMATICS Metric Spaces

Time Allowed: 1\frac{1}{2} hours.

Answer \underline{three} questions. Please write your answers in ink; pencil is acceptable for graphs and diagrams. Do not use red ink. Standard calculators will be provided. The marking scheme shown on each question is indicative only.

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0590313

1 (of 4). Suppose X is a non-empty set and that d : X × X → ℝ. What properties must d satisfy, in order for it to be a metric on X? If x_{0} ∈ X and r > 0, define the open ball B(x_{0}, r). [8]

Show that the function defined by

d_{0}(x, y) = \begin{cases}1 & if x ≠ y \\ 0 & if x = y \end{cases}

is a metric (the discrete metric) on X. [5]

If d is a metric on X and A ⊆ X, define the interior int(A), the exterior ext(A) and the boundary ∂A in the metric space (X, d). Show that A and its complement A^{c} have the same boundary. [8]

Define the terms open and closed, as applied to the set A in (X, d). Show that, under the discrete metric, all sets are both open and closed, and have empty boundary.

[9]

2 (of 4). Suppose (X, d) is a metric space, (x_{n})_{n∈ℕ} is a sequence in X and x ∈ X. What does it mean to say that x_{n} converges to x as n → ∞? Show that if, as n → ∞, x_{n} → x and x_{n} → y then x = y. [10]

Suppose a sequence (f_{n})_{n∈ℕ} converges to a limit f in the metric space C[a, b] (continuous, real-valued functions on the interval [a, b]), with the uniform metric

d_{∞}(f, g) = sup_{t∈[a,b]} |f(t) − g(t)| = max_{t∈[a,b]} |f(t) − g(t)|.

Show that f_{n} also converges pointwise to f; that is, that for each t ∈ [a, b] we have f_{n}(t) → f(t) in ℝ. [4]

Consider the following sequences in C[0, 1]:

g_{n}(t) = \frac{sin(nt)}{n};
h_{n}(t) = \frac{nt}{1 + (nt)^{2}} .

Indicate briefly why both of these sequences converge pointwise to 0, and determine whether or not each one converges to 0 in the uniform metric. [16]

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0590313

3 (of 4). Suppose (X, d) is a metric space and f : X → X is a function. What does it mean to say that a real number k ≥ 0 is a Lipschitz constant for f? Show that if f is a Lipschitz function (that is, one for which a Lipschitz constant exists) then f is continuous. [10]

If X is the interval [a, b] with the standard metric, f is differentiable on [a, b] and f′ is continuous on [a, b], use the Mean Value Theorem to show that the smallest Lipschitz constant for f is

max_{t∈[a,b]} |f′(t)|.  [5]

State without proof the Contraction Mapping Theorem [6] and use it to show that there is exactly one solution in the interval [0, 1] to the equation t = 1 − cos(1 − t). [6]

Starting from t_{0} = 0.25, use a calculator to find the next three steps in the iteration converging to this solution. [3]

4 (of 4). What is meant by a subsequence of a sequence (x_{n})_{n∈ℕ} in a metric space (X, d)? Show that if x_{n} → x ∈ X as n → ∞ then any subsequence of (x_{n})_{n∈ℕ} also converges to x. [9]

In ℝ with the standard metric, give examples of:

(a) a sequence with no convergent subsequence;

(b) a divergent sequence with a convergent subsequence

(and indicate why your examples have these properties). [8]

What does it mean to say that a metric space (X, d) is sequentially compact? Show that the following spaces are not sequentially compact:

(a) the open interval (0, 1) ⊆ ℝ with the standard metric;

(b) the set

{f : [0, 1] → ℝ : f is continuous and |f(t)| ≤ 1 for all t ∈ [0, 1]}

with the uniform metric d_{∞} (defined in Question 2). [13]

Page 3 (of 3) End of examination.

SOLUTIONS 0590313

1. The three properties of a metric are:

M1 d(x, y) ≥ 0 for all x, y ∈ X and d(x, y) = 0 if and only if x = y [figure: 2 Marks]

M2 d(x, y) = d(y, x) for all x, y ∈ X [figure: 2 Marks]

M3 d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X. [figure: 2 Marks]

The ball is defined by

B(x_{0}, r) = {x ∈ X : d(x_{0}, x) < r}.  [figure: 2 Marks]

M1 and M2 are clear from the definition. [figure: 1 Mark] M3 is clear in the case x = y, since then the LHS is zero and the RHS is nonnegative. In the case that x = y, so the LHS is 1, we must have either x ≠ z or z ≠ y, so the RHS is either 1 or 2, showing that M3 is true in this case too.

[figure: 4 Marks]

The interior of A is the set

int(A) = {x ∈ X : B(x, r) ⊆ A for some r > 0}  [figure: 2 Marks]

The exterior of A is the set

ext(A) = {x ∈ X : B(x, r) ⊆ A^{c} for some r > 0}  [figure: 2 Marks]

The boundary of A is the set

∂A = {x ∈ X : for all r > 0, B(x, r) intersects both A and A^{c}} [figure: 2 Marks]

A and A^{c} have the same boundary because the definition of ∂A is symmetric in A and A^{c}. [figure: 2 Marks]

A is open if it contains none of its boundary (A ∩ ∂A = ∅), [figure: 2 Marks] and closed if it contains all of its boundary (∂A ⊆ A). [figure: 2 Marks]

In the discrete metric d_{0}, we have

B(x_{0}, 1) = {x ∈ X : d_{0}(x_{0}, x) < 1} = {x_{0}}

If x ∈ A then B(x, 1) = {x} ⊆ A and if x ∈ A^{c} then B(x, 1) = {x} ⊆ A. It follows that int(A) = A, that ext(A) = A^{c} and that ∂A = ∅. [figure: 3 Marks] Since the boundary is empty, we have both A ∩ ∂A = ∅ and ∂A ⊆ A, so A is both open and closed. [figure: 2 Marks] Remarks. All familiar stuff. The definition of open and closed is unusual, of course, but this is the way I did it in lectures. If someone uses the more conventional approach, then of course they’ll get full credit.

[figure]

Total: 30 Marks

2. x_{n} converges to x as n → ∞ if for any ε > 0 there exists N ∈ ℕ such that if n > N then d(x_{n}, x) < ε. [figure: 3 Marks]

If both x_{n} → x and x_{n} → y then for any ε > 0 there exist M, N ∈ ℕ such that if n > M then d(x_{n}, x) < ε/2 and if n > N then d(x_{n}, y) < ε/2. If n > max(M, N) then

d(x, y) ≤ d(x, x_{n}) + d(x_{n}, y) < \frac{ε}{2} + \frac{ε}{2} = ε

since this is true for all ε > 0, d(x, y) ≤ 0. But d(x, y) ≥ 0 by definition, so d(x, y) = 0 and hence x = y. [figure: 7 Marks]

For any given t ∈ [a, b], we have

|f_{n}(t) − f(t)| ≤ sup_{t∈[a,b]} |f_{n}(t) − f(t)| = d_{∞}(f_{n}, f)

It follows that if f_{n} → f in the uniform metric, then f_{n}(t) → f(t) for each t. [figure: 4 Marks]

Since | sin(nt)|/n ≤ 1/n → 0, we have g_{n}(t) → 0 for any t; also

h_{n}(t) = \frac{1/(nt)}{1/(nt)^{2} + 1} → 0

for any t. [figure: 4 Marks] Calculate:

d_{∞}(g_{n}, 0) = sup_{t∈[0,1]} |f_{n}(t) − 0|

= sup_{t∈[0,1]} \frac{| sin(nt)|}{n}

≤ \frac{1}{n}

→ 0 as n → ∞

which shows that g_{n} → 0 in the metric d_{∞}. [figure: 5 Marks] Notice that h_{n}(t) = h_{1}(nt), and choose t_{0} ∈ (0, 1), so h_{1}(t_{0}) > 0. Since h_{n}(t_{0}/n) = h_{1}(t_{0}), we have

d_{∞}(h_{n}, 0) = sup_{t∈[0,1]} |h_{n}(t) − 0|

= sup_{t∈[0,1]} h_{n}(t)

≥ h_{n}(t_{0}/n) = h_{1}(t_{0})

This shows that d_{∞}(h_{n}, 0) ↛ 0 as n → ∞, so h_{n} does not converge to 0 in the metric d_{∞}. [figure: 7 Marks]

Remarks. All familiar ideas; but of course these particular examples are unseen.

[figure]

Total: 30 Marks

3. A Lipschitz constant is a constant k ≥ 0 with the property that d(f(x), f(y)) ≤ kd(x, y) for all x, y ∈ X. [figure: 3 Marks]

If f has a Lipschitz constant of 0, then it is evidently constant, and hence continuous. If f has a Lipschitz constant k > 0, then for any ε > 0 we let δ = ε/k, so if d(x, y) < δ then d(f(x), f(y)) ≤ kd(x, y) < kδ = ε, showing that f is uniformly continuous on X.

Let

k = max_{t∈[a,b]} |f′(t)|.

If s, t ∈ [a, b] and s ≠ t then, by the mean value theorem,

|f(s) − f(t)| = |f′(τ)||s − t| ≤ k|s − t|

where τ is some number between s and t. If s = t then |f(s)−f(t)| = 0 = k|s−t|, so k is indeed a Lipschitz constant for f. [figure: 7 Marks] To show that k is the minimal Lipschitz constant, choose some t_{0} such that |f′(t_{0})| = k. This means that

\frac{|f(t) − f(t_{0})|}{|t − t_{0}|} → k

as t → t_{0}. In particular, for any ε > 0 we can choose t such that

\frac{|f(t) − f(t_{0})|}{|t − t_{0}|} > k − ε

This shows that k−ε is not a Lipschitz constant for f, so k is the minimal Lipschitz constant. [figure: 5 Marks] The contraction mapping theorem states that if (X, d) is a complete metric space and F : X → X has a Lipschitz constant k < 1 then F has a unique fixed point in X. Moreover, for any x_{0} ∈ X the iterates (F^{(n)}(x_{0}))_{n∈ℕ} converge to the fixed point as n → ∞. [figure: 6 Marks]

Let f(t) = 1−cos(1−t). If t ∈ [0, 1] then cos(t) ∈ [0, 1], so 1−cos(1−t) ∈ [0, 1] also. This shows that f leaves invariant the complete metric space [0, 1].

We have f′(t) = sin(1 − t). On the interval [0, 1], sin is non-negative and increasing, so

max_{t∈[0,1]} | sin(1 − t)| = sin(1)

This shows that f has Lipschitz constant sin(1) < 1, and hence by the contraction mapping theorem a unique fixed in point in [0, 1]. [figure: 6 Marks]

Starting from t_{0} = 0.25, we have t_{1} = f(t_{0}) ≈ 0.268, t_{2} = f(t_{1}) ≈ 0.256 and t_{3} = f(t_{2}) ≈ 0.264. [figure: 3 Marks]

Remarks. Again, all familiar but an unseen example of a contraction.

[figure]

Total: 30 Marks

4. If (n_{k})_{k∈ℕ} is a strictly increasing sequence of natural numbers, then the sequence (x_{n_{k}})_{k∈ℕ} is a subsequence of (x_{n})_{n∈ℕ}. [figure: 3 Marks] Suppose x_{n} → x as n → ∞ and we have a subsequence (x_{n_{k}})_{k∈ℕ}. Then for any ε > 0 there exists N ∈ ℕ such that if n > N then d(x_{n}, x) < ε. Since (n_{k})_{k∈ℕ} is strictly increasing in ℕ, we have n_{k} ≥ k for all k. If follows that if k > N then n_{k} > N, so d(x_{n_{k}}, x) < ε, showing that x_{n_{k}} → x as k → ∞ [figure: 6 Marks]

(a) Let x_{n} = n. Then any subsequence (x_{n_{k}}) has x_{n_{k}} > k, so x_{n_{k}} → ∞ as k → ∞. This shows that there is no convergent subsequence. [figure: 4 Marks]

(b) Let x_{n} = (−1)^{n}. Then x_{2n} = 1 and x_{2n+1} = −1. These subsequences are trivially convergent, and the fact that they have different limits shows that (x_{n})_{n∈ℕ} itself is divergent. [figure: 4 Marks]

A metric space (X, d) is called sequentially compact if every sequence in X has a convergenbt subsequence. [figure: 3 Marks]

(a) Let x_{n} = 1/n. Since x_{n} → 0 as n → ∞ in ℝ, every subsequence of (x_{n})_{n∈ℕ} also converges to 0 in ℝ. If a subsequence were to converge in (0, 1), then it would also converge in ℝ, and would thus have two distinct limits in ℝ, which is impossible; it follows that no subsequence can converge in (0, 1).

[figure: 5 Marks]

(b) Let f_{n}(t) be a continuous function on [0, 1] which is zero outside the interval I_{n} = [1/(n+1), 1/n] and has a maximum value of 1 over I_{n}. Then, if m ≠ n |f_{m} − f_{n}| = |f_{n}| on I_{n}, so d_{∞}(f_{m}, f_{n}) ≥ 1. This shows that (f_{n})_{n∈ℕ} has no Cauchy subsequence, and hence no convergent subsequence. [figure: 5 Marks]

Remarks. Nothing at all new here, but the technical details are tricky.

[figure]

Total: 30 Marks