Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Assignment 1 Solutions

Let

A

and

B

be bounded subsets of a metric space

(X, d)

such that

A \cap B \neq \emptyset .

Show that

diam (A \cup B) \leq diam (A) + diam (B) .

What can you say if

A

and

B

are disjoint?

		Solution.
	The definition of $diam (A)$ is $diam (A) = sup {d (x, y) \| x, y \in A} .$ Assume $A \subseteq X$ and $B \subseteq X$ and $A \cap B = \emptyset$ and $diam (A) < \infty and diam (B) < \infty .$ To show: $diam (A \cup B) \leq diam (A) + diam (B) .$ To show: $diam (A) + diam (B)$ is an upper bound of ${d (x, y) \| x, y \in A \cup B} .$ To show: If $x, y \in A \cup B$ then $d (x, y) \leq diam (A) + diam (B) .$ Assume $x, y \in A \cup B .$ Case 1: $x, y \in A .$ Then $d (x, y) \leq diam (A) \leq diam (A) + diam (B) .$ Case 2: $x, y \in B .$ Then $d (x, y) \leq diam (B) \leq diam (A) + diam (B) .$ Case 3: $x \in A$ and $y \in B .$ Let $z \in A \cap B .$ Then $d (x, y) \leq d (x, z) + d (z, y) \leq diam (A) + diam (B) .$ Case 4: $x \in B$ and $y \in A .$ Let $z \in A \cap B .$ Then $d (x, y) \leq d (x, z) + d (z, y) \leq diam (A) + diam (B) .$ So $diam (A) + diam (B)$ is an upper bound of ${(x, y) \| x, y \in A \cup B} .$ So $diam (A \cup B) \leq diam (A) + diam (B) .$ $□$

Let

X = C [0, 1] = {f : [0, 1] \to ℝ | f is continuous} .

The supremum metric

d_{\infty} : X \times X \to ℝ_{\geq 0}

and the

L^{1}

metric

d_{1} : X \times X \to ℝ_{\geq 0}

are defined by

\begin{array}{rcl} d_{\infty} (f, g) & = & sup {| f (x) - g (x) | | x \in [0, 1]} and \\ d_{1} (f, g) & = & \int_{0}^{1} ∣ f (x) - g (x) ∣ d x . \end{array}

Consider the sequence

{f_{1}, f_{2}, f_{3}, \dots}

X

where

f_{n} (x) = n x^{n} (1 - x)

for

0 \leq x \leq 1 .

Determine whether ${f_{n}}$ converges in $(X, d_{1}) .$
Determine whether ${f_{n}}$ converges in $(X, d_{\infty}) .$

(You may use any standard results about limits of real sequences.)

		Solution.
	The graph of $y = x - x^{2}$ $\begin{matrix} \end{matrix}$ and $y = x^{n - 1}$ $\begin{matrix} \end{matrix}$ help to determine the graph of $y = n x^{n} (1 - x) = n x^{n - 1} (x - x^{2}) .$ $\begin{matrix} \end{matrix}$ Since $lim_{n \to \infty} n x^{n - 1} = 0$ for $x \in [0, 1)$ and $n x^{n - 1} = n$ for $x = 1$ and $x - x^{2} = 0$ for $x = 1,$ then $lim_{n \to \infty} n x^{n - 1} (x - x^{2}) = 0$ for $x \in [0, 1] .$ So the pointwise limit of ${f_{n}}$ is the zero function $f : [0, 1] \to ℝ$ given by $f (x) = 0 .$ ${f_{n}}$ converges in $(X, d_{1})$ if $lim_{n \to \infty} d_{1} (f_{n}, f) = 0 .$ ${f_{n}}$ converges in $(X, d_{\infty})$ if $lim_{n \to \infty} d_{\infty} (f_{n}, f) = 0 .$ Compute: $d_{1} (f_{n}, f) .$ $\begin{array}{rcl} d_{1} (f_{n}, f) & = & \int_{0}^{1} ∣ f_{n} (x) - f (x) ∣ d x \\ = & \int_{0}^{1} (n x^{n} (1 - x) - 0) d x \\ = & \int_{0}^{1} (n x^{n} - n x^{n + 1}) d x \\ = & (\frac{n x^{n + 1}}{n + 1} - \frac{n x^{n + 2}}{n + 2}) \|_{x = 0}^{x = 1} \\ = & (\frac{n}{n + 1} - \frac{n}{n + 2}) \\ = & \frac{n}{(n + 1) (n + 2)} . \end{array}$ So $\begin{array}{rcl} lim_{n \to \infty} d_{1} (f_{n}, f) & = & lim_{n \to \infty} \frac{n}{(n + 1) (n + 2)} \\ = & lim_{n \to \infty} \frac{1}{(1 + \frac{1}{n})} \frac{1}{(n + 2)} \\ = & 1 \cdot 0 \\ = & 0 . \end{array}$ So ${f_{n}}$ converges in $(X, d_{1}) .$ Compute $d_{\infty} (f_{n}, f) :$ To compute $d_{\infty} (f_{n}, f) = sup {∣ n x^{n} (1 - x) - 0 ∣ \| x \in [0, 1]}$ find the maximum of $f_{n} (x) = n x^{n} (1 - x)$ on the interval $[0, 1] .$ This maximum occurs at $x = 0$ or $x = 1$ or at a critical point. Since $\frac{d f_{n}}{d x} = \frac{d n x^{n} (1 - x)}{d x} = n^{2} x^{n - 1} - n (n + 1) x^{n} = n x^{n - 1} (n - n x - x)$ the critical points are at $x = 0$ and $x = \frac{n}{n + 1} .$ Since $f_{n} (0) = 0$ and $f_{n} (1) = 0$ and $f_{n} (\frac{n}{n + 1}) = n {(\frac{n}{n + 1})}^{n} (1 - \frac{n}{n + 1})$ the maximum of $f_{n}$ is ${(\frac{n}{n + 1})}^{n + 1} = {(1 - \frac{1}{n + 1})}^{n + 1} .$ So $\begin{array}{rcl} d_{\infty} (f_{n}, f) & = & sup {∣ f_{n} (x) - f (x) ∣ \| x \in [0, 1]} \\ = & sup {n x^{n} (1 - x) \| x \in [0, 1]} \\ = & {(1 - \frac{1}{n + 1})}^{n + 1} \end{array}$ and $\begin{array}{rcl} lim_{n \to \infty} {(1 - \frac{1}{n + 1})}^{n + 1} & = & lim_{n \to \infty} e^{log {(1 - \frac{1}{n + 1})}^{n + 1}} \\ = & lim_{n \to \infty} e^{(n + 1) log (1 - \frac{1}{n + 1})} \\ = & lim_{n \to \infty} e^{- (n + 1) (\frac{1}{n + 1} + \frac{1}{2} {(\frac{1}{n + 1})}^{2} + \frac{1}{3} {(\frac{1}{n + 1})}^{3} + \dots)} \end{array}$ since $- log (1 - x) = \int \frac{1}{1 - x} d x = \int (1 + x + x^{2} + \dots) d x = x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \dots .$ So $lim_{n \to \infty} d_{\infty} (f_{n}, f) = lim_{n \to \infty} {(1 - \frac{1}{n + 1})}^{n + 1} = e^{- (1 + \frac{1}{2} \cdot 0 + \frac{1}{3} \cdot 0^{2} + \dots)} = e^{- 1} .$ So ${f_{n}}$ does not converge in $(X, d_{\infty}) .$ $□$

Let

X

and

Y

be topological spaces. Let

A \subseteq X

and

B \subseteq Y .

Show that

\overline{A} \times \overline{B} = \overline{A \times B} .

Solution.

To show:

(a)	$\overline{A} \times \overline{B} \subseteq \overline{A \times B} .$
(b)	$\overline{A \times B} \subseteq \overline{A} \times \overline{B} .$

(a)

Assume

(x, y) \in \overline{A} \times \overline{B} .

To show:

(x, y) \in \overline{A \times B} .

To show:

(x, y)

is a close point of

A \times B .

Let

N

be a neighbourhood of

(x, y)

X \times Y .

By the definition of the product topology on

X \times Y

there exist

N_{X},

a neighbourhood of

x

X,

and

N_{y},

a neighbourhood of

y

Y,

such that

N_{x} \times N_{y} \subseteq N .

Since

x \in \overline{A}

there exists

a \in A

with

a \in N_{x} .

Since

y \in \overline{B}

there exists

b \in B

with

b \in N_{y} .

(a, b) \in N_{x} \times N_{y} \subseteq N

and

(a, b) \in A \times B .

(x, y)

is a close point of

A \times B .

(x, y) \in \overline{A \times B} .

\overline{A} \times \overline{B} \subseteq \overline{A \times B} .

(b)

To show:

\overline{A \times B} \subseteq \overline{A} \times \overline{B} .

Assume

(x, y) \in \overline{A \times B} .

To show:

(x, y) \in \overline{A} \times \overline{B} .

To show:

x \in \overline{A}

and

y \in \overline{B} .

Let

N_{x}

be a neighbourhood of

x \in X

and let

N_{y}

be a neighbourhood of

y \in Y .

Then

N_{x} \times N_{y}

is a neighbourhood of

(x, y) \in X \times Y .

Since

(x, y)

is a close point of

A \times B,

there exists

(a, b) \in A \times B

with

(a, b) \in N_{x} \times N_{y} .

a \in N_{x}

and

b \in N_{y}

and

a \in A

and

b \in B .

x

is a close point of

A

and

y

is a close point of

B .

x \in \overline{A}

and

y \in \overline{B} .

(x, y) \in \overline{A} \times \overline{B} .

$□$

Let

(X, d)

be a metric space and let

A

be a non-empty subset of

X .

Recall that for each

x \in X,

the distance from

x

A

d (x, A) = inf {d (x, a) | a \in A} .

Prove that $\overline{A} = {x \in X | d (x, A) = 0} .$
Prove that $| d (x, A) - d (y, A) | \leq d (x, y)$ for all $x, y \in X .$ [Hint: first show that $d (x, A) \leq d (x, y) + d (y, A) .]$
Deduce the function $f : X \to ℝ$ defined by $f (x) = d (x, A)$ is continuous.
Show that if $x \notin \overline{A}$ then $U = {y \in X | d (y, A) < d (x, A)}$ is an open set in $X$ such that $\overline{A} \subset U$ and $x \notin U .$

Solution.

(a)

To show:

(aa)	${x \in X \| d (x, A) = 0} \subseteq \overline{A} .$
(ab)	$\overline{A} \subseteq {x \in X \| d (x, A) = 0} .$

(aa)

Assume

x \in X

and

d (x, A) = 0 .

To show:

x \in \overline{A} .

Let

N

be a neighbourhood of

x

X .

Then there exists

ε \in ℝ_{> 0}

such that

B (x, ε) \subseteq N .

Since

d (x, A) = inf {d (x, a) | a \in A} = 0,

there exists

a \in A

such that

d (x, a) < ε .

Then

a \in B (x, ε) \subseteq N

and

a \in A .

x

is a close point of

A .

{x \in X | d (x, A) = 0} \subseteq \overline{A} .

(ab)

To show:

\overline{A} \subseteq {x \in X | d (x, A) = 0} .

Let

x \in \overline{A} .

x

is a close point of

A .

To show:

d (x, A) = 0 .

Let

ε \in ℝ_{> 0} .

Then

B (x, ε)

is a neighbourhood of

x

X .

Since

x

is a close point of

A

there exists

a \in A

such that

a \in B (x, ε) .

d (x, a) < ε .

d (x, A) < ε

for all

ε \in ℝ_{> 0} .

d (x, A) = 0 .

x \in {x \in X | d (x, A) = 0} .

\overline{A} \subseteq {x \in X | d (x, A) = 0} .

Thus

\overline{A} = {x \in X | d (x, A) = 0} .

(b)

Assume

x, y \in X .

To show:

(ba)	$d (x, A) - d (y, A) \leq d (x, y) .$
(bb)	$- (d (x, A) - d (y, A)) \leq d (x, y) .$

(ba)

Since

d (x, A)

is a lower bound of

{d (x, a) | a \in A},

a \in A

then

d (x, A) \leq d (x, a) .

Using

d (x, a) \leq d (x, y) + d (y, a),

a \in A

then

d (x, A) \leq d (x, y) + d (y, a) .

d (x, A)

is a lower bound of

{d (x, y) + d (y, a) | a \in A} .

Since

d (x, y) + d (y, A)

is the greatest lower bound of

{d (x, y) + d (y, a) | a \in A}

then

d (x, A) \leq d (x, y) + d (y, A) .

d (x, A) - d (y, A) \leq d (x, y) .

d (y, A) - d (x, A) \leq d (y, x) = d (x, y) .

- (d (x, A) - d (y, A)) \leq d (x, y) .

d (x, A) - d (y, A) \leq d (x, y)

and

- (d (x, A) - d (y, A)) \leq d (x, y) .

∣ d (x, A) - d (y, A) ∣ \leq d (x, y) .

(c)

To show: If

ε \in ℝ_{> 0}

and

x \in X

then there exists

δ \in ℝ_{> 0}

such that if

y \in X

and

d (x, y) < δ

then

d (f (x), f (y)) < ε .

Assume

ε \in ℝ_{> 0}

and

x \in X .

To show: There exists

δ \in ℝ_{> 0}

such that if

y \in X

and

d (x, y) < δ

then

d (f (x), f (y)) < ε .

Let

δ = ε .

To show: If

y \in X

and

d (x, y) < δ

then

d (f (x), f (y)) < ε .

Assume

y \in X

and

d (x, y) < δ .

To show:

d (f (x), f (y)) < ε .

By part (b),

d (f (x), f (y)) = ∣ d (y, A) - d (x, A) ∣ \leq d (x, y) < δ = ε .

f

is continuous.

(d)

Assume

x \notin \overline{A}

and let

U = {y \in X | d (y, A) < d (x, A)} .

To show:

(da)	$x \notin U .$
(db)	$U$ is open.
(dc)	$\overline{A} \subseteq U .$

(da)	Let $D = d (x, A) .$ Since $x \notin \overline{A}$ and, by part (a), $\overline{A} = {y \in X \| d (y, A) = 0}$ then $d (x, A) \neq 0 .$ So $D \neq 0 .$ We know $U = {y \in X \| d (y, A) < D} .$ Since $d (x, A) = D,$ $x \notin U .$
(db)	Since $U = f^{- 1} (ℝ_{< D}) = f^{- 1} ((- \infty, D))$ and $f$ is continuous, then $U$ is open.
(dc)	By part (a), $\overline{A} = {y \in X \| d (y, A) = 0} \subseteq {y \in X \| d (y, A) < D} = U .$ So $\overline{A} \subseteq U .$

$□$

Determine whether the following sequences of functions converge uniformly.

$f_{n} = e^{- n x^{2}}, x \in [0, 1];$
$g_{n} = e^{- x^{2} / n}, x \in [0, 1] .$
$g_{n} = e^{- x^{2} / n}, x \in ℝ .$

Solution.

Let

(X, ρ)

be a metric space and let

f_{n} : X \to ℝ,

n \in ℤ_{> 0},

be a sequence of functions from

X

ℝ .

Assume that

f : X \to ℝ

defined by

f (x) = lim_{n \to \infty} f_{n} (x)

is well defined.
The sequence

f_{1}, f_{2}, f_{3}, \dots

converges uniformly to

f

lim_{n \to \infty} (sup {ρ (f_{n} (x), f (x)) | x \in X}) = 0 .

(a)	Define $f_{n} : [0, 1] \to ℝ$ by $f_{n} (x) = e^{- n x^{2}} = e^{- {(\sqrt{n} x)}^{2}},$ for $n \in ℤ_{> 0} .$ Then $lim_{n \to \infty} e^{- n \cdot 0^{2}} = e^{- 0} = 1 and lim_{n \to \infty} e^{- n \cdot 1^{2}} = e^{- \infty} = 0$ and if $x \in (0, 1)$ then $lim_{n \to \infty} e^{- n x^{2}} = e^{- \infty} = 0 .$ $\begin{matrix} \end{matrix}$ Let $f : [0, 1] \to ℝ$ be given by $f (x) = {\begin{cases} 1, & if x = 0, \\ 0, & if x \in (0, 1] . \end{cases}$ Let $n \in ℤ_{> 0} .$ Then $\begin{array}{rcl} sup {ρ (f_{n} (x), f (x)) \| x \in X} & = & sup ({∣ e^{- n x^{2}} - 0 ∣ \| x \in (0, 1]} \cup {∣ 1 - 1 ∣}) \\ = & sup {e^{- n x^{2}} \| x \in (0, 1]} \\ = & 1 . \end{array}$ So $lim_{n \to \infty} (sup {∣ e^{- n x^{2}} ∣ \| x \in [0, 1]}) = lim_{n \to \infty} 1 = 1 .$ So $f_{1}, f_{2}, f_{3}, \dots$ is not uniformly convergent.
(b)	Define $g_{n} : [0, 1] \to ℝ$ by $g_{n} (x) = e^{- x^{2} / n} = e^{- {(x / \sqrt{n})}^{2}},$ for $n \in ℤ_{> 0} .$ Then $lim_{n \to \infty} e^{- 0^{2} / n} = e^{- 0} = 1 and lim_{n \to \infty} e^{- 1^{2} / n} = e^{- 1 / \infty} = e^{- 0} = 1$ and if $x \in (0, 1)$ then $lim_{n \to \infty} e^{- x^{2} / n} = e^{- 0} = 1 .$ $\begin{matrix} \end{matrix}$ Let $g : [0, 1] \to ℝ$ by given by $g (x) = 1 .$ Let $n \in ℤ_{> 0} .$ Then $\begin{array}{rcl} sup {ρ (g_{n} (x), g (x)) \| x \in X} & = & sup {∣ e^{- x^{2} / n} - 1 ∣ \| x \in [0, 1]} \\ = & ∣ e^{- 1^{2} / n} - 1 ∣ \\ = & 1 - e^{- 1 / n} . \end{array}$ So $\begin{array}{rcl} lim_{n \to \infty} (sup {∣ e^{- x^{2} / n} - 1 ∣ \| x \in [0, 1]}) & = & lim_{n \to \infty} 1 - e^{- \frac{1}{n}} \\ = & 1 - e^{- 1 / \infty} \\ = & 1 - e^{0} \\ = & 1 - 1 \\ = & 0 . \end{array}$ So $g_{1}, g_{2}, g_{3}, \dots$ uniformly converges to $g .$
(c)	Define $g_{n} : ℝ \to ℝ$ by $g_{n} (x) = e^{- x^{2} / n} = e^{- {(x / \sqrt{n})}^{2}},$ for $n \in ℤ_{> 0} .$ Let $x \in ℝ .$ Then $lim_{n \to \infty} e^{- x^{2} / n} = e^{- x^{2} / \infty} = e^{- 0} = 1 .$ Let $g : ℝ \to ℝ$ be given by $g (x) = 1 .$ Let $n \in ℤ_{> 0} .$ Then $\begin{array}{rcl} sup {ρ (g_{n} (x), g (x)) \| x \in X} & = & sup {∣ e^{- x^{2} / n} - 1 ∣ \| x \in ℝ} \\ = & ∣ e^{- \infty^{2} / n} - 1 ∣ \\ = & ∣ e^{- \infty} - 1 ∣ \\ = & ∣ 0 - 1 ∣ \\ = & 1 . \end{array}$ So $lim_{n \to \infty} sup {ρ (g_{n} (x), g (x)) \| x \in X} = lim_{n \to \infty} 1 = 1 .$ So $g_{1}, g_{2}, \dots$ is not uniformly convergent.

$□$

Let

X

be the set of all real sequences with finitely many non-zero terms with the supremum metric: if

x = (x_{i})

and

y = (y_{i})

then

d (x, y) = sup {| x_{i} - y_{i} | | i \in ℤ_{> 0}} .

For each

n \in ℕ,

let

x^{n} = (1, 1 / 2, 1 / 3, \dots, 1 / n, 0, 0, \dots) .

Show that ${x^{n}}$ is a Cauchy sequence in $X .$
Show that ${x^{n}}$ does not converge to a point in $X .$ (So $X$ is not complete.)

Solution.

Let

X = {(x_{1}, x_{2}, \dots) | x_{i} \in ℝ and all but a finite number of x_{i} are zero}

and define

d : X \times X \to ℝ_{> 0}

d (x, y) = sup {∣ x_{i} - y_{i} ∣ | i \in ℤ_{> 0}} .

Let

f_{n} = (1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}, 0, 0, \dots, 0) .

(a)

Show that

{f_{n}}

is a Cauchy sequence in

X .

To show: If

ε \in ℝ_{> 0}

then there exists

N \in ℤ_{> 0}

such that if

m, n \in ℤ_{> 0}

and

m > N

and

n > N

then

d (f_{m}, f_{n}) < ε .

Assume

ε \in ℝ_{> 0} .

To show: There exists

N \in ℤ_{> 0}

such that if

m, n \in ℤ_{> 0}

and

m > N

and

n > N

then

d (f_{m}, f_{n}) < ε .

Let

N = \frac{1}{ε} .

Assume

m, n \in ℤ_{> 0},

m > N

and

n > N

and

m < n .

To show:

d (f_{n}, f_{m}) < ε .

\begin{array}{rcl} d (f_{n}, f_{m}) & = & sup {1 - 1, \frac{1}{2} - \frac{1}{2}, \dots, \frac{1}{m} - \frac{1}{m}, \frac{1}{m + 1}, \frac{1}{m + 2}, \dots, \frac{1}{n}, 0 - 0, 0 - 0, \dots} \\ = & \frac{1}{m + 1} \\ < & \frac{1}{N + 1} \\ = & \frac{1}{\frac{1}{ε} + 1} \\ = & \frac{ε}{ε + 1} \\ < & \frac{ε}{1} \\ = & ε . \end{array}

{f_{n}}

is a Cauchy sequence in

X .

(b)

The limit of

{f_{1}, f_{2}, \dots}

is the sequence

f = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots) .

k \in ℤ_{> 0}

the

k^{th}

entry of

f

\frac{1}{k}

which is not equal to

0 .

So all entries of

f

are nonzero.
So

f \notin X .

{f_{1}, f_{2}, \dots}

is a Cauchy sequence in

X

which does not converge to a point in

X .

X

is not complete.

$□$

Let

X

be a nonempty set and let

(Y, d)

be a complete metric space. Let

f : X \to Y

be an injective function and define

d_{f} (x, y) = d (f (x), f (y))

for

x, y \in X .

Explain briefly why $d_{f}$ is a metric on $X .$
Show that $(X, d_{f})$ is a complete metric space if $f (X)$ is a closed subset of $Y .$

Solution.

(a)

To show:

(aa)	If $x, y \in X$ then $d_{f} (x, y) = d_{f} (y, x) .$
(ab)	If $x \in X$ then $d_{f} (x, x) = 0 .$
(ac)	If $x, y \in X$ and $d_{f} (x, y) = 0$ then $x = y .$
(ad)	If $x, y, z \in X$ then $d_{f} (x, y) \leq d_{f} (x, z) + d_{f} (z, y) .$

(aa)	Assume $x, y \in X .$ To show: $d_{f} (x, y) = d_{f} (y, x) .$ $d_{f} (x, y) = d (f (x), f (y)) = d (f (y), d (x)) = d_{f} (y, x) .$
(ab)	Assume $x \in X .$ To show: $d_{f} (x, x) = 0 .$ $d_{f} (x, x) = d (f (x), f (x)) = 0 .$
(ac)	Assume $x, y \in X$ and $d_{f} (x, y) = 0 .$ To show: $x = y .$ Since $0 = d_{f} (x, y) = d (f (x), f (y))$ and $d$ is a metric, then $f (x) = f (y) .$ Since $f : X \to Y$ is injective and $f (x) = f (y)$ then $x = y .$
(ad)	Assume $x, y, z \in X .$ To show: $d_{f} (x, y) \leq d_{f} (x, z) + d_{f} (z, y) .$ $\begin{array}{rcl} d_{f} (x, y) & = & d (f (x), f (y)) \\ \leq & d (f (x), f (z)) + d (f (z), f (y)) \\ = & d_{f} (x, z) + d_{f} (z, y) . \end{array}$

(b)

Assume

f (x)

is a closed subset of

Y .

Since

f : X \to Y

is injective and

f : X \to f (X)

is surjective, then

f : X \to f (X)

is bijective.
Since

d_{f} (x, y) = d (f (x), f (y))

for

x, y \in X

then

f : (X, d_{f}) ⟶ (f (X), d)

is an isometry.
To show:

(X, d_{f})

is complete.
To show:

(f (X), d)

is complete.
To show: If

z_{1}, z_{2}, \dots

is a Cauchy sequence in

f (X)

then

z_{1}, z_{2}, \dots

converges with

lim_{n \to \infty} z_{n}

in f(X).
Assume

z_{1}, z_{2}, \dots

is a Cauchy sequence in

f (X) .

Then

z_{1}, z_{2}, \dots

is a Cauchy sequence in

Y .

Since

Y

is complete

z = lim_{n \to \infty} z_{n}

exists with

z \in Y .

Since

z

is a close point of

z_{1}, z_{2}, \dots,

then

z

is a close point of

f (X) .

z \in f (X) .

f (X)

is complete.
So

(X, d_{f})

is complete.

$□$

Let

f : ℝ_{\geq 0} \to ℝ_{\geq 0}

be given by

f (x) = \frac{2}{2 + x} .

Show that $f$ defines a contraction mapping $f : ℝ_{\geq 0} \to ℝ_{\geq 0} .$
Fix $x_{0} \geq 0$ and $x_{n + 1} = f (x_{n})$ for all $n \geq 0 .$ Show that the sequence ${x_{n}}$ converges and find its limit with respect to the usual metric on $ℝ .$

Solution.

(a)

A function

f : X \to X

is a contraction mapping if there exists

α \in (0, 1)

such that if

x, y \in X

then

d (f (x), f (y)) \leq α d (x, y) .

Let

f : ℝ_{\geq 0} \to ℝ_{\geq 0}

be given by

f (x) = \frac{2}{2 + x} .

Let

α = \frac{1}{2} .

To show: If

x, y \in ℝ_{> 0}

then

d (f (x), f (y)) \leq α d (x, y) .

Assume

x, y \in ℝ_{> 0} .

To show:

∣ f (y) - f (x) ∣ \leq α ∣ y - x ∣ .

\begin{array}{rcl} ∣ f (y) - f (x) ∣ & = & ∣ \frac{2}{2 + x} - \frac{2}{2 + y} ∣ \\ = & 2 ∣ \frac{2 + y - (2 + x)}{(2 + x) (2 + y)} ∣ \\ = & \frac{2}{(2 + x) (2 + y)} ∣ y - x ∣ \\ \leq & \frac{2}{4} ∣ y - x ∣ \\ = & \frac{1}{2} ∣ y - x ∣ . \end{array}

f

is a contraction mapping.

(b)

The Banach fixed point theorem gives that the sequence

x_{n + 1} = f (x_{n})

for

x_{0} \in X

converges to the (unique) fixed point of

f .

In our case the fixed point is

p \in ℝ_{> 0}

such that

p = f (p) = \frac{2}{2 + p} .

p^{2} + 2 p = 2

and

p = \frac{- 2 \pm \sqrt{4 + 4 \cdot 2}}{2} = - 1 \pm \sqrt{3} .

Since

p \in ℝ_{> 0}

then

p = - 1 + \sqrt{3} .

$□$

Let

X

be a connected topological space. Let

f : X \to ℝ

be continuous with

f (X) \subseteq ℚ .

Show that

f

is a constant function.

		Solution.
	To show: If $f$ is not a constant function then $X$ is not connected. Assume $f$ is not a constant function. Let $a, b \in f (X)$ with $a < b .$ Let $z \in ℝ,$ $z \notin ℚ$ with $a < z < b .$ Let $A = f^{- 1} ((- \infty, z))$ and $B = f^{- 1} ((z, \infty)) .$ Then $\begin{matrix} A \neq \emptyset since a \in A, \\ B \neq \emptyset since b \in B, \\ A \cap B = \emptyset since (- \infty, z) \cap (z, \infty) = \emptyset . \end{matrix}$ Since $z \in ℚ$ then $z \notin f (X)$ and since $(- \infty, z) \cup (z, \infty) = ℝ - {z}$ then $A \cup B = X .$ So $X$ is not connected. $□$

Show that

X = {(x, y) \in ℝ^{2} | x y = 0}

is not homeomorphic to

ℝ .

		Solution.
	Let $\begin{array}{rcl} X & = & {(x, y) \in ℝ^{2} \| x y = 0} \\ = & {(x, y) \in ℝ^{2} \| x = 0} \cup {(x, y) \in ℝ^{2} \| y = 0} . \end{array} \begin{matrix} \end{matrix}$ Assume that $f : X \to ℝ$ is a homeomorphism. Let $a = f ((0, 0)) .$ Then $f : X - {(0, 0)} \to ℝ - {a}$ is a homeomorphism. So $X - {(0, 0)}$ and $ℝ - {a}$ have the same number of connected components. Since $X - {(0, 0)}$ has 4 connected components and $ℝ - {a}$ has 2 connected components, this is a contradiction. So $X$ is not homeomorphic to $ℝ .$ $□$

Notes and References

These are a typed copy of Assignment 1 Solutions from a series of handwritten lecture notes for the class Metric and Hilbert Spaces.

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