Chapter 3: Closed and Open sets

Question 1

Sequence 1/n

in (0,1]
[0,1]

Answer 1

Sequence has limit 0, in {0,1} not (0,1]

Question 2

DEF 3.1: CLOSED SUBSET

Answer 2

Let X be a metric space, and let A be a subset of
X. We say that A is a closed subset of X if whenever we have a
sequence x_1, x_2, . . . in A which converges to a limit x ∈ X, then the limit x also lies in A.

Question 3

PROP 3.2: convergent sequences in R…

Answer 3

Let (x_n) be a convergent sequence in R with limit
x.
If x_n > 0 for all x then x > 0. Therefore [0,∞) is closed in R.

Question 4

LEMMA 3.3: closed ball and subsets

Answer 4

A closed ball B[a, r] in a metric space (X, d) is a

closed subset

Question 5

Example: closed ball?

[a, b] = B[(a+b)/2,(b−a)/2]

Answer 5

For a < b, the closed interval [a, b] = B[(a+b)/2,(b−a)/2] in
R is a closed ball, so it is closed in the sense of Def. 3.1

Question 6

SHOWING A SUBSET OF A METRIC SPACE IS NOT CLOSED

Answer 6

To show that a subset F of a metric space is not closed, we just
need to produce one sequence within F with a limit outside F.

Question 7

EXAMPLE:
F = (0, ∞).
IS F closed?

Answer 7

In R, let F = (0, ∞). Let x_n =1/n
. Then x_n ∈ F for
all n but x_n → 0 ∈/ F. Therefore F is not closed.

Question 8

EXAMPLE:

the subset F = {(x,y)∈ R²: x > 0}
of R²

is it closed?

Answer 8

In (R², d₂), the subset F = {(x,y)∈ R²: x > 0}
of R²
is not closed. Let xₙ = ( 1/n, 0). Then xₙ ∈ F for all n but
xₙ → (0, 0) ∈/ F.

Question 9

EXAMPLE: Is Q closed?

Answer 9

Q is not closed in R because the sequence 1, 1.4, 1.41, 1.414, · · ·
has limit √2 NOT IN Q but each term, having a finite decimal expansion,
is rational.

Question 10

EXAMPLE:

let F be the set of polynomial
functions from [0, b] to R.

is it closed in (C[0, b], d∞). ?

Answer 10

Let b > 0 and, in C[0, b], let F be the set of polynomial
functions from [0, b] to R. In Example 2.14, we saw a sequence
of elements of F converging to e
x under d∞. As e^x NOT IN F,
F is not
closed in (C[0, b], d∞).

Question 11

EXAMPLE:
Let
F = {f ∈ C[0, 1] : f(1) = 1}.
Show that F is closed when the metric is d_∞ but not when the
metric is d_1.

Answer 11

Solution Let (f_n) be a sequence of elements of F that converges to
some f ∈ C[0, 1] under d∞. Then fn(1) = 1 for all n. To show
that F is closed we must prove that f ∈ F, that is f(1) = 1. By
Proposition 2.15, fn → f pointwise. In particular, f_n(1) → f(1),
and since fn(1) = 1 for all n this means that f(1) = 1. Thus f ∈ F,
as required.

By Example 2.16, the sequence f_n(x) = x
n
converges to the zero
function f(x) = 0 under d1. Here each fn ∈ F, because fn(1) = 1,
whereas f /∈ F because f(1) = 0. Thus F is not closed.

Question 12

DEF 3.10: OPEN SET

Answer 12

A subset A of a metric space is open if for each
a ∈ A there is r > 0 such that B(a, r) ⊆ A.

A set is open if
every point of the set can be surrounded by an open ball that is also
contained in the set

Question 13

LEMMA 3.11:

open ball

Answer 13

An open ball B(x, t) in a metric space (X, d) is an
open subset in the sense of 3.10 (open set)

PROOF:

Question 14

open interval (a,b) = B((a+b)/2, (b−a)/2)

is open or closed set?

Answer 14

For a < b, the open interval (a,b) = B((a+b)/2, (b−a)/2) in
R is an open subset of R.

…..

Question 15

EXAMPLE:
In (R², d₂) let
F = {(x, y) : x > 0}.
Show that F is open. (We saw in Ex. 3.6 that F is not closed. )

Answer 15

Solution Let (x, y) ∈ F, and set r = x BIGGER THN 0. Let (a, b) ∈ B((x, y), r).
Then
|a − x| ≤ √ [(a − x) + (b − y)] LESS THAN x
so −x LESS THN a − x
LESS THN x.
Adding x, 0 LESS THN a(LESS THN2x).

Thus (a, b) ∈ F.
Therefore B((x, y), r) ⊆ F and hence F is open.

Question 16

EXAMPLE:
for any interval of the
form [a, b), [a, b] or (a, b].

is subset of R an open subset?

Answer 16

eg (0, 1] ⊂ R is not an open subset.

Suppose it is
open.

Then there is an open ball B(1, r) ⊆ (0, 1]. But B(1, r) =
(1 − r, 1 + r) always contains 1 + r/2 ∈/ (0, 1]. This is a contradiction
so (0, 1] is not open.

Question 17

DEF: COMPLEMENT

Answer 17

the complement of

A in X is the set X\A = {x ∈ X : x /∈ A}.

Question 18

THM 3.15: open/closed subsets and complements

Answer 18

A subset A of a metric space X is open if and only
if the complement X \A is closed. (Applying this to the complement
X\A, which has complement A, we get that X is closed if and only
if the complement X \ A is open.)

Question 19

PROPERTIES: closed vs open

Answer 19

subsets may be open but not closed,
closed but not open,
neither open nor closed,

or both open and
closed.

All these can be seen inside the real line: (0, 1) is open but
not closed, [0, 1] is closed but not open, [0, 1) is neither closed nor
open, and the whole real line is both open and closed as a subset of
itself.

**sets can be neither
closed nor open!!!
**A common error is to say that a set is open because
it is not closed. Sets are not doors!

Question 20

DEF: unions and intersections for pairs of sets

Answer 20

Let X be a set and let Ai
, i ∈ I be subsets of X, indexed by some
set I..

∪[i∈I]
of
Ai 
= {x : x ∈ Ai
for some i ∈ I},
∩[i∈I]
of
Ai
 = {x : x ∈ Ai
for all i ∈ I}.

Question 21

de Morgan Laws

Answer 21

X \ ∪Aᵢ= ∩(X \ Aᵢ),
X \ ∩Aᵢ = ∪(X \ Aᵢ).
So taking complement turns unions into intersections and vice versa.

Question 22

PROP 3.17

properties of open sets

union,
intersection

Answer 22

Let (X, d) be a metric space. Then:

*i) X and ∅ are open subsets of X.
*ii) The union of any number of open subsets of X is again open
*iii) Let A_1, A_2, . . . , A_n be a finite collection of open sets in a metric
space X. Then A_1 ∩ A_2 ∩ · · · ∩ A_n is also open.

PROOF:

Question 23

PROP 3.18

properties of open sets

union,
intersection

Answer 23

Let (X, d) be a metric space. Then:

*i) X and ∅ are closed subsets of X.
*ii) The intersection of any number of closed subsets of X is again
closed.
*iii) Let A1, A2, . . . , An be a finite collection of closed sets in a metric
space X. Then A1 ∪ A2 ∪ · · · ∪ An is also closed.

PROOF:

Question 24

FINITENESS of properties of intersections and unions

Answer 24

in R, the intersection of the open sets (−1/n,1/n) is {0} which
is not open and the union of the closed sets [ 1/n, 1] is (0, 1], which is not closed.

Question 25

LEMMA 3.11:

open ball

Proof ***

Answer 25

An open ball B(x, t) in a metric space (X, d) is an
open subset in the sense of 3.10 (open set)

PROOF:
Let a ∈ B(x, t). 
Thus d(a, x) < t. 
Let r = t −d(a, x) > 0
and let y ∈ B(a, r). 

Thus d(a, y) < r. Then
d(x, y) ≤ d(x, a) + d(a, y)
< d(x, a) + r = t.

Thus y ∈ B(x, t). Therefore B(a,r) ⊆ B(x, t), and so B(x, t) is
open

Question 26

LEMMA 3.3: closed ball and subsets

***PROOF

Answer 26

A closed ball B[a, r] in a metric space (X, d) is a
closed subset

PROOF:

Let x_n → x be a convergent sequence in X with each x_n∈B[a, r]. Then

d(a, x) ≤ d(a, x_n) + d(x_n, x) ≤r + d(x_n, x),
so

d(xn, x) + r − d(a, x) ≥ 0.
But d(xn, x) → 0 so, by the algebra of limits,
d(xn, x) + r − d(a, x) → r−d(a, x).

By Proposition 3.2,
r−d(a, x) ≥ 0 so d(a, x) ≤ r, that is x ∈ B[a, r].

Thus B[a, r] is closed.