Chapter 3: Closed and Open sets Flashcards
Sequence 1/n
in (0,1]
[0,1]
Sequence has limit 0, in {0,1} not (0,1]
DEF 3.1: CLOSED SUBSET
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.
PROP 3.2: convergent sequences in R…
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.
LEMMA 3.3: closed ball and subsets
A closed ball B[a, r] in a metric space (X, d) is a
closed subset
Example: closed ball?
[a, b] = B[(a+b)/2,(b−a)/2]
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
SHOWING A SUBSET OF A METRIC SPACE IS NOT CLOSED
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.
EXAMPLE:
F = (0, ∞).
IS F closed?
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.
EXAMPLE:
the subset F = {(x,y)∈ R²: x > 0}
of R²
is it closed?
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.
EXAMPLE: Is Q closed?
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.
EXAMPLE:
let F be the set of polynomial
functions from [0, b] to R.
is it closed in (C[0, b], d∞). ?
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∞).
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.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.
DEF 3.10: OPEN SET
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
LEMMA 3.11:
open ball
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:
open interval (a,b) = B((a+b)/2, (b−a)/2)
is open or closed set?
For a < b, the open interval (a,b) = B((a+b)/2, (b−a)/2) in
R is an open subset of R.
…..
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. )
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.
EXAMPLE:
for any interval of the
form [a, b), [a, b] or (a, b].
is subset of R an open subset?
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.
DEF: COMPLEMENT
the complement of
A in X is the set X\A = {x ∈ X : x /∈ A}.
THM 3.15: open/closed subsets and complements
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.)
PROPERTIES: closed vs open
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!
DEF: unions and intersections for pairs of sets
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}.
de Morgan Laws
X \ ∪Aᵢ= ∩(X \ Aᵢ),
X \ ∩Aᵢ = ∪(X \ Aᵢ).
So taking complement turns unions into intersections and vice versa.
PROP 3.17
properties of open sets
union,
intersection
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:
PROP 3.18
properties of open sets
union,
intersection
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:
FINITENESS of properties of intersections and unions
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.
