Description
Exercises 3.2.7 (b), 3.2.11, 3.3.1, 3.3.5, 4.2.5 (c), 4.3.3 (a), 4.3.9, 4.4.3,
4.4.12, 4.5.7.
Exercise 3.2.7.
Given AC R, let L be the set of all limit points of A.
(a) Show that the set L is closed.
(b) Argue that if x is a limit point of AUL, then x is a limit point of A. Use this observation to furnish a proof for Theorem 3.2.12.
(b) Argue that if x is a limit point of AUL, then a is a limit point of A. Use this observation to furnish a proof for Theorem 3.2.12.
Proof. Suppose I is a limit point of AUL. Let > 0. Then since a is a limit point of AUL, there exists y EAUL such that yx and
If yЄ A, then
so z is a limit point of A.
– <<
If y A, then yЄ L. Therefore, y is a limit point of A, so there exists a € A such that |ay| < xy.
Therefore, ax and
> |k − x| + |h − x| > |x − h| + |k − y| 5 |x − h + h − v | = |x − v|
+
Again, this means that x is a limit point of A.
Thus, we see that if x is a limit point of AUL, then x is a limit point of A. Therefore, the set of limit points of AUL is contained in L; in particular, AUL contains all of its limit points, so A=AUL is closed, which was all that remained to complete the proof of Theorem 3.2.12.
b) If a is a limit point of AUL, then a is a limit point of A.
Proof. Suppose is a limit point of AUL. Let (c) be a sequence converging to z. Either infinitely many elements of the sequence come from A or infinitely many come from L. In the first case, we can find a subsequence of (c) that lies in A. Since the sequence converges, the subsequence converges to z, and so r is a limit point. In the second case, we can find a subsequence that lies in L, but since L is closed, a E L and is a limit point of A.
Prove Theorem 3.2.12
Proof. From the observation above, the closure of A contains all of its limit points and is therefore closed.
Suppose B is a closed subset with ACBCA. We will show B = A.
Let x A. Then, x E A or x Є L, where L is the set of limit points of A. If x E A, then x E B. If x L, then there is a sequence () in A that converges to z. However, the sequence is also in B, and since B is closed, it contains the limit. Thus, Є B. This shows AC B, and so BA. Thus, A is the smallest closed set that contains A.
Exercise 3.2.11.
(a) Prove that AUB=AUB.
(b) Does this result about closures extend to infinite unions of sets?
(a)
AUB AUB
=
It is clear that ACAUBACAUB so
AUB AUB
AUB
A= 0)
To show inclusion the other way, Let x E AUB such that x (otherwise it would be trivial). If x A then for some > 0, V(x) but V(x) (AUB) implies Ve(a) B for all > 0. This shows € B. By same argument, if x 4 B then a A so xe AUB. This shows
So, we get the required result.
AUBCAUB
(b)
Take sets
AUB AUB
An
[1-4]
where n ≥ 2. Since each An is closed set. Then An
=
∞
An
=
n=2
However
n=2
0,1
–
=(0, 1)
[0,1] so we get
UA-U0,1-0,1) = [0, 1]
n=2
n=2
=
It shows this does not extend to union of infinite sets.
Exercise 3.3.1.
Show that if K is compact. and nonempty, then sup K and inf K both exist and are elements of K.
Let K be a non empty compact set. Then by Theorem 3.3.4 K must be closed and bounded. Since K is bounded, a = sup K and 3 = inf K exists since K is non-empty.
For any e > 0, a — e is not supremum so there exists x E K such that a — < x < a. If a then a is limit point and K being closed contains a.
Similarly, € is not infimum so there exists y such that < y < ,81 — € and
if y /3, then is limit point so )3 E K as K is closed. .
Exercise 3.3.5.
Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample.
- The arbitrary intersection of compact sets is compact.
- The arbitrary union of compact sets is compact.
- A be arbitrary, and let K be compact.. Then, the intersection A fl K is compact.
- If F1 D F2 D F D F4 D is a. nested sequence of nonempty closed sets,
then the intersection MI F 0.
The arbitrary intersection of compact set is compact.
Since arbitrary intersection of closed sets is closed, arbitrary intersection of compact sets are also closed. By definition, compact set is bounded and arbitrary intersection of compact is contained in each (bounded) compact set. Therefore being bounded and closed, it must be compact.
(b)
The arbitrary intersection of compact sets may riot be compact. Take in [n, n 1] which are compact for each n. Then
OG
Uin = [1, (x))
n=1
which is not bounded so not compact.
Take A (0, 1) and K = [0, 1] then A n K = (0, 1) which is not compact since it is not closed.
Take Fn = [n, oc) then Fn is closed as it contains it’s every limit point. Also F1D F2 D F3 D . . . but
h.° Fn
n=1
as we can prove any x > 1 cannot lie on the set (using Archimedean property we may cho n)
Exercise 4.2.5.
Use Definition 4.2.1 to supply a proper proof for the following limit statements.
- lirn„.2(3x + 4) = 10.
- l oo zs = 0.
- linl„,2(x2 + x — 1)
- „.31/r = 1/ .
- c) xlim(x2 +x — 1) = 5.
Proof. Let E > 0 be given. Choose 6 = min(1, e/6). If D < x — 21 < 8, then Ix — 21 < 1, or 1 < x < 3. Thus, 4 < x + 3 < 6, and
Also,
— 21 < <
6 + 31′
SO
| However, Ix — 211x + 3 | 21Ix ± 3
X2 + X – 6 |
< E.
= 1X2 ± X – 1 – 51, SO < |
1 2x — 1— 5
Exercise 4.3.3.
(a) Supply a proof for Theorem 4.3.0 using the charac-
terization of continuity.
(b) Give another proof of this theorem using the sequential characterization of continuity (from Theorem 4.3.2 Olin.
Theorem 4.3.9 (Composition of Continuous Functions). Given f : A-41,
and g : B R, assume that the range f (A) (x) : x E Al is contained in
the domain B so that the composition go f(x) (f (x)) is defined on A.
If f is continuous at e E A, and if is continuous at PO E B. then 9o f is continuous at c.
Proof. Exercise 4.3.3. ❑
Exercise 4.3.9.
Assume h : R —> R is continuous on R and let K h(x) CI} Show that K is a closed set.
Let h R be continuous function and K = {x E II.: h(s) = 0} then K
is closed set.
Proof For continuous function h, and Vs (c) IT K 0 for all (5 > 0, For all
E > 0, there exists 6 > 0 such that whenever x E Va(c) then h(x) E V,(h(c)).
If h(c) 0 0 then there exists some e > 0 such that h(x) E VE(h(c))
f (x) 0 0, This contradicts that V8(c) n K 0 since it says for every S there exists x E (c) such that f (x) = 0.
So h(e) = 0 c E K so K. is closed as it contains its every limit point.
Exercise 4.4.3.
Show that fix) = 1/x2 is uniformly continuous on the set [1, cc) but not on the set (0,1].
| Proof Let E > 0 be given. If x, y 1 E [1, cc) then < 1 and ____ 1
2 sy y Choose S = €/2. Then, for x, y E [1, ac), if x — yl < 6, then |
< 1. |
| 1 1
x2 — y2 |
Ix — y2
X2y2 |
ix y
YI______________ x2y2 |
( \
= X — Y 1 1 2 + 2 2X Y < 26 = 2€12 = 6.
sy x y
Thus, f is uniformly continuous on [1, Do).
To show f is not uniformly continuous on (0, 1], we use the Sequential
Criterion for Absence of Uniform Continuity. Let ar, = —1, and b„ 1 = _ .
2n
3
Then, Ian —bid =4n 0. However,
2
f (a„) — f 0,31 = In2 — 4n2I = 3n2 > 3. Hence, f is not uniformly continuous on (0, 1],
Exercise 4A.12.
Review Exercise 4.4.11, and then determine which of the following statements is true about a continuous function defined on R:
- f (B) is finite whenever B is finite.
- f -1(K) is compact whenever K is compact.
- f -1(A) is bounded whenever A is bounded.
- f -1(F) is closed whenever F is closed.
(a) f-1(B) is finite whenever B is finite.
False. Take function f : N -> {O, 1} by f(x) = x (mod 2) then f-1({0, 1}) = N is infinite.
(b) If K is compact then I-1(K) is also compact.
False. Take f(x) = sinx then f : R -> [-1, 1] where 1-1, 1] is compact and R is not compact,
2 (c) f -1(A) is bounded whenever A is bounded. r False. Take above example, f(x) = sin x.
| (d) f -1(F) is closed whenever F is closed.
True. By continuity if R\ F is open so f-1(R\F) = |
(a) (b) (c) may be false. Consider the example f(x) = 1 for all x c R. Let B = K = A, then f -1(B) = f -1(K) = f—‘ (A) – R which is clear not finite, not compact, not bounded.
(d) is true. To show f-1(F) is closed, it suffices to show that f-1(F)e is open. We show that
f-1(F)c = (Fe). Then because F’ is open, by definition of Exercise 4.4.11, f-l(F)c = f-1(Fe) is
also open.
Now we show f-1(F)’ = f-1(Fc). First we show f-1(F)e c f-1 (Fe). Let a E ri(F)c, then a 0 f-1(F), i.e. f (a) 0 F, Therefore f (a) c F’ , hence a E f-1(F°).
Then we show f-1(Fe) c f-1(F)e. Let a e f-1(Fc), then f(a) e Fe. Thus f(a) F and a 0 f-1(F). Therefore a r f (F)` , this implies f -1 c f-1(F)c.
Now we have f -1 (F)e = f-1(Fe).
Exercise 4.5.7.
Let f be a continuous function on the closed interval [0,1] with range also contained in [0,1]. Prove that f must have a fixed point; that is, show 1(x) = x for at. least one value of x E [0, 1].