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Unformatted text preview: Prof. Girardi Math 554 Fall 2010 09.23.10 Exam 1
MARK BOX PROBLEM POINTS 

instructions INSTRUCTIONS: (1) For the ﬁll—inthe—blank/ box problems, write your directly on this exam paper. (2) For the proof part of the proof pmblems, write a neat formal proof on (the front side only
of) your homemade Blue Book. Start each new problem on a new page. Do NOT recopy
the statement of the problem but on each page, in the upper right hand cornor, put the
problem number. (3) When ﬁnished, hand in this exam paper and your proofs on your lined paper. Please put
all in order and staple. Do your scratch work on the scratch paper provided and do not
hand it in. (4) The MARK BOX indicates the problems along with their points. This test is copied 2sided.
Check that your copy of the exam has all of the problems. (5) You may not use any electronic device books, or personal notes. (6) During this exam, do not leave your seat. If you have a question, raise your hand. When
you ﬁnish: turn your exam over, put your pencil down, and raise your hand. (7) This exam covers (from Introduction to Real Analysis by William F. Trench, Free Edition 1):
Sections 1.1 and 1.2. . Honor Code Statement
I understand that it is the responsibility of every member of the Carolina community to uphold and maintain
the University of South Carolina’s Honor Code.
As a Carolinian, I certify that I have neither given nor received unauthorized aid on this exam.
Furthermore, I have not only read but will also follow the above Instructions. Signature : (frank/m 11/15?”th 1 Cl lel Exam 1 Bid, 1. Fields.
121. A ﬁeld F is a set F on which the operations of (J) metal and mull/QM ‘cm/ are deﬁned so that every pair of elements from F has a unique sum and product, both in F, with
the following properties. TMM (A) ml): +a. am echebcu EWWﬂlML
' (ela)c:o.(lao a: ‘
adad‘jxﬁ 6, (E) ll“ 1b. An ordered ﬁeld is a ﬁeld F along with a relation < which has the following properties. (F) do L2. F ml} 45!" (g) I! <x<lo Cwol l)‘ C , Klee 0x 4 ‘" l1”; £31;va / ’ l
1:: Mb a+c<lorc v’e ‘ 0c <loe , (H) lJML
1c. An ordered ﬁeld is complete provided it has the following property. /
All. 1‘3 LOW/VIM dill—5*)”ﬂ A35 0. ‘ Cl
r“ 1d. Give an example of a ﬁeld S for which it is impossible to deﬁne an order relation < that would
V
Lqub‘w lam/é make it an ordered. ﬁeld. No proof necessary but do give how the 2 operations indicated in part (laJ) are deﬁned.
‘ C’ —_ ' + c = x . 2 = or =
phi o Wﬂﬁ ”(1% “Ii 1 3+3 ml lg (“B/K k /X
li+h=§5¥x=v . ,,
3 <2 d” 3.. , 5: 1e. Give an example of an ordered ﬁeld that is not a complete ordered ﬁeld. No proof necessary but
do clearly indicate why the ﬁeld is not complete. i 6:) be [Eff Wu (54" ”if. l{’ c
, a , 7, r+ l
/ Wolff 'll'kﬂ M S ’ l’l/V E Q l X A 3 Woblem Ins/fl/‘a/lloﬂ/f Clasg de/Ul: 2. Let R be the set of real numbers, lli be the extended real numbers and S C R. 2a. Another term for the supremum of S, i.e. for sup S, is ﬁnd? W v. S gr lu’l’s .
2b. Another term for the inﬁmum of S, i.e. for the inf S is wlést \O‘Nﬂ’ lDGWV‘Ol " S W LS . 2c. Let S be a nonempty set that is bounded above. Then the supS is the unique real number ﬂ 6 R
such that (1) for each a: E S, .___%—_—_X '4‘ _ _________. f” .
(2) supS = 00 if and only if US W1“ “ M lDOleM OJW’I‘T: _
(3) supS z —00 if and only if _—g_g : . __ . 2e. Let S be a nonempty set that is bounded below. Then the inf S is the unique real number oz 6 R
such that 4
(1) foreachmES, og —?C III X8<D<~‘l—8 (2 ) if a > 0, then 3.735 E S such that 1
4
A [ﬁlldﬂwﬂowafw M X£*°<1l’5
2f. We know that inf S e R. More speciﬁcally, C, p i l l ,I V
(1)infS’ElRifandonlyif Q M WWW»! 7W Owﬁ‘ berm/wide lat/{WT '  .r 0
(2) inf S = —oo if and only if S S M "T‘WMiwiru (Mi/L "Yet?f l‘JWM/x M low ‘
' \. i r ’ .
(3) infS= 00 if and only if 5" é ‘ ?rol3lcm Iﬂsbiroul’lﬁmz 3 WWé % I,’ _‘y: 5 3. Fill in the chart. No proof necessary. I—I—
__—I
II_ 22/; __—
_Imn
__ TS
x
I
0"
/\
H
’{I‘
:2
l
(y
/\
s l“
p
$3
(/3
i l
”13‘
\N
L) x
\_
{a
\M ”is, S: PM} VF”, 4. Let r be a positive real number. Let {mn}§°=0 be a sequence of real numbers satisfying Iran“ — mnl S 7‘ lacn —— mn_1 for each n E N . (4) Let Pn be the proposition that mm Wind. Inch—mg s W1 lml—mol . . (Pn) Show that Pn is true for each n E N by math induction. Use strong induction only if you must do so. 5. In class we proved Archimedean’s Property, which is the following.
If p > 0 and e > 0, then there exists n E Z such that me > p. Provide a proof of Archimedean’s property.
6. Let T be a nonempty subset of R that is bounded above and let ,6 = sup T.
Let c be a positive constant (i.e. c 6 R and c > 0). Define the subset cT by
cT := {ctEletGT}. Show that
sup (CT) = c,8. (6) . 1%on Inspirahm : 53th \2.3 ’hram whiLmré 4. Let r be a positive real number. Let {xn}r°f=0 be a sequence of real numbers satisfying
lmn+1 —— mnl S r Ixn — asnnll for each n E N . (4)
Let Pn be the proposition that ‘
lmn — 23111] S r”‘1 lacl — 3:0] . . ' (Pn) Show that Pn is true for each n E N by math induction. Use strong induction only if you must do so. ”Problem IﬂSflmllm " dial in Cl 153) Very slow/lg ./ Archimedean’s Property Archimedean’s Property
If p > 0 and e > 0, then there exists n E Z such that 718 > p. (AP)
Proof. Note that (AP) can symbolically be written as
(Vp>0) (V5>0) (ElnEZ) [ne>p]. We shall show that (AP) holds by contradiction. Thus, assume that (AP) does not hold. So we
are assuming that ' (3p > 0) (35 > 0) (Vn E Z) [n5 5 p] » (1)
and we want to ﬁnd a contradiction. Let p > 0 and e > 0 be as given in (1), thus
(Vn E Z) [718 S p] . (2) Consider the set
S := {zeR:x=ne,n€Z} = {neelenEZ}. Then S is nonempty (e.g., 8 E S). Also, (2) implies that S is bounded above by p. So by the
completeness axiom of IR, S has a supremum in R. Let [3 := supS .
By deﬁnition of supremum, if s E S then s 3 ﬂ and so
ianN,thennegﬂ. (3)
If n E N then 71 + 1 E N and so (3) implies
ianN,then (n+1)e$ﬁ,
which is equivalent to (by some algebra)
ianN,thenna_<_ﬂ—s. (4) Now (4) implies that ﬂ — a is a_n upper bound of S. However, 3 is the least upper bound of S. This
is a contradiction. So (AP) holds. El l l
h m
\1 Prolaluem Insp‘mtm
U) SISMI‘ch‘ +0 WWI/E § [I 4L— [0 (2) (Magda 03 an Em... 1a: a; Supremum of a Set Let T be a nonempty subset of litthat is bounded above and let ,6 = sup T.
Let c be a positive constant (i.e. c E R and c > 0). Deﬁne the subset cT by cT := {ctER:tET} .
Show that
sup (cT) = cﬂ . (WTS) Proof. Let the givens be given. Note that ﬂ 6 1R since T is nonempty and bounded above. We will
show that equation (WTS) holds.
Step 1: First we will show that cﬁ is _a_n upper bound of cT. Fix :1: 6 CT. By deﬁnition of the set cT, there exists t E T such that x = ct . (1)
Since t E T and ﬂ = supT
t g ,6 . (2)
Since c > O, inequality (2) implies that
at < cﬂ . (3)
Combining (l) and (3) now give
:1: < cﬂ So cﬂ is indeed ﬂ upper bound of cT.
Step2: Next we show that cﬂ is the least upper bound of cT.
Fix 5 > 0. Way # 1 for Step 2 <WTS cﬂ — 5 is not an upper bound of cT so WTF .1; 6 cT s.t. cﬂ  a < $2>
Note that % > 0 since 5 > 0 and c > 0. Thus, since ﬂ = sup T, there is a t5 6 T such that 5
ﬂ _ Z < t5  ‘ (4)
Since c > 0, inequality (4) implies that
cﬂ — E < cte . By deﬁnition of the set cT, since tE E T we know that ctE 6 CT.
Way # 2 for Step 2 <Since c > o, WTS cﬁ — as is not an upper bound of cT so WTF :25 5 0T s.t. c5 — ca < z,.>
Since ﬂ = sup T, there is a t5 6 T such that )6 — 5 < t5 . (5)
Since 0 > 0, inequality (5) implies that
cﬂ — c5 < ct5 . By deﬁnition of the set cT, since t5 6 T we know that ate 6 cT.
Step 3: By deﬁnition of supremum of a set cﬁ = sup(cT). El ...
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 Girardi

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