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標題: Supremum [打印本頁]

作者: kingwinner 時間: 10-2-25 12:14 PM 標題: Supremum

If A and C are subsets of R, let AC={ac: a E A, c E C}. If A and C are bounded and A and C consist of strictly positive elements only, prove that sup AC = supA supC.

thx

[ 本帖最後由 kingwinner 於 10-4-3 07:15 AM 編輯 ]

作者: Naozumi 時間: 10-2-25 01:45 PM

如果A同 C有negative number,咁咪唔o岩lor?

作者: kingwinner 時間: 10-2-26 01:03 AM

係 wo...
咁如果係 positive 點 prove?

作者: kingwinner 時間: 10-4-3 07:21 AM

仲未識做呢題, 暫時我只係諗到一半...

A, C bounded => supA, supC exist (by least upper bound axiom)
0<a<=supA for all a E A
0<c<=supC for all c E C
=> 0<ac<=supA supC for all ac E AC
=> supA supC is an upper bound of AC
=> supAC<=supA supC

有冇人識點 prove 埋另外o個個方向?
thx!

作者: Gnoehc 時間: 10-4-3 09:50 PM

I think we can just prove by definition of supremum.
For any epsilon > 0, exsits a x in A such that
    $\\sup A-\\epsilon < x \\leq \\sup A$
while for same choice of epsilon, exists a y in B such that
    $\\sup B-\\epsilon < y \\leq \\sup B$
Since two supremums are assumed to be positive, the epsilon can be chosen small enough to make left hand side of these two inequality to be positive, thus on multiplication, we get (let A' = sup A, B' = sup B)
    $A'B'-(A'+B')\\epsilon < A'B'-(A'+B')\\epsilon + \\epsilon^2 < xy \\leq A'B'$
this shows that A'B' is the supremum (by the definition).

作者: kingwinner 時間: 10-4-4 12:27 PM

唔明最後o個 part 點解 prove 到 supA supB = supAB, 可以解釋下 ma?

thx

作者: Gnoehc 時間: 10-4-4 01:40 PM

原帖由 kingwinner 於 10-4-4 12:27 PM 發表
唔明最後o個 part 點解 prove 到 supA supB = supAB, 可以解釋下 ma?

thx

The definition of supremum is that for any epsilon > 0, there exists a x in X such that
$\\sup X - \\epsilon < x \\leq \\sup X,$
i.e. any number slightly smaller than sup X cannot be a upper bound.

while we have shown that for any epsilon > 0, exists a x in X, y in Y such that
$\\sup X \\sup Y - \\epsilon < xy \\leq \\sup X \\sup Y$
This shows that sup X sup Y is the supremum of the set $\\{xy:x\\in X,y\\in Y\\}=XY$

作者: kingwinner 時間: 10-4-5 04:11 AM

兩個問題:

1) ..."the epsilon can be chosen small enough to make left hand side of these two inequality to be positive"
So it is not true for ANY epsilon > 0...

2) You showed that there exist x E X,y E Y such that
supA supB - (sup A + supB) ε < xy

Why is this the same as supA supB - ε < xy?

Thx!

[ 本帖最後由 kingwinner 於 10-4-5 11:54 AM 編輯 ]

作者: Gnoehc 時間: 10-4-5 01:20 PM

原帖由 kingwinner 於 10-4-5 04:11 AM 發表
兩個問題:

1) ..."the epsilon can be chosen small enough to make left hand side of these two inequality to be positive"
So it is not true for ANY epsilon > 0...

2) You showed that there exist x ...

1) You may insist on talking any epsilon. There is still a good reason why we don't need so.
Observe that both elements a and b are strictly positive, it is reasonable to choose such small epsilon to ensure left hand side to be positive. (since there is no such negative a!)

2) recall that sup A and sup B are constants, if you like you can set at the very begining that
$\\sup A - \\frac{\\epsilon}{\\sup A+\\sup B}<a\\leq \\sup A,\\exists a\\in A$
(similar to B) and then multiply each other, the coefficient of epsilon turns out to be 1.

[ 本帖最後由 Gnoehc 於 10-4-5 01:32 PM 編輯 ]

作者: kingwinner 時間: 10-4-5 04:28 PM

原帖由 Gnoehc 於 10-4-5 01:20 PM 發表

1) You may insist on talking any epsilon. There is still a good reason why we don't need so.
Observe that both elements a and b are strictly positive, it is reasonable to choose such small epsilo ...

1) To ensure that the LHS is positive, we are forced to take only small epsilons, so we have only proved that "for all small ε>0, there exist a E A, b E b such that supA supB - ε < ab"

But what we need to prove is that "for ALL ε>0, there exist a E A, b E b such that supA supB - ε < ab" (in other words, for ALL ε>0, supA supB - ε is not an upper bound of AB). How can we prove this?

thx

[ 本帖最後由 kingwinner 於 10-4-5 04:34 PM 編輯 ]

作者: Gnoehc 時間: 10-4-5 05:32 PM

1) I am sorry I prove it carelessly. (I find no definition of supremum that requires small epsilon only=.=)

OK now after choosing suitable epsilon such that the following holds,
    $0<\sup A\sup B - \epsilon$
it immediately tells us
    $\begin{cases}\displaystyle 0<\sup A - \frac{\epsilon}{\sup A+\sup B}\\\displaystyle 0<\sup B - \frac{\epsilon}{\sup A+\sup B}\end{cases}$
(simple algebra manipulation)
then we go through the mentioned argument again, we have for this special range of epsilon,
    $\sup A\sup B-\epsilon <ab\leq \sup A\sup B,\exists ab\in AB$
holds.

Next we consider a larger epsilon such that
    $\sup A\sup B - \epsilon \leq 0$
for this range of epsilon the statement
    $\sup A\sup B-\epsilon <ab\leq \sup A\sup B,\exists ab\in AB$
holds obviously because ab is always positive, thus we conclude for all epsilon > 0, the following statement is true
    $(\forall \epsilon>0)(\exists ab\in AB)(\sup A\sup B-\epsilon <ab\leq \sup A\sup B)\\\mathop{\iff}\limits^{\text{charactarization}}_{\text{of supremum}} \sup(AB)=\sup A\sup B$

[ 本帖最後由 Gnoehc 於 10-4-5 06:07 PM 編輯 ]

作者: Gnoehc 時間: 10-4-5 06:31 PM

Well because I touched supremum and infimum half year ago..., I remember that the difficulty can be eased if we know some basic results.
1. If C' = sup C,  then there exists a sequence {c_n} in C such that lim c_n = C'
2. If x_n and y_n are convergent sequences with respectively limit x and y, then
   lim x_ny_n = (lim x_n)(lim y_n) = x y

Having these two propositions, we focus on the statement.
Define A' = sup A, B' = sup B, then clearly for any ab in AB,
   ab <= A'B',
i.e. A'B' is an upper bound of ab.

Now by proposition 1., exists two convergent sequences {a_n in A} and {b_n in B} such that lim a_n = A', lim b_n = B',
by proposition 2., lim (a_n)(b_n) = A'B'.

We conclude that A'B' = sup (AB)

[ 本帖最後由 Gnoehc 於 10-4-5 08:45 PM 編輯 ]

作者: Naozumi 時間: 10-4-6 03:30 AM

#12的conclusion有誤, 你證明了 lim an bn = A'B',但並不表示 sup anbn = A'B'
If sup A = A' ==>有convergent sequence an tends to A,但是converse並不正確

作者: Gnoehc 時間: 10-4-6 12:34 PM

原帖由 Naozumi 於 10-4-6 03:30 AM 發表
#12的conclusion有誤, 你證明了 lim an bn = A'B',但並不表示 sup anbn = A'B'
If sup A = A' ==>有convergent sequence an tends to A,但是converse並不正確

Ya in addtion I have said that A'B' is an upperbound, these two add up to the conclusion.

作者: Naozumi 時間: 10-4-6 05:28 PM

我唔太明點樣加埋A'B'係upperbound可以得出#12的結論,煩請指點一下

作者: Gnoehc 時間: 10-4-6 05:50 PM

原帖由 Naozumi 於 10-4-6 05:28 PM 發表
我唔太明點樣加埋A'B'係upperbound可以得出#12的結論,煩請指點一下

The reason is as follows. The condition C' is an upper bound means that
$c_n\\leq C'$
on the other hand,
$\\lim_{n\\to\\infty}c_n=C'\\implies (\\forall \\epsilon > 0)(\\exists N\\in\\mathbb{N}})(n>N\\implies C'-\\epsilon <c_n)$
That's the definition of supremum. (again, any slightly smaller number cannot be an upper bound)

[ 本帖最後由 Gnoehc 於 10-4-6 05:52 PM 編輯 ]

作者: Naozumi 時間: 10-4-6 09:41 PM

o,明白了,原本limit有此妙用的,我真的想不到,有新野學到了,Thx a lot!

作者: kingwinner 時間: 10-4-7 06:30 AM

1) I am sorry I prove it carelessly. (I find no definition of supremum that requires small epsilon only=.=)

唔該哂先!

Actually, I think it ENOUGH to prove only for small epsilons...

因為其實只要 prove 到:
For all small ε>0 (say for all 0<ε<ε' for some ε', where ε' is some number that guarantees BOTH supA-ε>0 and supB-ε>0), there exist a E A, b E b such that supA supB - ε < ab
i.e. for all 0<ε<ε', supA supB - ε is not an upper bound of AB 就已經足夠, 對嗎?
因為 for all 0<ε<ε', supA supB - ε is not an upper bound of AB, 咁好明顯 supA supB - ε for ε≥ε' 都肯定唔會係 upper bound 啦, 對嗎?

[ 本帖最後由 kingwinner 於 10-4-7 08:04 AM 編輯 ]

作者: Naozumi 時間: 10-4-7 02:22 PM

還有一個問題不明白,請指點

而家證左
Sup(an bn) = Sup an Sup bn
但係而家係証 Sup(AB) = Sup A Sup B
an同 bn並不是A和B的任意element,

應如何解決此問題呢?

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