Re: [分析] i無理,{ni-m:n,m€N}稠密問題
※ 引述《znmkhxrw (QQ)》之銘言：
: 首先強調一下我想要討論的是n,m€natural numbers
: --------------------------------------------
: 當i是正的無理數時
: 已經知道{ni+m:n,m€整數} is dense in R
: 而證法有很多，之前也有跟板友討論過
: 但是
: 這些證法我只能推得{ni+m:n,m€整數} is dense in R
: 並無法推得{ni-m:n,m€正整數} is dense in R
: 我想知道的是，若是正整數的話，依然稠密嗎？
: 若是，如何證明？
: 這邊就不附上"{ni+m:n,m€整數} is dense in R"的證明了 有需要再補上
: 謝謝
讓我練習一下ow o
Let E = {ni-m : n, m in N}, i in R, i not in Q, i > 0
F = {ni+m : n, m in Z}
L(E) is the set of all limit point of E
C(E) is the closure of E = E union L(E)
Need to prove C(E) = R.
(Thm) C(F) = R (proven)
(Coro) L(F) = R
(pf) By continued fraction of i, 0 is in L(F).
then for any a in F, a = ni+m+0 is in L(F).
(Prop) If r, s in L(E), then
(1) r-1 in L(E)
(2) r+1 in L(E)
(3) kr in L(E) for any k in N
(4) r+s in L(E)
(pf) (1)(3)(4) is trivial.
(2) Let {r_j} in E, r_j -> r.
Since {r_j = ni-1: |r_j - r| < 1, n in N } is finite
we can remove them. thus {r_j+1} in E and r_j+1 -> r+1
(Lemma) Given r in R, then one or two of r, -r is in L(E)
(pf) Let {r_j} in F, r_j -> r.
Since {r_j = ni+0: |r_j - r| < 1, n in Z}
{r_j = 0i+m: |r_j - r| < 1, m in Z}
{r_j = ni+m: |r_j| < 1+|r|, n, m in Z, mn > 0}
are all finite set, we can remove them.
Thus we have mn < 0 for all r_j
Then either infinite r_j have m<0, n>0,
thus those {r_j} in E and r in L(E)
or infinite r_j have m>0, n<0,
thus those {-r_j} in E and -r in L(E)
or both.
(Thm) C(E) = R
(pf) By Lemma, 0 is in L(E). Thus by Prop Z is in L(E).
Given M in N, M > 1, we have -1/M = -1 + (M-1)( 1/M)
and 1/M = 1 + (M-1)(-1/M)
Hence by lemma and prop,
both 1/M and -1/M is in L(E), so does Q.
Now Q in L(E) means Q in C(E) and R = C(Q) in C(C(E)) = C(E)
由於 Prop 的關係，證明中最大的問題是 1/M in L(E)
原文底下Ricci大的回覆有提示可以考慮正負的情況(?) 就是那個lemma
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嗯嗯ow o
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沒寫清楚...qw q 已更正
※ 編輯: Desperato (140.112.25.105), 10/11/2017 07:49:22
推
10/11 16:05,
8年前
, 1F
10/11 16:05, 1F
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