推 Vulpix : 吸一口。 01/16 17:11
※ 引述《XII (Mathkid)》之銘言：
: ※ 引述《Desperato (Farewell)》之銘言：
: : 1. Show that if R^n = Union W_i, i = 1, 2, 3, ...
: : W_i are subspaces of R^n
: : Then R^n = W_i for some i
: Fact.
: If V is a vector space over field F, k in Z^+ with k<|F|,
: and W_1,..,W_k are proper subspaces of V,
: then V≠∪_{i=1~k} W_i.
: proof.
: Suppose V=∪_{i=1~k} W_i.
: W.O.L.G. we may assume W_i is not contained in ∪_{j≠i} W_j.
: There are w_i in W_i-∪_{j≠i} W_j for i=1,..,k.
: Since k≧2, consider a*w_1+w_2 for a in F.
: Since k<|F|, by Pigeonhole Principle,
: there are distinct a, b in F s.t.
: u=a*w_1+w_2 and v=b*w_1+w_2 in W_n for some n.
: Since u-v=(a-b)*w_1 in W_1-∪_{j≠1} W_j, we have n=1.
: So w_2=u-a*w_1 in W_1, a contradiction (w_2 not in W_1).
: QED
: By above Fact. R^n=W_i for some i
: : 雖然這題擺明可以用measure做，但線代...?
: 其實用鴿籠原理就可以了..
這個證法蠻強大的XD
我寫一次有 R 的版本好了
If V is a vector space over field F,
I is an index set with Card(I) < Card(F),
{W_i}_(i in I) is a set of proper subspaces of V,
then V≠∪_{i in I} W_i.
(pf)
Suppose V = ∪_{i in I} W_i.
Also |I| = 1 is trivial, so we may assume |I| > 1.
By Zorn's Lemma, there is some subset I' in I
such that V = ∪_{i in I'} W_i,
and no proper subset of I' satisfied this property.
Replace I by this I'.
Hence, for each i in I, there is some w_i in W_i,
such that w_i is not in ∪_{j in I, j≠i} W_j.
Now consider elements { a*w_1 + w_2 | a in F }
Since |F| > |I|, there is some a≠b in F
such that both a*w_1 + w_2 and b*w_1 + w_2 are in W_i for some i
Hence (a-b)*w_1 is in W_i, we get i = 1.
But then w_2 = (a*w_1+w_2) - w_1 is in W_1, a contradiction.
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取 I = N, F = R 就完成了
另外，我發現這個證明從頭到尾沒用到 V 是 finite dimensional
但說實話這超過我的理解範圍了XD
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我想到的證明不是這個方向
所以也打一次好了，但比起 XII 的證明弱很多
Show that if R^n = ∪_{i in N} W_i,
W_i are subspaces of R^n,
Then R^n = W_i for some i.
(pf)
Induction on n.
n = 1, the only subspace of R^1 is {0}, so the statement clearly holds.
Assume for R^(n-1) the statement holds but for R^n it fails.
That is, R^n = ∪_{i in N} W_i,
but all W_i are proper subspaces of R^n.
Consider the set of subspaces
U_t = { (v_t)^⊥ | v_t = (1, t, 0, ..., 0) }
Then each t in R gives an U_t of dimension n-1
U_t = U_s if and only if t = s
Hence there is some U = U_t such that U≠W_i for all i
Now consider R^n∩U = (∪_{i in N} W_i)∩U = ∪_{i in N} (W_i∩U)
If dim(W_i) = n-1 and W_i∩U = W_i, then W_i = U, a contradiction
Thus dim(W_i∩U) < n-1, each W_i∩U is a proper subspace of R^n∩U
Apply a rotation f: R^n∩U -> R^(n-1)
We prove that R^(n-1) = ∪_{i in N} f(W_i∩U)
where all f(W_i∩U) are proper subspaces of R^(n-1)
which contradicts to the induction hypothesis.
我覺得這個證明比較有線代感(?)
uncountable 的部分出現在選 U 的地方
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嗯嗯ow o
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