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# 證明無限維度的 B-space 的每一組基底都不可數

If X is a Banach space with dim(X) = ∞ , prove that every basis of X contains uncountably many elements. Why do we need to require X to be Banach ?ps. Banach space 是指一個有 norm 的向量空間使得在這個向量空間上的每一個 Cauchy 數列都收斂.

### 2 Answers

- EricLv 61 decade agoFavorite Answer
Proof. Suppose X, with field F = R or C, had a basis {xn ∈ X: n ∈ N}. Define for N ∈ NAN = {a1x1 + ... + aNxN: a1,...,aN ∈ F}the set of linear combinations of x1,...,xN. Clearly AN is nowhere dense since any δ-neighborhood of a1x1 + ... + aNxN contains a1x1 + ... + aNxN + (δ/2)xN+1, which is not in AN. Furthermore,X = ∪N∈N AN. This contradicts the Baire category theorem, which asserts that a complete space is not a countable union of nowhere dense sets. ∎We require X to be Banach because there are infinite-dimensional normed vector spaces with bases of countably many elements. For example, consider the space of sequences with only finitely many nonzero elements (as a subspace of ℓ∞) endowed with the norm║{an}║ = supn |an|.This has {(1,0,0,0,...), (0,1,0,0,...), (0,0,1,0,...),...} as a basis, yet(1,0,0,0,...), (1,1/2,0,0,...), (1,1/2,1/4,0,...)is a Cauchy sequence that does not converge.

2006-10-23 10:01:52 補充：

1.先排好基底元素 x1,x2,...

An = 前 n 個基底元素的線性組合

2.是的

- LLv 71 decade ago
你的 A_n 的意思是基底內任取 n 個元素的線性組合的聯集 ? 還是把基底元素順序排好取前 n 個元素的線性組合 ?

2006-10-23 00:33:08 補充：

最後那個例子收斂到

(1,1/2,1/4,1/8,1/16,...) @@?