Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $V$ be a complete Hausdorff locally convex topological vector space over the field $\mathbb{K}$.

Let $B$ be a subset of $V$ satisfying

.

Linearly Independent: For all functions $f$ in $\mathbb{K}^B$, if $\displaystyle\sum_{b \in B} f(b) \cdot b = 0$, then $f$ is identically zero.

Spanning Set: For all vectors $v$ in $V$, there exists a function $f$ in $\mathbb{K}^B$ such that $\displaystyle\sum_{b \in B} f(b) \cdot b = v$.

.

Let $C$ be another subset of $V$ satisfying the above conditions with $B$ replaced with $C$.

Does it follow that $|B| = |C|$?

.

(I know such 'bases' don't always exist, but when they do, do they give a unique dimension?)

share|improve this question
2  
Are you implicitly assuming the series are absolutely convergent? Otherwise you should not be working with subsets B but nets into V. Also, you left out an existence condition: for which functions f are you supposing the series converges (absolutely?)? Presumably not "all functions". –  KConrad Aug 20 '10 at 1:33
    
No, but I am working with nets into V (for the potentially uncountable sums), so why should I not be working with subsets B ? Also, I'm not supposing the generalized series converges for any functions f . (although it obviously does for functions of finite support) –  Ricky Demer Aug 20 '10 at 1:48
    
So in the first condition you mean to say «for all functions $f$ in $\mathbb K^B$ such that the sum $\sum_{b\in B} f(b)b$ converges, if the sum is zero, then $f$ is zero»? –  Mariano Suárez-Alvarez Aug 20 '10 at 2:01
1  
Even in R^n you don't sum a set of vectors but a sequence of vectors, since, after all, maybe v_m = v_n for some m < n. You can't express that idea if you sum over subsets, and you don't want to refuse such a possibility. For a net s : I ---> V such that {j < i} is finite for all i in I, you can ask if the series sum_{i in I} v_i converges, where v_i = s(i). For any function f : I ---> K (not B --> K for a subset B) you can ask if sum_{i in I} f(i)v_i converges. Without nets the problem makes no sense if you try to give a precise meaning to your sums. Otherwise it looks like symbol-pushing. –  KConrad Aug 20 '10 at 2:09
    
If s : I ---> V is a net such that {j < i} is finite for all i in I, then I would define sum_{i in I} s(i) = v to mean: for any open set U around v there is some i_0 in I such that for all i > i_0, sum_{j < i} s(i) lies in U. (Note the sum over j < i has finitely many terms, so we define an infinite series in terms of approximating finite series.) If you instead would like to use a sum over a subset B, please tell us exactly what the equation sum_{b in B} b = v means. –  KConrad Aug 20 '10 at 2:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.