Why is the infinite dimensional vector space with only finitely many nonvanishing components incomplete?

269 Views Asked by At

Define a complex vector space $V$ such that any element $\{a_i\}=(a_1,a_2,\dots)\in V$ has only finitely many components $a_i\ne 0$. The inner product is defined as

$$(\{a_i\},\{b_j\})=\sum_i^\infty a_i^* b_i$$

Can we find an example to show that $V$ is incomplete?

One example I have is this sequence $\langle i\rangle=(0,0,\dots,1/i,0,0,\dots)$, i.e., the $i$-th element of the sequence has only one nonzero component, which is the $i$-th component and whose value is $1/i$. But this sequence should converge to $0$....

I need help! Thanks!

1

There are 1 best solutions below

5
On

Consider the sequence $\{a^{(k)}\}_{k=1}^\infty$ where $a^{(k)} = \{a_{i}^{(k)}\}_{i=1}^\infty$, and we define $$ a_i^{(k)} = \begin{cases} 1/i & i \leq k\\ 0 & i > k \end{cases} $$ So, we have $$ a^{(1)} = \{1,0,0,0,0,\dots\}\\ a^{(2)} = \{1,1/2,0,0,0,\dots\}\\ a^{(3)} = \{1,1/2,1/3,0,0,\dots\}\\ \vdots $$ Why is this a Cauchy sequence in $V$? How do we know that the sequence doesn't converge?