Proving the space which consists sequences of finitely many non-zero terms is incomplete with respect to $2$ -norm

Question

In order to prove the space is not complete w.r.t to the given norm, we have to find a Cauchy sequence in the space but it does not converge to an element in the given set.

My attempt was taking $(Z^n) = ( (1,0,0, ... ) , (1, 1/2, 0, 0, ... ), ....)$ I tried to prove this is Cauchy and ended up with,

here $m>n$,

$\lVert Z^n - Z^m \rVert \leq \dfrac {(m-n)^{1/2}}{n}$

how could I make this arbitrarily small? ($<\epsilon$)

here the $n$th term of the sequence would be $Z^n = (1,1/2,1/3,...,1/n,0,0,0...)$ So when $n \rightarrow \infty$, $z^n$ has infinitely many non zero terms, would that be enough to say the space is incomeplete?

Answer 1

As you say, if $m>n$, $$ Z^m-Z^n=\Big(0,\ldots,0,\frac1{n+1},\ldots,\frac1m,0,\ldots\Big). $$ So $$ \|Z^m-Z^n\|_2^2=\sum_{k=n+1}^m\frac1{k^2}\leq\int_n^\infty\frac1{x^2}\,dx=\frac1n. $$ So the sequence is Cauchy: given $\varepsilon>0$, choose $n_0$ with $n_0\geq\frac1{\sqrt \varepsilon}$. Then for any $n,m\geq n_0$ you have $\|Z^n-Z^m\|<\varepsilon$.

Your space embeds naturally in $\ell^2(\mathbb N)$. There it is easy to check that $Z^n\to Z$, where $Z=(1,\frac12,\frac13,\ldots)$. Since $Z$ is not in the space $c_{00}$ of finitely supported sequences and since limits are unique, the sequence $\{Z^n\}$ does not converge in $c_{00}$.

Answer 2

It is enough because of the following argument: Convergence in 2-norm implies convergence of each coordinate. It follows that if your sequence converges in the norm them the limit has to be $(1,\frac 1 2,\frac 1 3, \frac 1 4,....)$, but this last sequence has infinitely many non-zero terms.