# Thread: precompactness in sequence spaces

1. ## precompactness in sequence spaces

I have 2 questions about precompactness in sequence spacesDoh)
1. Is A = { $\displaystyle (x_{n} : \sum_{n=1}^{\infty} \sqrt{n} |x_{n} | \leq 1$ } is precompact in $\displaystyle c_{0}$ = set of all sequences converging to 0 equipped with the sup norm ?
2. Is A= { $\displaystyle (x_{n} : n x_{n} \rightarrow 0$ as $\displaystyle n \rightarrow \infty$ } in $\displaystyle l_{1}$ ?

2. 1. We notice that if $\displaystyle (x_n)_n\in A$ we have $\displaystyle |x_n|\leq\frac 1{\sqrt n}\, \forall n$. Given an $\displaystyle \varepsilon>0$, we can find a $\displaystyle N$ such that $\displaystyle \frac 1{\sqrt N}\leq \varepsilon$. We can conclude because $\displaystyle \left[0,1\right]^{N-1}$ is precompact.
2. Let $\displaystyle e^{(n)}$ the sequence defined by $\displaystyle e^{(n)}_k=\begin{cases}1& \mbox{if } n=k\\ 0& \mbox{if } n\neq k \end{cases}$.
We notice that $\displaystyle e^{(n)}\in A$ and if $\displaystyle m\neq n,\: ||e^{(n)}-e^{(m)}||_{l^1}=\sqrt 2$. We conclude that $\displaystyle A$ is not precompact.