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Let $Y=\{e_n \mid n\in N\}\subset (\ell_p,d_p),\; 1\leq p< \infty$ where $e_n^k = \delta_{kn}$, Kronecker delta. Then show that $Y$ is neither totally bounded nor compact in $X=(\ell_p,d_p),\; 1\leq p< \infty$

It is not compact because $(e_n)_n$ is a sequence in $Y$ that doesn't have any convergent subsequence since $d(e_i,e_j)=2^{\frac1p} > \frac{2^{\frac1p}}{2}\; \forall i\neq j$. So $Y$ is not sequentially compact and hence not compact.

So, any subsequence of $(e_n)$ is not Cauchy and hence not convergent. Hence, $Y$ is not sequentially compact and hence is not compact.

But how do I show that $Y$ is not totally bounded?

Reason why I struggle to show this is because see here for definition of total boundedness for a subset of metric space

Here, the centres can lie in $X$. So that is what is troubling me. Because if the centres were to lie in $Y$ then we can not find $n\in N$ such that $$Y\subseteq \bigcup_{i=1}^{n} B_{\tfrac{2^{\frac1p}}{2}}(e_i)$$ which would mean it is not totally bounded.

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  • $\begingroup$ First, show that any open ball of radius $1/2$ contains at most one member of $Y$. $\endgroup$ Oct 23, 2019 at 9:28

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$Y$ is complete becasue it is a closed subset of a complete space. A basic theorem on compactness in metric spaces says that a metric space is compact iff it is compelte and totally bounded. Hence $Y$ cannot be totally bounded.

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