# Show that $Y=\{e_n | n\in N\}\subset (\ell_p,d_p),\; 1\leq p< \infty$ is neither totally bounded nor compact

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.

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

$$Y$$ 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.