# How to show that this union is relatively compact using total boundedness

Edit I've made a mistake in the formulation. There should be in inclusion, not an equality.

Let $E$ be a Banach space. Let $\varnothing\neq K_j\subset E$ be compact for all $j\ge1$ such that $K_{j+1}\subset\{x+y: x\in K_j~\mbox{and } y\in E~\mbox{ such that } \|y\|\le \eta_j\}$ where $(\eta_j)$ a sequence of strictly positive real numbers such that $\sum_j\eta_j$ converges in $\mathbb{R}$.

How can I show that $\cup_{j\ge1}K_j$ is relatively compact using total boundedness? I have been able to prove this using a diagonal argument, but the proof is quite messy. I feel like using total boundedness is easier since the closure of this union is complete.

-

We have $K_{j+1}=K_j+B(0,\eta_j\}$ so by induction $K_j=K_0+B\left(0,\sum_{k=1}^j\eta_k\right)$. This gives $$K:=\bigcup_{j=1}^{+\infty}K_j=K_0+B\left(0,\sum_{j=1}^{+\infty}\eta_j\right)\supset B\left(x_0,\sum_{j=1}^{+\infty}\eta_j\right),$$ where $x_0\in K_0$ and $B(x,r):=\{y\in E,\lVert x-y\rVert<r\}$. In particular, if $E$ is an infinite dimensional space $K$ is not relatively compact.

So with this construction, $K$ is relatively compact if and only if $E$ is finite dimensional.

-
Indeed, if I read this right, $K_2$ is already not relatively compact. –  Nate Eldredge Oct 11 '12 at 12:50

This is counterexample for the original question.

Consider $K_1=\{0\}$ and $\eta_j=j^{-2}$, then $$\bigcup\limits_{j=1}^{\infty} K_j= \mathrm{Ball}\left(0,\sum\limits_{j=1}^\infty j^{-2}\right)= \mathrm{Ball}\left(0,\frac{\pi^2}{6}\right)$$ If $E$ is infinite dimensional any ball is not relatively compact.

This is answer to the edited question.

Fix $\varepsilon>0$. We know that $K_j$ is a compact for all $j\in\mathbb{N}$, hence for all $j\in\mathbb{N}$ there exist $\{x_{j,l}:l=1,\ldots,N_j\}$ such that $$K_j\subset \bigcup\limits_{l=1}^{N_j}\mathrm{Ball}\left(x_{j,l},\frac{\varepsilon}{2}\right)$$ Since the series $\sum\limits_{j=1}^\infty\eta_j$ converges there exist $m\in\mathbb{N}$ such that $\sum\limits_{j=m}^\infty\eta_j<\frac{\varepsilon}{2}$. Then for all $j>m$ \begin{align} K_j&\subset K_m+\mathrm{Ball}(0,\eta_m)+\ldots+\mathrm{Ball}(0,\eta_{j-1})\\ &\subset K_m+\mathrm{Ball}\left(0,\sum\limits_{i=m}^j\eta_j\right)\\ &\subset K_m+\mathrm{Ball}\left(0,\sum\limits_{i=m}^\infty\eta_j\right)\\ &\subset K_m+\mathrm{Ball}\left(0,\frac{\varepsilon}{2}\right)\\ &\subset \bigcup\limits_{l=1}^{N_m}\mathrm{Ball}\left(x_{m,l},\frac{\varepsilon} {2}\right)+\mathrm{Ball}\left(0,\frac{\varepsilon}{2}\right)\\ &\subset \bigcup\limits_{l=1}^{N_m}\left(\mathrm{Ball}\left(x_{m,l},\frac{\varepsilon}{2}\right)+\mathrm{Ball}\left(0,\frac{\varepsilon}{2}\right)\right)\\ &\subset \bigcup\limits_{l=1}^{N_m}\mathrm{Ball}(x_{m,l},\varepsilon) \end{align} Since the last inclusion holds for all $j>m$ we conclude $$\bigcup\limits_{j=m+1}^\infty K_j\subset \bigcup\limits_{l=1}^{N_m}\mathrm{Ball}(x_{m,l},\varepsilon)$$ Hence \begin{align} \bigcup\limits_{j=1}^\infty K_j&=\left(\bigcup\limits_{j=1}^m K_j\right)\cup\left(\bigcup\limits_{j=m+1}^\infty K_j\right)\\ &\subset\left(\bigcup\limits_{j=1}^m \bigcup\limits_{l=1}^{N_j}\mathrm{Ball}\left(x_{j,l},\frac{\varepsilon}{2}\right)\right)\cup\left(\bigcup\limits_{l=1}^{N_m}\mathrm{Ball}(x_{m,l},\varepsilon)\right)\\ &\subset\left(\bigcup\limits_{j=1}^m \bigcup\limits_{l=1}^{N_j}\mathrm{Ball}(x_{j,l},\varepsilon)\right)\cup\left(\bigcup\limits_{l=1}^{N_m}\mathrm{Ball}(x_{m,l},\varepsilon)\right)\\ &=\bigcup\limits_{j=1}^m \bigcup\limits_{l=1}^{N_j}\mathrm{Ball}(x_{j,l},\varepsilon) \end{align} Thus, for each $\varepsilon>0$ we found finite $\varepsilon$-net $\{x_{j,l}:l=1,\ldots,N_j,\;j=1,\ldots,m\}$ for the set $\bigcup\limits_{j=1}^\infty K_j$. Hence it is relatively compact.

-