Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
1  
Indeed, if I read this right, $K_2$ is already not relatively compact. –  Nate Eldredge Oct 11 '12 at 12:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.