I read that a subset $B \subseteq X$ is totally bounded if:
$$ \forall \epsilon > 0, \exists N \in \mathbb{N}, \exists a_1,...,a_N \in X \quad \quad s.t.B \subseteq \bigcup_{n=1}^{N}B_{\epsilon}(a_n) $$
where $B_{\epsilon}$ means a ball of radius $\epsilon$...
From the same text, it gave two definitions of compactness:
- a set is compact iff any open cover has a finite sub-cover
- a set is compact when given any cover of balls (of possibly unequal radii), there is a finite sub-collection of them that cover the set
I read many threads that differentiate totally bounded from compactness using concepts of completeness (i.e. Cauchy complete).
But given only these definitions, the only difference I see is that totally bounded requires a $\textit{uniform}$ $\epsilon$ radius for all balls, while compactness allows different values of $\epsilon$ for the radii...
How do these definitions apply to for example $[0,1)$ or $[0,1] \cap \mathbb{Q}$? Is uniformity of the radii of the ball covers essential to differentiate totally bounded and compactness?
A subset of a metric space is compact iff it is totally bounded and complete. Any subset of a totally bounded set is also totally bounded, but a subset of a compact set need not be compact. $[0,1)$ and $[0,1]\cap \mathbb Q$ are both totally bounded because they are subsets of the compact set $[0,1]$. However they are not compact because they are not complete. (A subset of a complete metric space is complete if it is closed and it is easy to see that these two sets are not closed in $\mathbb R$).