Let ($\mathbb{R}$,|.|) be the real metric space.
Let $a\lt b \in \mathbb{R}$.
Show that $([a,b],|.|)$ is totally bounded ( knowing that (X,d) is totally bounded if $\forall \varepsilon \gt0 $ , $\exists x_{1},…,x_{n}\in X$ such that $ X\subseteq \bigcup_\limits{i\in {1,…,n}} B(x_{i},\varepsilon) $ ).
Let $\varepsilon \gt0$ there exists $n\in \mathbb{N}$ such that $n\varepsilon \gt (b-a) $
Now $\forall k\in {0,…,n} $ Let $x_{k}=a+k\frac{(b-a)}n$. Let's show that $[a,b]\subseteq \bigcup_\limits{k\in {0,…,n}} B(x_{k},\varepsilon)$.
Let $x\in[a,b]$.
If $x=a$ we have $x\in B(x_{0},\varepsilon)$ And if $x=b$ we have $x\in B(x_{n},\varepsilon)$ .. I'm stuck at the point where $x\in]a,b[$ .. Any help please.
Every closed interval in $\mathbb{R}$ is totally bounded
metric-spaces
Best Answer
if $x\in (a,b)$ take the smallest $x_i$ such that $x\leq x_i$. It follows $x_i-\frac{b-a}{n}<x\leq x_i$ and so $|x_i-x| = x_i-x < \frac{b-a}{n} < \epsilon$.