Let $ [a,b] \subseteq \mathbb{R} $ be a non-degenarate closed interval

Question

Let $ [a,b] \subseteq \mathbb{R} $ be a non-degenarate closed interval . For $ \epsilon >0 $ prove that there is a partition $ P $ of $ [a,b] $ such that $ || P|| < \epsilon $.

Answer: Let $P: \ a=x_{0} <x_{1} <x_{2} <...........< x_{n}=b $ be the partition of $ [a,b] $.

||P||=$ max \{|x_{i}-x_{i-1}| \}$ , i=1,2,.........,n.

Now how to proceed? Any help please ?

You can always find $N\in \Bbb N$ such that $(b-a)/N<\epsilon$, just cut $[a,b]$ into $N$ equal parts — Nick, Jul 01 '17 at 09:52
"Any idea is there ?" Funny, I was going to ask you exactly this... — Did, Jul 02 '17 at 07:59

score 2 · Answer 1 · answered Jul 01 '17 at 10:25

From a different perspective: If $P:x_0<x_1<\ldots<x_n$ is a partition, then $P':x_0<\frac{x_0+x_1}2<x_1<\frac{x_1+x_2}2<\ldots <x_n$ is a partition with $\|P'\|=\frac 12\|P\|$. As there exist some partition at all, this implies that the infimum of $\|P\|$ where $P$ runs over all partitions must be $0$.

score 1 · Answer 2 · answered Jul 01 '17 at 09:54

1

Let $n$ be so large that $\frac{b-a}{n} < \epsilon$. Let $x_i = a + \frac{i}{n}(b-a)$, for $i=0,..,n$.

Then $|x_i - x_{i-1}| = \frac{b-a}{n}< \epsilon$, for $i=1,..,n$.

answered Jul 01 '17 at 09:54

Epiousios

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Let $ [a,b] \subseteq \mathbb{R} $ be a non-degenarate closed interval

2 Answers2