Covering an interval by a finite number of arbitrary small intervals

Question

Let $\epsilon>0$, how to prove the existence of a finite number of reals $x_1,\dots,x_k\in [0,1]$ such that $$[0,1]\subset \bigcup_{i=1}^{k}\left[x_{i}-\epsilon,x_{i}+\epsilon\right].$$

My attempt:

If $\epsilon>\frac{1}{2}$, then $[0,1]\subset \left[\frac{1}{2}-\epsilon,\frac{1}{2}+\epsilon\right].$

If $\frac{1}{2}> \epsilon>\frac{1}{3}$, then $[0,1]\subset \left[\frac{1}{3}-\epsilon,\frac{1}{3}+\epsilon\right]\cup \left[\frac{2}{3}-\epsilon,\frac{2}{3}+\epsilon\right]$...

Is this the right path?

Answer 1

Let $k$ be an integer greater than $\frac{1}{\epsilon}$.

Define your $x_i$ as follows.

$$x_1= \frac {\epsilon}{ 2}, x_2= \frac{3\epsilon}{ 2}, x_3=\frac { 5\epsilon}{2},...,x_k= \frac {(2k-1)\epsilon}{2}$$
Then the interval $[0,1]$ will be covered by the union.

$$ [0,1]\subset \bigcup_{i=1}^{k}\left[x_{i}-\epsilon,x_{i}+\epsilon\right].$$

Answer 2

Your intuition is ok. Given $\epsilon$ you divide $[0,1]$ into intervals of length $2\epsilon$. Like toilet paper. If you need a number $k$ of such intervals to cover $[0,1]$ then $k$ verifies $1<k\times 2\epsilon$, i.e. you'll need at least $1/(2\epsilon)$.

Once you have that, just distribute your $k$ numbers uniformly on $[0,1]$, as follows: $$x_i = 2 k\epsilon-\epsilon.$$

It should be now clear to show that every number on $[0,1]$ is at distance less than $\epsilon$ to one of these reals, proving your claim.