Question on an index family and its max

Question

Trying to find the maximum of $$A = \bigcup_{n\geqslant 1} \left[\frac{1}{n},2-\frac{1}{n}\right]$$

(I'm pretty sure the answer is 2 since the set is bracketed, but the solution manual for my book says there is none!!)

Answer 1

For all $n \ge 1$, we have $2\notin [1/n,2-1/n]$ because $2-1/n<2$, so $2\notin A$.

Answer 2

It is simple enough to see that the sets increase in size if $n_1 > n_2$ , and you can see that the right limit of every set in $A$ is below $2$, then you can prove easily that $\sup A = 2$ but the maximum doesn't exist.

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One way could be to prove that

• $\forall n > 0 $ it is true that $2-\frac{1}{n} < 2 $

• $\forall \epsilon >0 \exists n >0 / \ 2-\frac{1}{n}> 2-\epsilon $

Answer 3

$A=(0,2)$, indeed if $x\in A$, then there exists $n\geqslant 1$ such that $\frac{1}{n}\leqslant x\leqslant 2-\frac{1}{n}$, this means that $x\in (0,2)$. If now $x\in (0,2)$, let $n=1+\max\left(\left\lfloor\frac{1}{2-x}\right\rfloor,\left\lfloor\frac{1}{x}\right\rfloor\right)$, then $x\in \left[\frac{1}{n},2-\frac{1}{n}\right]\subset A$. Thus $\max A$ does not exist because if it existed, you would have $\max A<1+\frac{\max A}{2}$ and $1+\frac{\max A}{2}\in A$ because it is in $(0,2)$. But $\sup A$ exists and $\sup A=2$.