2

Let $\mathbb{R} \supset E \neq \varnothing$. Put $\alpha = \sup E $. Then, for all $n \in \mathbb{N}$, $\alpha - \frac{1}{n} $ is not an upper bound of $E$, but $\alpha + \frac{1}{n}$ is an upper bound.

Solution

We have $\frac{1}{n} > 0 $ for all $n \in \mathbb{N}$. Suppose there is some $n_0 \in \mathbb{N}$ such that $\alpha - \frac{1}{n_0}$ is an upper bound of $E$. By definition, we have

$$ \alpha \leq \alpha - \frac{1}{n_0} < \alpha \implies \alpha < \alpha $$

Contradiction. $\alpha + \frac{1}{n}$ is obvious an upper bound of $E$ since $\alpha < \alpha + \frac{1}{n} $ for all $n$.

Community
  • 1

1 Answers1

0

If $\alpha-\frac{1}{n}$ is an upper bound for $E$, then $x\le\alpha-\frac{1}{n}$, for all $x\in E$. But you can't use $\alpha$ for $x$, because you don't know that $\alpha\in E$.

However, the property of the least upper bound guarantees that, for every $\varepsilon>0$, there is $x\in E$ with $\alpha-x<\varepsilon$. If $\varepsilon=\frac{1}{n}$, you have $$ \alpha-\frac{1}{n}<x\le\alpha-\frac{1}{n} $$ which is a contradiction.

However, also your reasoning is correct, once you have proved that the set of upper bounds of $E$ (supposing it's not empty) has a minimum element (namely the least upper bound).

For the second part, note that if $b$ is an upper bound of $E$ and $c>b$, then $c$ is an upper bound of $E$ (so your reasoning is correct).

egreg
  • 238,574