Let $\epsilon > 0$ be any positive number. Some $a \in A$ with $a > x - \epsilon$.

Question

Let $A \subset \mathbb{R}$ be nonempty set that is bounded above. Suppose that $x$ is the least upper bound for $A$. Now let $\epsilon > 0$ be any positive number. Explain why you know that there must be some $a ∈ A$ with $a > x - \epsilon$.

What I have so far: Since $\epsilon$ is greater than $0$ then I assume that $\epsilon > A \subset \mathbb{R}$. When we subtract the least upper boundaries or the min of the set from the $\max (\epsilon(x - \epsilon))$ you should get some value that is less than a so $x =$ least upper bound $\epsilon > 0$ $a \in A$ $A \subset \mathbb{R}$ $x - \epsilon > 0$ but $< a$ so $x - \epsilon < a$. Is my reasoning correct?

Answer 1

If there was no $a\in A$ such that $a>x-\varepsilon$ You have $$x-\varepsilon\geq a$$ for all $a\in A$ and thus $x-\varepsilon$ would be an upper bound for $A$ contradicting that $x$ is the least upper bound.

Answer 2

The idea is that $x-\epsilon < x$.

The definition for $x$ being an upper bound of $A$ means (along with $x$ being an upper bound of $A$) is that anything less than $x$ is not an upper bound of $A$.

$x-\epsilon < x$ so $x-\epsilon$ is not an upper bound of $A$.

The definition of $y$ being an upper bound of $A$ is that $y$ is equal to or larger then any $a$ in $A$. Since $x-\epsilon$ is not an upper bound of $A$ this is not true for $x-\epsilon$.

So it is not true that every $y$ is equal to or larger than any $a$ in $A$, there must be at least one $a \in A$ so that $x-\epsilon$ is smaller than $a$.