Fake proof that every nonempty subset of $\mathbb R$ has a maximum element

Question

Let $A \subseteq \mathbb{R}$ be nonempty and let sup(A)=$\alpha$. Then we have $$(\forall \epsilon > 0)(\exists x \in A)~x>\alpha-\epsilon$$ Note that $x > \alpha-\epsilon$ for all $\epsilon > 0$ implies $x \geq \alpha$. Additionally, $x \leq \alpha$ since $\alpha$ is an upper bound of A. Thus we get $x = \alpha$, which is not necessarily true. But I don't know where goes wrong.

Answer 1

The point is the order of the quantifiers matters. For each $\epsilon$, there is an $x$ which depends on $\epsilon$ such that $x>\alpha-\epsilon$. We’re not saying there is an $x$ such that for every $\epsilon$, $x>\alpha-\epsilon$.

Answer 2

Your error is here : for all $\varepsilon$, there exists an $x$ that is greater than $\alpha-\varepsilon$. Here, $x$ depends on $\varepsilon$ because the "there exists" comes after : you find $x$ after having chosen $\varepsilon$. Change $\varepsilon$ and the $x$ you found may not be greater than $\alpha-\varepsilon$.

For instance : let $\alpha=10$ and $\varepsilon=0.1$. Then $x=9.9999$ satisfies $x>\alpha-\varepsilon$.

Now let $\varepsilon=0.01$, then the same $x$ is suitable.

But let $\varepsilon=0.0000000000001$ and now your $x$ is not suitable, you have to find another $x$.