Edit #1: I apologize to everyone who took their time to read and answer my question. In my book, the second definition actually states that there exists an $x$ that is GREATER than $k-\epsilon$, rather than equal to. The definitions were indeed not equivalent because of this typo.
Edit #2: The universe is the set of real numbers. I should have been more clear about this.
First of all, let's define the upper bound of a set:
k is an upper bound of set A $\iff \text{upperBound}(k, A) \iff \forall x\in A:k\geq x$
Next, I propose two definitions of least upper bound of a set:
Definition 1:
k is THE LEAST upper bound of set A $\iff \text{leastUpperBound}(k, A) \iff \text{upperBound}(k, A) \land \forall k_2(\text{upperBound}(k_2, A)\implies k\leq k_2)$
Definition 2:
k is THE LEAST upper bound of set A $\iff \text{leastUpperBound}(k, A) \iff \text{upperBound}(k, A) \land \forall \epsilon >0 \exists x\in A:x=k-\epsilon$
I have tried and failed to use the rules of first-order logic to prove the equivalence of the two definitions. Help is much appreciated!
Best Answer
Def 2 correctly reads as: $k$ is an upper bound of $A$ and $$ \forall\varepsilon >0,\quad \exists x\in A,\quad x>k-\varepsilon. $$
Def 1 implies Def 2. Let $\sup A =: s$ according to Def 1. Suppose there exists $\varepsilon_0 >0$ such that $x\leqslant s-\varepsilon_0$ for every $x\in A$. Then $s-\varepsilon_0$ is an upper bound of $A$, a contradiction.
Def 2 implies Def 1. Let $\sup A =: s$ according to Def 2. Let $t$ be an upper bound of $A$, that is, $x\leqslant t$ for every $x\in A$. Suppose $s>t$. Then by Def 2, there exists $x\in A$ such that $x>t$, a contradiction. Thus, $s\leqslant t$.