Given the definition of least upper bound as:
Definition A: Let S be a set of real numbers, then
- a is an upper bound for S if $x \leq a \space \forall x \in S$, and
- b is a least upper bound for S if b is an upper bound, and $b \leq a$ whenever a is an upper bound for S.
We can prove that this is equivalent to the following definition:
Definition B: Let $S \neq \emptyset$. Then b is a least upper bound for S iff
- $x \leq b \space \forall x \in S$, and
- $\forall \epsilon > 0 \space \exists x \in S \space | \space x > b - \epsilon$
Proof: Let b be a l.u.b. Then (1) holds. Suppose (2) is not true. Then $ \exists \space \epsilon > 0 \space \forall \space x \in S, \space x \leq b - \epsilon$. This contradicts that b is the l.u.b. Next, assume (1) and (2). Then b is an upper bound from (1). If for some b', b' < b then from (2) b' is not an upper bound for S. Hence b is the least upper bound.
Now, let $ S = [0,1) \cup \{ a \} \cup \{ b \}$, with the usual ordering for real numbers on $[0,1)$, and $a \geq x \space \forall x \in [0,1)$ and $b \geq x \space \forall x \in [0,1)$. Let $X = [0,1) \subset S$. We can see that both a and b are upper bounds for X. Using definition B we can see that for every $\epsilon > 0, a - \epsilon \in X$ and also $\epsilon > 0, b - \epsilon \in X$. Therefore a and b are both least upper bounds for X.
I was told this by my analysis lecturer a while ago as proof that, while partially ordered sets have a unique least upper bound, subsets of partially ordered sets can have more than one least upper bound, but I never fully understood how it could be true since by definition A, either $a \leq b$ or $b \leq a$.
My question is: why does it seem as if the two equivalent definitions give opposite results?