Which definition of the least upper bound is right

Question

I have encountered two definitions of the least upper bound.

Let $A$ be a set of real numbers. The two definitions are

First definition

$k$ is an upper bound of $A$ iff (1) $\forall \ x \in A$, $x \leq k$ and (2) if $c$ is an upper bound of $A$, $c \geq k$.

Second definition

$k$ is an upper bound of $A$ iff (1) $\forall \ x \in A$, $x \leq k$ and (2) $\forall \ \epsilon > 0$, there is an element $x$ in $A$ such that $k - \epsilon < x \leq k$.

Are they essentially the same? I'm inclined to say yes, but I'm not sure.

Answer 1

yes, they are equivalent.

Both definitions agree on the upper bound part.

On the least upper bound the statements sound different but say the same thing.

For the first definition no upper bound is allowed to get less than $k$, so $k$ is the least upper bound.

For the second one, $k-\epsilon$ is not allowed to be an upper bound due to the fact that the set has an element greater than $k-\epsilon.$

Answer 2

Yes, they are equivalent.

Suppose $k$ satisfies definition $1$, given $\epsilon >0$, if $k \in A$, then we can pick $x$ to be $k$. Suppose $k \notin A$, and suppose there is no $x \in A$, that satisfy $k-\epsilon < x \le k$, then $k-\epsilon$ is an upper bound of $A$ but $k - \epsilon < k$ which is a contradiction. Hence if $k$ satisfies definition $1$, it satisfies definition $2$.

Suppose $k$ satisfies definition $2$, if $c$ is an upper bound of $A$. suppose on the contrary that $c < k$. then $k-c>0$, try to get a contradiction to prove that the two definitions are equivalent.