An element $u$ is an upper bound of $E$ if and only if $t>u$ implies $t\notin E$

Question

Let $S$ be an ordered field and $S \supset E\neq \varnothing$. Then, the following are equivalent:

1. $u \in S$ is an upper bound of $E$.

2. $t \in S$ and $t > u$ implies $t \notin E $.

My Try:

Assume $(1)$ holds. Let $ t \in S$ with $t > u$ and suppose $t \in E$. Then by definition, we must have $u \geq t$. An evident contradiction. Hence, $t \notin E$.

Conversely, suppose $(2)$ holds and say $u$ is not an upper bound of $E$. In particular, we can find some $e \in E$ with $e > u$. But, an application of $(2)$ implies that $ e \notin E$. Another contradiction.

Is this correct? I know this is trivial result, but I would like to know whether I am going through the correct path. Thanks for any advice in advance.

Answer 1

The proof is correct. If you wanted to condense it (not necessarily a good idea), the following could be pointed out.

1. says that for all $t\in S$, the implication $t\in E\implies t\le u$ holds.
2. says that for all $t\in S$, the implication $t>u \implies t\notin E$ holds.

Since $t>u \implies t\notin E$ is the contrapositive of $t\in E\implies t\le u$ they are logically equivalent. Both say the same thing: "either $t\notin E$ or $t\le u$".