# I Bounded above definition

1. Apr 10, 2016

### Incand

Is the subset $E$ necessary in the following definition? It doesn't seem to serve any purpose at all and could've been written with $S$ directly? Isn't $E$ just another ordered set since it's a subset of $S$?

Definition:
Suppose $S$ is an ordered set, and $E \subset S$. If there exists a $\beta \in S$ such that $x \le \beta$ for every $x \in E$, we say that $E$ is bounded above, and call $\beta$ and upper bound of $E$.

Definition of order:
Let $S$ be a set. An order on $S$ is a relation, denoted by $<$, with the following properties:
(i) If $x\in S$ and $y \in S$ then one and only one of the statements
$x<y,\; \; \; x=y, \; \; \; y < x$ is true.
(2) If$x,y,z \in S$, if $x<y$ and $y< z$ then $x<z$.

2. Apr 10, 2016

### Samy_A

$E$ as a subset of $S$ makes sense.
The upper bound $\beta$ isn't necessarily an element of $E$.

Example: the real interval ]0,1[. It is bounded above as a subset of $\mathbb R$.

3. Apr 10, 2016

### Incand

Thanks! That would explain it.

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