- #1
Incand
- 334
- 47
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##.
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##.