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I Bounded above definition

  1. Apr 10, 2016 #1
    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. jcsd
  3. Apr 10, 2016 #2

    Samy_A

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    ##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##.
     
  4. Apr 10, 2016 #3
    Thanks! That would explain it.
     
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