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14yamaka
 3 years ago
Best ResponseYou've already chosen the best response.0It means that a set does have elements, and is not an empty set.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0I find the phrase "nonempty" absent in the definition of supremum, but it is present in the definition of completeness axiom. Can you explain why?

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Should I write down the definitions here?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1I know what they are. I'm sure in the context, it's clear that the set for which you're finding the sup is not empty. If it were, the sup wouldn't be defined.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Is it " If it were" or " If it weren't"

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1If a subset of the real number, \( A \subset \mathbb{R} \), were empty, then it would have no sup.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Hence \[ \sup A \] is defined if and only if \( \emptyset\neq A \subset \mathbb{R} \) and \( A \) has an upper bound.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1the least upper bound is the supremum. Different words, same thing. Likewise for glb and inf

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1I am. The lub = least upper bound. The lub is the supremum. Same thing.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0And why is it absent in the definition of upperbound

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1upper bound or least upper bound?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Give me the definition you're talking about. But in any case, we say the empty set has no upper or lower bound; we just don't define it if the set is empty.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Definition of Upper Bound: Let S be a set of real numbers. If there is a real number b such that \[x\le b\] for every x in S, then b is called an upper bound for S and we say that S is bounded above by b

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0My point is why not say "let S be a non empty set of real numbers"

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Very technically, you are correct. Because by this definition, every real number is an upper bound of the empty set. And that leads to a clash with the completeness axiom because there is no least upper bound, no sup of the empty set. So to avoid this problem, we should say the empty set has no upper bound. agreed.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0But I copied that definition from T.M. Apostol, of course someone before me would have pointed this out, if it were not acceptable.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1It could also be clear in the context in which this definition is given that the subset is nonempty. But in any case, whatever this book is, it isn't the word of god. Take it from me that, yes: this only makes sense if the subset is nonempty. And move on.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry my computer crashed, and it got restarted.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Also take a look at this definition of neighbourhood of a point (Its from the lecture note of our prof) Let \[c\in \mathbb{R}\]. A subset S of R is said to be a neighbourhood of c, if there exists an open interval (a,b) such that \[c\in (a,b)\subset S\]

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0I mean I would like to state it as "Let c∈R. A non empty subset S of R is said to be a neighbourhood of c, if there exists an open interval (a,b) such that c∈(a,b)⊂S"

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1An open interval is by definition necessarily nonempty, so in this case, I would argue it's not necessary. To be more explicit, an open interval is a subset \[ T = \{ x \  \ a < x < b \} \] where \( a < b \) and therefore \( T \) is not empty.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Have a look at this http://en.wikipedia.org/wiki/Empty_set#Extended_real_numbers The first line.. that starts with "since the empty set........."

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Right, whatever that article is talking about "Extended Reals" or whatever else you want to call it, that's not the Real Numbers as well usually define them.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Okay,... I am getting you.... Thanks.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0what did you mean by that, I am sorry I didn't understand that...
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