2bornot2b
  • 2bornot2b
What is the significance of the phrase "non empty"?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
It means that a set does have elements, and is not an empty set.
2bornot2b
  • 2bornot2b
I 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
  • 2bornot2b
Should I write down the definitions here?

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JamesJ
  • JamesJ
I 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
  • 2bornot2b
Is it " If it were" or " If it weren't"
2bornot2b
  • 2bornot2b
I didn't get you ...
JamesJ
  • JamesJ
If a subset of the real number, \( A \subset \mathbb{R} \), were empty, then it would have no sup.
JamesJ
  • JamesJ
Hence \[ \sup A \] is defined if and only if \( \emptyset\neq A \subset \mathbb{R} \) and \( A \) has an upper bound.
JamesJ
  • JamesJ
the least upper bound is the supremum. Different words, same thing. Likewise for glb and inf
JamesJ
  • JamesJ
I am. The lub = least upper bound. The lub is the supremum. Same thing.
2bornot2b
  • 2bornot2b
And why is it absent in the definition of upperbound
JamesJ
  • JamesJ
upper bound or least upper bound?
2bornot2b
  • 2bornot2b
Just upper bound
JamesJ
  • JamesJ
Give 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
  • 2bornot2b
Definition 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
  • 2bornot2b
My point is why not say "let S be a non empty set of real numbers"
JamesJ
  • JamesJ
Very 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
  • 2bornot2b
But I copied that definition from T.M. Apostol, of course someone before me would have pointed this out, if it were not acceptable.
JamesJ
  • JamesJ
It could also be clear in the context in which this definition is given that the subset is non-empty. 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 non-empty. And move on.
2bornot2b
  • 2bornot2b
Sorry my computer crashed, and it got restarted.
2bornot2b
  • 2bornot2b
Also 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
  • 2bornot2b
I 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
  • JamesJ
An open interval is by definition necessarily non-empty, 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
  • 2bornot2b
Have 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
  • JamesJ
Right, 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
  • 2bornot2b
Okay,... I am getting you.... Thanks.
JamesJ
  • JamesJ
**we [not well]
2bornot2b
  • 2bornot2b
what did you mean by that, I am sorry I didn't understand that...

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