Can the maximum be the supremum?
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Can the maximum be the supremum?
Note that if S has a maximum, then the maximum must be the supremum: indeed, if t∈X is such that s≤t for all s∈S, then in particular m∈S, so m≤t, proving that m satisfies the conditions to be the supremum.
How do you prove something is the supremum?
Suppose that A ⊂ R is a set of real numbers. If M ∈ R is an upper bound of A such that M ≤ M′ for every upper bound M′ of A, then M is called the supremum of A, denoted M = sup A. If m ∈ R is a lower bound of A such that m ≥ m′ for every lower bound m′ of A, then m is called the or infimum of A, denoted m = inf A.
How do you prove Infimum and Supremum?
Similarly, given a bounded set S ⊂ R, a number b is called an infimum or greatest lower bound for S if the following hold: (i) b is a lower bound for S, and (ii) if c is a lower bound for S, then c ≤ b. If b is a supremum for S, we write that b = sup S. If it is an infimum, we write that b = inf S.
Is sup and Max the same?
A maximum is the largest number WITHIN a set. A sup is a number that BOUNDS a set. A sup may or may not be part of the set itself (0 is not part of the set of negative numbers, but it is a sup because it is the least upper bound). If the sup IS part of the set, it is also the max.
How do you prove an upper bound?
An upper bound which actually belongs to the set is called a maximum. Proving that a certain number M is the LUB of a set S is often done in two steps: (1) Prove that M is an upper bound for S–i.e. show that M ≥ s for all s ∈ S. (2) Prove that M is the least upper bound for S.
How do you prove the greatest lower bound?
Prove that the inf of S = (1,5] is 1. 2 satisfies 1 1 + ǫ is not a lower bound, proving that 1 is the greatest lower bound.
How do you prove something is an upper bound?
How do you prove an upper bound is the least upper bound?
Let S be a non-empty set of real numbers.
- A real number x is called an upper bound for S if x ≥ s for all s ∈ S.
- A real number x is the least upper bound (or supremum) for S if x is an upper bound for S and x ≤ y for every upper bound y of S.
How do you prove upper bounds?
How do you prove a lower bound of a set?
To say “a is a lower bound of S” is to say that a≤s for all s∈S. To prove this, you show exactly that: start with an arbitrary s∈S, use the definition of membership of S, then via some logical path, prove that a≤s.
Do all bounded sequences have a supremum?
Every nonempty subset of that is bounded above has a supremum.
Can a set not have a supremum?
A set of natural numbers indeed doesn’t have a supremum among natural numbers. Some extension of a set of natural numbers is required. Thanks.
How do you prove a set has the least-upper-bound property?
It is possible to prove the least-upper-bound property using the assumption that every Cauchy sequence of real numbers converges. Let S be a nonempty set of real numbers. If S has exactly one element, then its only element is a least upper bound.
Is there always a supremum?
Yes, depends on the partial ordering you define. A partial ordering for which a bounded set have sup and inf is called complete. I think there might be something wrong with your example. You show that your partial order isn’t a total order, but a complete order is different.
