High School Doubt regarding least upper bound?

[Alpharup]

I am using Spivak Calculus. I have a general doubt regarding the definition of least upper bound of sets.
Let A be any set of real numbers and A is not a null set. Let S be the least upper bound of A.
Then by definition "For every x belongs to A, x is lesser than or equal to S"

Let M be an upper bound such that M is less than S and M does not belong to A(Assertion 1)

Then, it leads to contradiction because A is bounded above by M and M is less than least upper bound S, which means an upper bound is less than least upper bound.
So, we have either M=S or M>S. So, Assertion 1 is false.

Then by converse of Assertion 1, if N is less than S, it may belong to A but can't be an upper bound or

if N is less than S, it may not belong to A. (Assertion 2)

Is my logic right?

 

[Svein]

As long as your "or" and "and" are taken as logical operators and not English conjunctions, your loigic seems OK.

 

[Mark44]

I am using Spivak Calculus. I have a general doubt regarding the definition of least upper bound of sets.
Let A be any set of real numbers and A is not a null set. Let S be the least upper bound of A.
Then by definition "For every x belongs to A, x is lesser than or equal to S"

Let M be an upper bound such that M is less than S and M does not belong to A(Assertion 1)

Then, it leads to contradiction because A is bounded above by M and M is less than least upper bound S, which means an upper bound is less than least upper bound.
So, we have either M=S or M>S. So, Assertion 1 is false.

My take on this is that S really wasn't the least upper bound.

Consider the set $A = \{x \in \mathbb{R} | 0 \le x \le 1\}$; i.e., the open interval (0, 1).
I claim that 1.5 is the lub for A. I later notice that 1 is not an element of A, and 1 < 1.5. This doesn't contradict anything, as all I have really done here is to mistakenly identify 1.5 as the lub.

 

[HallsofIvy]

I am using Spivak Calculus. I have a general doubt regarding the definition of least upper bound of sets.
Let A be any set of real numbers and A is not a null set. Let S be the least upper bound of A.
Then by definition "For every x belongs to A, x is lesser than or equal to S"

That's a property that every upper bound has, not just the least upper bound. To uniquely refer to the "least upper bound you must add "if M is an upper bound on A then S\le M

Let M be an upper bound such that M is less than S and M does not belong to A(Assertion 1)

This is not an "assertion" because it is not a complete statement. You have defined M to have certain properties but have not said anything about it.
And "let M be an upper bound such that M is less than S" violates the definition of least upper bound. Such an M does not exist.

Then, it leads to contradiction because A is bounded above by M and M is less than least upper bound S, which means an upper bound is less than least upper bound.
So, we have either M=S or M>S. So, Assertion 1 is false.

Then by converse of Assertion 1, if N is less than S, it may belong to A but can't be an upper bound or
if N is less than S, it may not belong to A. (Assertion 2)

Is my logic right?

A number that is less than the least upper bound is cannot be an upper bound by definition of "least"!

 

[jbriggs444]

A number that is less than the least upper bound is cannot be an upper bound by definition of "least"!

That seems to have omited one word and should have read:

A number that is less than the least upper bound is cannot be a least upper bound by definition of "least"!

Edit: Doh, I misunderstood.

 

[HallsofIvy]

No, it should be exactly what I said! Please read it again!