Upper bound

tronter

Let [itex] E [/itex] be a nonempty subset of an ordered set; suppose [itex] \alpha [/itex] is a lower bound of [itex] E [/itex] and [itex] \beta [/itex] is an upper bound of [itex] E [/itex]. Prove that [itex] \alpha \leq \beta [/itex].

So do I just use the following definition: Suppse [itex] S [/itex] is an ordered set, and [itex] E \subset S [/itex]. If there exists a [itex] \beta \in S [/itex] such that [itex] x \leq \beta [/itex] for every [itex] x \in E [/itex], then [itex] \beta [/itex] is an upper bound for [itex] E [/itex], and similarly for lower bound?

arildno

Indeed, so it is fairly trivial.

However, you should exert yourself a bit more for your own sake:
WHY is the condition of non-emptiness of E crucial?

mathman

Easy proof: Let x be an element of E. a<=x<=b. Therefore a<=b.

tronter

If [itex] E [/itex] was empty, then it would not have a upper or lower bound.

arildno

Incorrect!
Try again..

tronter

If [tex] E [/tex] was empty, then there would be no order relation? Or the upper bound and lower bound would be the same?

John Creighto

Didn't someone above prove the original problem via transitivity?

tronter

no but he asked why E must be non empty.

arildno

Nope.

The trick is to rewrite the definitions of upper bounds and lower into EQUIVALENT expressions that makes it clear why, for an empty subset a lower bound can be greater than an upper bound, say:

Now, the standard def. of the upper bound for some set E is that U is an upper bound for E if and only if for every element x in E, we have x<=U.

But, this can be rephrased equaivalently as:

A number U is the upper bound of E if and only if there exists NO elements x in E so that x>U

Similarly can the lower bound L be rephrased.

Now, let us say that E is an empty subset of a set of integers, for simplicity.

Since there are no elements in E at all, there certainly don't exist any elements in E that are greater than 1, so 1 is certainly an upper bound for E.

But, 2 is definitely a LOWER bound of E, by a similar argument.

But 2>1, so a lower bound of E is greater than an upper bound of E.

John Creighto

The result was proved in post #3 via transitivity. As for why E must be not empty why would we define a lower bound for an empty subset of the real numbers?

arildno

Why do maths at all?

Why develop precision in our thinking and understand why conditions stated were NECESSARY for the proposition to be true?

John Creighto

While, I haven't decide water I agree with the first definition, can you prove to me it is logically equivalent to the second?

John Creighto

There are clearly a multitude of reasons and I don't consider why not sufficient.

arildno

In the last one, remember if there can't exist any elements greater than U, then for whatever existing elements in E, they must be equal or less than U. But thereby, the last one implies the first one.

Now, the other way around:
Suppose that whatever elements is in E IS less than or equal to U. Then there can't be any elements in E that are greater than U, meaning the first implies the second statement.

They are equivalent statements, and contrapositives of each other.

arildno

Incoherent and irrelevant remark.

adrian_durham

Or, equivalently, the transitivity proof that goes:

"Easy proof: Let x be an element of E. a<=x<=b. Therefore a<=b."

clearly doesn't work if there is no actual x in E. Maybe it isn't entirely obvious, but you really have to have an actual x, here. You aren't doing something like "IF x WERE an element of E, then...," but rather are actually getting an x from E to use. That requires the x to actually exist (i.e. for E to be nonempty).

arildno

That is indeed why you get the proof working for non-empty sets.

The prposition (i.e that lower bounds are less than or equal to greater bounds) is not meaningless, but just false, for the empty set.