Is Sup P Always Less Than or Equal to Sup Q When P Is a Subset of Q?

[mateomy]

If $P\,\subset\,Q\,\subset\,\mathbb{R},\,P\,\neq\,emptyset$ and P and Q are bounded above, show that sup P $\leq$ sup Q.

I can visualize the reality of this but I can't put it down nicely. This is what I've done so far:

Assume P is a subset of Q. Then sup P $\in$ Q. If Q is a subset of $\mathbb{R}$ then sup Q $\in\,\mathbb{R}$.

I don't know how convincing this is so any pointers would be appreciated.

 

[algebrat]

If $P\,\subset\,Q\,\subset\,\mathbb{R},\,P\,\neq\,emptyset$ and P and Q are bounded above, show that sup P $\leq$ sup Q.

I can visualize the reality of this but I can't put it down nicely. This is what I've done so far:

Assume P is a subset of Q. Then sup P $\in$ Q. If Q is a subset of $\mathbb{R}$ then sup Q $\in\,\mathbb{R}$.

I don't know how convincing this is so any pointers would be appreciated.

If $P=Q=(0,1)$, then $\sup P\notin Q$, but you claim $\sup\in Q$.

Try chopping defintion of supremum into two parts,

1. If $x\in A$, then $x\le\sup A$
2. If $r<\sup A$, then there is $x\in A$ such that $r<x$.

In words, we can remember this as, least upper bound is

1. an upper bound
2. the least such. I.e., any smaller number is not an upper bound.

Play with that, for $P$ and $Q$.

 

[mateomy]

So if I say something along the lines of:

If x $\in$ P then x $\in$ Q since we know that P $\subset$ Q. By the properties of upper bounds we can say that x < supQ, which supQ is also an upper bound of P. Since supP is the least upper bound of P we have supP < supQ.

It feels sloppy.

 

[algebrat]

So if I say something along the lines of:

If x $\in$ P then x $\in$ Q since we know that P $\subset$ Q. By the properties of upper bounds we can say that x < supQ,

Should say, $x\le\sup Q$.

which supQ is also an upper bound of P.

Correct, with my correction.

Since supP is the least upper bound of P we have supP < supQ.

It is not clear how you came to this conclusion. Also, strict inequality would be false, since we could possibly have $\sup P=\sup Q$

It feels sloppy.

Well, it is. Keep trying.

 

[algebrat]

which supQ is also an upper bound of P. Since supP is the least upper bound of P we have supP < supQ.

Oh, I just reread it, since supQ is an upper bound for P, and supP is the least such one, we must have supP≤supQ. So you basically had it.

 

[mateomy]

Thanks for the help. This stuff is frustrating because I can visualize it and see why its true but putting it down on the paper is another beast altogether.

 

[algebrat]

Thanks for the help. This stuff is frustrating because I can visualize it and see why its true but putting it down on the paper is another beast altogether.

The devil is in the details.

People complain about proofs in math, but the alternative is 20 page papers for english, or 10 page technical papers in science. You want to find your niche in life, everybody's different.

In math, it is far too tempting to assume something that seems intuitive, that may for some strange reason in fact be false. Only when you go to try to prove it might we come closer to the fact that it is false. Sometimes, it takes a great deal of effort, alternately trying to prove or find a counterexample before we may know something better.

So it is a good skill if it is for you, and like Mr Miyagi basically said, it takes a lot of practice.

 

[algebrat]

Try chopping defintion of supremum into two parts,

1. If $x\in A$, then $x\le\sup A$
2. If $r<\sup A$, then there is $x\in A$ such that $r<x$.

In words, we can remember this as, least upper bound is

1. an upper bound
2. the least such. I.e., any smaller number is not an upper bound.

There is one more form that I like, which replaces the second condition with something like it's contrapositive, if for all x in A, r is at least x, then r is at least sup A.