Is Proofing Supremum and Infimum as Easy as It Seems?

[playboy]

Hi.

Can somebody please check my work, its this dumb proof in the textbook which is the most obvious thing.

Let S and T be nonempty bounded subsets of R with $S \subseteq T$.
Prove that $inf T \leq inf S \leq sup S \leq sup T$.

I first broke it up into parts and tried to prove each part.

1. $sup S \leq sup T$
2. $inf S \leq sup S$
3. $inf T \leq inf S$

also, define:

sup T = a
sup S = b
inf S = c
inf T = d

1. Prove that $sup S \leq sup T$
By Definition, let X be a nonempty subset of R and let X be bounded above. Thus, m = sup X iff $m \geq x$. for all $x \in X$

Since S is bounded above, S has a supremum and say, sup S = b. But, $S \subseteq T$, and thus, $b \in T$.

But, by definition, a = supT iff a is (greater than or equal too) all other elements in T

Thus, $a \geq b$ and thus, $sup T \geq sup S$

2. Prove that $inf S \leq sup S$
By Definition, inf S;
c = inf S iff c is (less than or equal too) all elements in S.
and, sup S is
b = supS iff b is (greater than or euqal too) all elements in S.
Thus, since $c \leq s$ and $b \geq s$, where s is (all elements in S), $c \leq b$ as required.

3. This would be the same proof as 1, but using the definition of infimum instead.

So can somebody please check this proof and if I am doing it properly?

 

[JasonRox]

Hi.

Can somebody please check my work, its this dumb proof in the textbook which is the most obvious thing.

Let S and T be nonempty bounded subsets of R with $S \subseteq T$.
Prove that $inf T \leq inf S \leq sup S \leq sup T$.

I first broke it up into parts and tried to prove each part.

1. $sup S \leq sup T$
2. $inf S \leq sup S$
3. $inf T \leq inf S$

also, define:

sup T = a
sup S = b
inf S = c
inf T = d

1. Prove that $sup S \leq sup T$
By Definition, let X be a nonempty subset of R and let X be bounded above. Thus, m = sup X iff $m \geq x$. for all $x \in X$

Since S is bounded above, S has a supremum and say, sup S = b. But, $S \subseteq T$, and thus, $b \in T$.

But, by definition, a = supT iff a is (greater than or equal too) all other elements in T

Thus, $a \geq b$ and thus, $sup T \geq sup S$

2. Prove that $inf S \leq sup S$
By Definition, inf S;
c = inf S iff c is (less than or equal too) all elements in S.
and, sup S is
b = supS iff b is (greater than or euqal too) all elements in S.
Thus, since $c \leq s$ and $b \geq s$, where s is (all elements in S), $c \leq b$ as required.

3. This would be the same proof as 1, but using the definition of infimum instead.

So can somebody please check this proof and if I am doing it properly?

Looks good to me.

I don't think a proof of 2 is required. That's the definition. I hate examples that say prove this when it's the definition.

 

[playboy]

Thanks for checking.
I hate these proofs also, its so obvious and all I am doing is just restating the definition.

 

[JasonRox]

Thanks for checking.
I hate these proofs also, its so obvious and all I am doing is just restating the definition.

I hate them too, but sometimes doing proofs like this help with understanding and remembering definitions. Nevertheless, still annoying.

 

[Hurkyl]

Just about every step is invalid.

By Definition, let X be a nonempty subset of R and let X be bounded above. Thus, m = sup X iff $m \geq x$. for all $x \in X$

That's not the definition of sup.

Since S is bounded above, S has a supremum and say, sup S = b. But, $S \subseteq T$, and thus, $b \in T$.

Why?

But, by definition, a = supT iff a is (greater than or equal too) all other elements in T

Again, that's not the definition.

By Definition, inf S;
c = inf S iff c is (less than or equal too) all elements in S.
and, sup S is
b = supS iff b is (greater than or euqal too) all elements in S.

Again, those are not the definitions.

Thus, since $c \leq s$ and $b \geq s$, where s is (all elements in S), $c \leq b$ as required.

This doesn't make sense. Specifically, "where s is (all elements in S)".

 

[JasonRox]

Yeah, you're right that isn't the definition of the supremum.

The mistake I made.

 

[playboy]

In words, the definition of Supremum is "In mathematics, the supremum of an ordered set S is the least element that is greater than or equal to each element of S. Consequently" (Wikipedia)

My Text book says: (and I am writing it out word-for-word)

Let S be a nonempty subset of R. If S is bounded above, then the least upper bound of S is called a supremum and is denoted by supS. thus m = sup S iff
a)$m \geq s$, for all $s \in S$, and
b)if $m' < m$, then there exists $s' \in S$ such that $s' > m'$.

Ill redo my proof and post it up again.

In my proof, I just skiped part b) since i thought it had no significance

 

[matt grime]

You skipped half the of the definition because you thought it had no significance? a) just states that something is an upper bound, without b the it is not the sup.

 

[playboy]

ok.
So a) means its the upperbound.
Suppose that I put in the full definition, that is, both a) and b), then would the proof be correct?

1. Prove that $sup S \leq sup T$

By Definition, let X be a nonempty subset of R and let X be bounded above. Thus, m = sup X iff $m \geq x$ for all $x \in X$ AND if $m' < m$, then there exists $x' \in X$ such that $x' > m'$.

Since S is bounded above (as stated in the question), S has a supremum and say, sup S = b. But, since $S \subseteq T$ and b = sup S, this must mean that $b \in T$.

But, by definition, a = supT iff $a \geq t$ for all $t \in T$ AND if $a' < a$, then there exists $t' \in T$ such that $t' > a'$.

Thus, $a \geq b$ and so, $supT \geq sup S$ as required.

Is this better now? I mean, i put in the actuall definition.

 

[Hurkyl]

At the risk of sounding like a broken record...

Since S is bounded above (as stated in the question), S has a supremum and say, sup S = b. But, since $S \subseteq T$ and b = sup S, this must mean that $b \in T$.

Why?

 

[playboy]

I was affraid that one was coming.

Actually, now that I think of it, that statement might not be true.
Say S and T were the same sets, that is, S = T, and supS could lie outside of S or inside of S, depending if the set has a maxmimum or not. Suppose supS lies outside the set S, then $sup S \notin T$.

Ill come back to this, I am going to give it more thought.

 

[matt grime]

Right, you're making this way too hard.

I'm just going to write out the upper half of the question. By writing it out with the addition of what 'sup means' (ie what we need to show) I practically prove it for you.

We have A, B non-empty and bounded A<=B.

We want to show that sup(A)<=sup(B).

Since sup(A) is the least upper bound sup(A)<=U for any upper bound U we need only show that...

 

[playboy]

Sorry I took so long for a reply, i had a cold.

Anyways, I ended up seeing how my professor proved this.

In all honesty, he just wrote out the definitions which took only a few lines, and not a hudge page...Something similar to your's matt grime.

Proof's can be difficult to understand sometimes.

 

[matt grime]

I think in this case it is more that 'knowing what the professor wants you to write' is difficult to understand. I find this a lot in the early stages of proof. For some reasons, because students are new to proofs, teachers equate this with being new to 'thinking mathematically' and this ask questions which are very easy. It thus is difficult to produce a good proof since the proof really is too close to just writing out the definitions again.

Here is an question that was on a midterm for a course I was supervising (think 'taking problem classes on'):

If A and B are sets show that A is a subset of AuB.

Now, that requires no proof, how can it require proof, you'd have to have a very strange idea of what AuB is not to see that that is true. And since the students had the correct idea of AuB, that is the set of elements of A together with the set of elements of B, how do you 'prove' the obvious?

In case you are wondering, they were supposed to write was

Since (x in A) is true, then (x in A)OR(x in B) is true, hence (x in AuB) is true.

This strikes me as being a poncy way of saying: duh!