Proving the Infimum Axiom: Real Analysis Exercise Solution

[STEMucator]

Homework Statement

So I have started my long journey through N.L. Carothers Real Analysis and my intention is to work through every single exercise along the way.

The first problem : http://gyazo.com/ddef0387f04d789c660548c08796585d

Homework Equations

The Attempt at a Solution

Suppose A is bounded below, we want to show A has an infimum.

That is, we want to show $\exists m \in ℝ$ such that :

(i) $m$ is a lower bound for A.
(ii) If $x$ is any lower bound for A, then $x ≤ m$.

So suppose (i) is satisfied ( Since A is bounded below ), we want to show (ii) holds, so we consider the set $-A = \{ -a \space | \space a \in A \}$.

Since A is bounded below by m, -A is bounded above by -m. Since -A is a nonempty subset of real numbers with an upper bound, $\exists s \in ℝ \space | \space -m ≤ s$.

Hence sup(-A) = -m ( i.e, -m is the least upper bound of -A ) so that -sup(-A) = m ( i.e, m is the greatest lower bound of A ), but this is precisely what we desire. Thus -sup(-A) = inf(A) = m as desired.

That was my go at it. Any thoughts to make this better? I feel as if my argument wasn't as strong as it should be near the end.

 

[LCKurtz]

Homework Statement

So I have started my long journey through N.L. Carothers Real Analysis and my intention is to work through every single exercise along the way.

The first problem : http://gyazo.com/ddef0387f04d789c660548c08796585d

Homework Equations

The Attempt at a Solution

Suppose A is bounded below, we want to show A has an infimum.

That is, we want to show $\exists m \in ℝ$ such that :

(i) $m$ is a lower bound for A.
(ii) If $x$ is any lower bound for A, then $x ≤ m$.

So suppose (i) is satisfied ( Since A is bounded below ), we want to show (ii) holds, so we consider the set $-A = \{ -a \space | \space a \in A \}$.

A is bounded below so it has a lower bound m. That doesn't mean m is the greatest lower bound nor that it satisfies ii.

Since A is bounded below by m, -A is bounded above by -m. Since -A is a nonempty subset of real numbers with an upper bound, $\exists s \in ℝ \space | \space -m ≤ s$.

No. Nothing says -m is a least upper bound for -A. The least upper bound s will satisfy $s\le -m$.

Hence sup(-A) = -m ( i.e, -m is the least upper bound of -A ) so that -sup(-A) = m ( i.e, m is the greatest lower bound of A ), but this is precisely what we desire. Thus -sup(-A) = inf(A) = m as desired.

That was my go at it. Any thoughts to make this better? I feel as if my argument wasn't as strong as it should be near the end.

Even if you did have that -m is the l.u.b. of -A, you would still have to prove i and ii hold for it. Back to the drawing board. You might think about showing - sup(-A) works for the g.l.b of A.

 

[STEMucator]

A is bounded below so it has a lower bound m. That doesn't mean m is the greatest lower bound nor that it satisfies ii.

Okay, so suppose that m is a lower bound of A. ( So (i) is satisfied and we want (ii) to be satisfied at the end ). So we consider the set -A.

No. Nothing says -m is a least upper bound for -A. The least upper bound s will satisfy $s ≤ -m$

Since m is a lower bound for A, -m is an upper bound for -A. Since -A is a nonempty subset of real numbers with an upper bound, $\exists s \in ℝ \space | \space s ≤ -m$.

So sup(-A) = s so that 's' is the least upper bound for -A. Hence -sup(-A) = -s is a lower bound for -A.

Since -s ≥ m and m is a lower bound for A, this tells us that -s is the greatest lower bound for A and hence inf(A) = -sup(-A) = -s.

This seems to make sense ( hopefully ).

 

[LCKurtz]

Okay, so suppose that m is a lower bound of A. ( So (i) is satisfied and we want (ii) to be satisfied at the end ). So we consider the set -A.

Since m is a lower bound for A, -m is an upper bound for -A.

It's easy, but you should show this, not just state it.

Since -A is a nonempty subset of real numbers with an upper bound, $\exists s \in ℝ \space | \space s ≤ -m$.

Sure, but there may be lots of such s's. What do you really want for s?

So sup(-A) = s so that 's' is the least upper bound for -A.

Why? See above.

Hence -sup(-A) = -s is a lower bound for -A.

Again, you should show this.

Since -s ≥ m and m is a lower bound for A, this tells us that -s is the greatest lower bound for A

No it doesn't. You have to show that if n is any lower bound for A then $n\le m$.

and hence inf(A) = -sup(-A) = -s.

This seems to make sense ( hopefully ).

You have a ways to go yet.

 

[STEMucator]

Okay, so suppose that m is a lower bound of A. ( So (i) is satisfied and we want (ii) to be satisfied at the end ). So we consider the set -A.

Since m is a lower bound for A, -m is an upper bound for -A.

It's easy, but you should show this, not just state it.

Since $m ≤ a, \space \forall a \in A, \space -m ≥ -a, \space \forall (-a) \in (-A)$. So that -m is an upper bound for -A.

Since -A is a nonempty subset of real numbers with an upper bound, ∃s∈R | s≤−m.

Sure, but there may be lots of such s's. What do you really want for s?

I want this s so I can show that -A has a supremum. Not just an upper bound ( sup axiom ).

So sup(-A) = s so that 's' is the least upper bound for -A.

Why? See above.

Above.

Hence -sup(-A) = -s is a lower bound for A.

Again, you should show this.

Since -s ≥ m and m is a lower bound for A, would that not show that inf(A) = -s = -sup(-A) ( Even stronger than a lower bound as -s is the greatest lower bound )?

Since s ≤ -m → -s ≥ m and m is a lower bound for A, this tells us that -s is the greatest lower bound for A

No it doesn't. You have to show that if n is any lower bound for A then n≤m.

The point of the question is to show A has an infimum using the fact that A is bounded below ( by m ).

So I assumed A had a lower bound m. Then I considered -m as the upper bound of -A. Then using the sup axiom I found sup(-A) = s so that inf(A) = -s so A has an infimum. I don't seem to see where this goes wrong?

 

[LCKurtz]

The point of the question is to show A has an infimum using the fact that A is bounded below ( by m ).

So I assumed A had a lower bound m. Then I considered -m as the upper bound of -A. Then using the sup axiom I found sup(-A) = s

That is not what your argument said. You said "Since -A is a nonempty subset of real numbers with an upper bound, ∃s∈R | s≤−m". You said nothing about sup(-A). You haven't written what you mean.

so that inf(A) = -s so A has an infimum. I don't seem to see where this goes wrong?

That is the idea of the argument. But your writeup is not very good. And the copy function on PF makes it very difficult to respond appropriately.

 

[STEMucator]

I would have assumed you had knowledge of the upper axiom, which is perhaps the reason my argument didn't seem too precise.

Is it okay afterwards though, considering I'm allowed to at least use the upper axiom without proof ( heavily implied by the author ).

 

[LCKurtz]

I would have assumed you had knowledge of the upper axiom, which is perhaps the reason my argument didn't seem too precise.

I am not the student here, you are. I have been trying to get you to understand what's wrong with your argument. You don't seem to know how to write up a good argument. I could show you how but you don't seem to actually address the issues I raise.

 

[STEMucator]

I am not the student here, you are. I have been trying to get you to understand what's wrong with your argument. You don't seem to know how to write up a good argument. I could show you how but you don't seem to actually address the issues I raise.

I apologize for not including the axiom in relevant equations. I've addressed some of your concerns; if it's clear that I have the right idea, then feel free to expand on things.

 

[LCKurtz]

I apologize for not including the axiom in relevant equations. I've addressed some of your concerns; if it's clear that I have the right idea, then feel free to expand on things.

This thread has gotten too disjointed to follow. At this point, if you would like to post your careful, most complete, best solution, with all the steps, I will be happy to respond to it. Write it up as if you were going to hand it in for grading. It will be later this evening before I can be back, so take your time and do it right.

 

[STEMucator]

The question : http://gyazo.com/ddef0387f04d789c660548c08796585d

Useable info : Sup Axiom.

Suppose $A$ is a nonempty subset of $ℝ$ bounded below by $n$, we want to show $A$ has an infimum.

Consider the set from the hint : $-A = \{-a \space | \space a \in A \}$

We want to show that $n$ which is a lower bound for $A$, is an upper bound for $-A$. So, since $n ≤ a, \space \forall a \in A \Rightarrow -n ≥ -a, \space \forall a \in (-A)$. So since $(-a) \in A$, we conclude $n ≥ -a$ and so $n$ is an upper bound for $-A$.

Now, since $-A$ is bounded above by $n$ and $-A$ is a nonempty subset of the reals, we know by the supremum axiom that $\exists m \in ℝ \space | \space sup(-A) = m$

Now, we want to show $-m$ is such that :

(i) $-m$ is a lower bound for $A$.
(ii) If $n$ is any lower bound for $A$, then $n≤m$.

To prove (i) let's assume the contrary that $-m$ is not a lower bound for $A$. Then $\exists a \in A \space | \space -m ≥ a \Rightarrow m ≤ -a$. Since $(−a) \in (-A)$, this contradicts the fact that $sup(-A) = m$ from earlier and hence $-m$ must be a lower bound for $A$.

I'm having some trouble with (ii) for some reason. I assumed the contrary that $n≥m$, but I'm having a hard time continuing with it.

 

[Dick]

The question : http://gyazo.com/ddef0387f04d789c660548c08796585d

Useable info : Sup Axiom.

Suppose $A$ is a nonempty subset of $ℝ$ bounded below by $n$, we want to show $A$ has an infimum.

Consider the set from the hint : $-A = \{-a \space | \space a \in A \}$

We want to show that $n$ which is a lower bound for $A$, is an upper bound for $-A$. So, since $n ≤ a, \space \forall a \in A \Rightarrow -n ≥ -a, \space \forall a \in (-A)$. So since $(-a) \in A$, we conclude $n ≥ -a$ and so $n$ is an upper bound for $-A$.

Now, since $-A$ is bounded above by $n$ and $-A$ is a nonempty subset of the reals, we know by the supremum axiom that $\exists m \in ℝ \space | \space sup(-A) = m$

Now, we want to show $-m$ is such that :

(i) $-m$ is a lower bound for $A$.
(ii) If $n$ is any lower bound for $A$, then $n≤m$.

To prove (i) let's assume the contrary that $-m$ is not a lower bound for $A$. Then $\exists a \in A \space | \space -m ≥ a \Rightarrow m ≤ -a$. Since $(−a) \in (-A)$, this contradicts the fact that $sup(-A) = m$ from earlier and hence $-m$ must be a lower bound for $A$.

I'm having some trouble with (ii) for some reason. I assumed the contrary that $n≥m$, but I'm having a hard time continuing with it.

You are making this complicated enough that it's making my eyes hurt. Probably because you are trying to prove everything by contradiction. You don't have to. If A has a lower bound consider the set of all lower bounds of A, call it L. Show L is bounded above. So L has a sup. So?

 

[LCKurtz]

The question : http://gyazo.com/ddef0387f04d789c660548c08796585d

Useable info : Sup Axiom.

Suppose $A$ is a nonempty subset of $ℝ$ bounded below by $n$, we want to show $A$ has an infimum.

Consider the set from the hint : $-A = \{-a \space | \space a \in A \}$

We want to show that $n$ which is a lower bound for $A$, is an upper bound for $-A$.

That isn't even true. The same n that is a lower bound for A is not an upper bound for -A.

So, since $n ≤ a, \space \forall a \in A \Rightarrow -n ≥ -a, \space \forall a \in (-A)$. So since $(-a) \in A$, we conclude $n ≥ -a$ and so $n$ is an upper bound for $-A$.

No. Since each element of -A is -a for some a in A, and $n\le a$ you have $-n\ge -a$ which tells you that -n is an upper bound for -A

Now, since $-A$ is bounded above by $n$ (you mean $-n$) and $-A$ is a nonempty subset of the reals, we know by the supremum axiom that $\exists m \in ℝ \space | \space sup(-A) = m$

Now, we want to show $-m$ is such that :

(i) $-m$ is a lower bound for $A$.
(ii) If $n$ is any lower bound for $A$, then $n≤m$.

To prove (i) let's assume the contrary that $-m$ is not a lower bound for $A$. Then $\exists a \in A \space | \space -m ≥ a \Rightarrow m ≤ -a$. Since $(−a) \in (-A)$, this contradicts the fact that $sup(-A) = m$ from earlier and hence $-m$ must be a lower bound for $A$.

I'm having some trouble with (ii) for some reason. I assumed the contrary that $n≥m$, but I'm having a hard time continuing with it.

Like Dick has observed, you don't need any indirect argument. Just like the argument above where a lower bound of n for A gives an upper bound of -n for -A you should be able to show that an upper bound m for -A gives rise to a lower bound -m for A and you should be able use the fact that you have a least upper bound for -A gives rise to a greatest lower bound for A.

 

[STEMucator]

You are making this complicated enough that it's making my eyes hurt. Probably because you are trying to prove everything by contradiction. You don't have to. If A has a lower bound consider the set of all lower bounds of A, call it L. Show L is bounded above. So L has a sup. So?

For (ii), suppose that A is bounded below by n. We want to show n ≤ -m. Consider the set $L = \{x \in ℝ \space | \space x ≤ a, \space \forall a \in A \}$ which is the set of all lower bounds of A.

Clearly $n, -m \in L$ by hypothesis and (i) so that $L$ is not empty.
---------

EDIT : I realized after reading LC's comment that there is a very easy way to do this without really putting effort into (i) or (ii).

Since m is the least upper bound of -A, -m is the greatest lower bound for A. That is $m ≥ -a \Rightarrow -m ≤ a$ so that -m is the greatest lower bound of A.

 

[CAF123]

I think the statement you are trying to prove is the Completeness property for infima. If you know the Completeness Property for suprema and the Reflection principles, then I believe this can be done in a couple of steps.

 

[LCKurtz]

EDIT : I realized after reading LC's comment that there is a very easy way to do this without really putting effort into (i) or (ii).

Since m is the least upper bound of -A, -m is the greatest lower bound for A. That is $m ≥ -a \Rightarrow -m ≤ a$ so that -m is the greatest lower bound of A.

But you can't just declare that and call it a proof. You have to get your hands dirty with both i and ii. To prove them for A you have to use the corresponding properties for sup and -A and use them to get the results you want for A.

 

[STEMucator]

You are making this complicated enough that it's making my eyes hurt. Probably because you are trying to prove everything by contradiction. You don't have to. If A has a lower bound consider the set of all lower bounds of A, call it L. Show L is bounded above. So L has a sup. So?

But you can't just declare that and call it a proof. You have to get your hands dirty with both i and ii. To prove them for A you have to use the corresponding properties for sup and -A and use them to get the results you want for A.

Back from a long work shift, let's do math. I did (i) by contradiction did I not? Anyway, i'll try them both without any contradiction.

Since m is the least upper bound of -A, -m is a lower bound for A. That is $m≥−a \Rightarrow −m≤a$.

Argh! I'm literally getting stuck with the reasoning here. My issue is that I said m was an lower bound for A at the beginning of my proof, but now I've just shown -m is also a lower bound?

 

[LCKurtz]

Argh! I'm literally getting stuck with the reasoning here. My issue is that I said m was an lower bound for A at the beginning of my proof, but now I've just shown -m is also a lower bound?

That is part of the reason it is so difficult to help you. What is m and where is it defined? That's why I asked you to give a complete argument from the beginning, which you attempted in post #11. In post #13 I indicated some problems with your argument. You have not acknowledged that you understood my corrections so I don't know if you even read them or whether you understood them if you did read them. Now we are back to you being confused about m or -m.

I'm sorry, but I am tiring of this thread and I think I have said all I want to say.

 

[Dick]

Back from a long work shift, let's do math. I did (i) by contradiction did I not? Anyway, i'll try them both without any contradiction.

Since m is the least upper bound of -A, -m is a lower bound for A. That is $m≥−a \Rightarrow −m≤a$.

Argh! I'm literally getting stuck with the reasoning here. My issue is that I said m was an lower bound for A at the beginning of my proof, but now I've just shown -m is also a lower bound?

The proof is even easier if you don't use the hint. L, the set of lower bounds, has an upper bound, any element of A is an upper bound. That means L has a sup, call it n. Doesn't n have all of the properties you need to be an inf of A?

 

[STEMucator]

The question : http://gyazo.com/ddef0387f04d789c660548c08796585d

Useable info : Sup Axiom.

Suppose $A$ is a nonempty subset of $ℝ$ bounded below, we want to show $A$ has an infimum.

Consider the set of lower bounds of A : $L = \{x \in ℝ \space | \space x ≤ a, \space \forall a \in A \}$. Notice L is not empty because A is bounded below. Since $x ≤ a$, L is bounded above for any $a \in A$.

Now, L is a nonempty subset of real numbers, which is bounded above by at least one element ( Since A is nonempty ), so we know sup(L) exists by the sup axiom, call it n.

n is in the set of lower bounds of a, so it satisfies (i). Now since $x ≤ n ≤ a$ ( From each respective inequality throughout the proof ) (ii) is satisfied. So we can write inf(A) = n.

 

[Dick]

The question : http://gyazo.com/ddef0387f04d789c660548c08796585d

Useable info : Sup Axiom.

Suppose $A$ is a nonempty subset of $ℝ$ bounded below, we want to show $A$ has an infimum.

Consider the set of lower bounds of A : $L = \{x \in ℝ \space | \space x ≤ a, \space \forall a \in A \}$. Notice L is not empty because A is bounded below. Since $x ≤ a$, L is bounded above for any $a \in A$.

Now, L is a nonempty subset of real numbers, which is bounded above by at least one element ( Since A is nonempty ), so we know sup(L) exists by the sup axiom, call it n.

n is in the set of lower bounds of a, so it satisfies (i). Now since $x ≤ n ≤ a$ ( From each respective inequality throughout the proof ) (ii) is satisfied. So we can write inf(A) = n.

Well, sure. If I understand what you mean by i) and ii). Now it really wouldn't hurt to try to do it using the hint considering -A. All you should have to do is change '+' into '-' and '<' into '>' and apply definitions. Not turn it into a hash of confusing proofs by contradiction.

 

[STEMucator]

http://gyazo.com/ddef0387f04d789c660548c08796585d

Useable info : Sup Axiom.

Suppose $A$ is a nonempty subset of $ℝ$ bounded below by $n \in ℝ$, we want to show $A$ has an infimum.

Consider the set from the hint : $-A = \{-a \space | \space a \in A \}$

Since n is a lower bound for $A$, -n is an upper bound for $-A$. Proof : $n ≤ a, \space \forall a \in A$ so that $-n ≥ -a, \space \forall -a \in -A$. Thus -n is an upper bound for -A.

***Sorry I just have to stop here and ask something, could I not just use your same idea of considering the set of all upper bounds of -A right now? Anyway back to having no knowledge of that.

Since -A is a nonempty subset of the reals and is bounded above, we know sup(-A) exists by the sup axiom, call it m.

Since -a ≤ m, we know a ≥ -m so that -m is a lower bound for A, so -m satisfies (i). Since n is another lower bound of A, we want n ≤ -m ≤ a, but that means -n ≥ m ≥ -a which we know is true from prior inequalities in the proof, so (ii) is true. Thus inf(A) = -m.

 

[Dick]

http://gyazo.com/ddef0387f04d789c660548c08796585d

Useable info : Sup Axiom.

Suppose $A$ is a nonempty subset of $ℝ$ bounded below by $n \in ℝ$, we want to show $A$ has an infimum.

Consider the set from the hint : $-A = \{-a \space | \space a \in A \}$

Since n is a lower bound for $A$, -n is an upper bound for $-A$. Proof : $n ≤ a, \space \forall a \in A$ so that $-n ≥ -a, \space \forall -a \in -A$. Thus -n is an upper bound for -A.

***Sorry I just have to stop here and ask something, could I not just use your same idea of considering the set of all upper bounds of -A right now? Anyway back to having no knowledge of that.

Since -A is a nonempty subset of the reals and is bounded above, we know sup(-A) exists by the sup axiom, call it m.

Since -a ≤ m, we know a ≥ -m so that -m is a lower bound for A, so -m satisfies (i). Since n is another lower bound of A, we want n ≤ -m ≤ a, but that means -n ≥ m ≥ -a which we know is true from prior inequalities in the proof, so (ii) is true. Thus inf(A) = -m.

It's kind of late here so I might not be reading every line in detail. But yes, that's about how easy it is. No proof by contradiction needed, right?

 

[STEMucator]

It's kind of late here so I might not be reading every line in detail. But yes, that's about how easy it is. No proof by contradiction needed, right?

Yes, if I explicitly mention some inequalities along the way while I'm using my definitions, then no contradiction is needed at all.

After seeing it the easy way with L, this way with -A just sort of fell together when I realized how important the inequalities themselves actually were. The ex with L helped me actually think about how the quantities were being manipulated for some reason and how to actually prove (i) and (ii).

Any comments on *** though out of question? Is my hunch okay?

 

[Dick]

Yes, if I explicitly mention some inequalities along the way while I'm using my definitions, then no contradiction is needed at all.

After seeing it the easy way with L, this way with -A just sort of fell together when I realized how important the inequalities themselves actually were. The ex with L helped me actually think about how the quantities were being manipulated for some reason and how to actually prove (i) and (ii).

Any comments on *** though out of question? Is my hunch okay?

Yes, the easy way is to consider the set of upper bounds and call it U. But you might need the inf axiom for that. Works just like L. It avoids even mentioning -A. That's the clever way. Using -A is the obvious way. They both work.