Your proof works, but you do need to take note that since it is essential to the assertion.Originally Posted by AlexP
I think it is time for you to move on to chapter 2.

Your proof works, but you do need to take note that since it is essential to the assertion.Originally Posted by AlexP
I think it is time for you to move on to chapter 2.
ok, I understand that. This is chapter 2, actually. Algebraic properties of the real number system. What type of material do you think I should be working on now? The rest of chapter 2 is on absolute value, the completeness of , and supremums.
Completeness of and supremums and infemums are central to real analysis. Concentrate on that until you understand them well.Originally Posted by AlexP
Let . Show in detail that the set has lower bounds, but no upper bounds. Show that .
Since every real number is either positive, zero, or negative, and consists of all nonnegative real numbers, the set of all negative real numbers consists of all the lower bounds of . has no upper bounds because given an arbitrarily large positive real number, one cannot find a real number such that , since, for example, if , we can still form the positive real number (thus an element of ) which is larger.
because 0 is a lower bound and for any lower bound of S, call it , .
is a lower bound of because every real number is either positive, negative, or zero. Zero is clearly less than any positive number, and equal to itself, thus 0 is less than or equal to all and is a lower bound.
S consists of all positive real numbers and zero, so clearly since all negative real numbers are less than 0, and 0 is equal to itself, 0 is greater than or equal to all lower bounds of S and .
I think you have the gist of this one, but the wording seems a bit unclear at points.Originally Posted by AlexP
One might address it in this way.
Let . Then is clearly a lower bound for by construction. It is also clear that any negative number is a lower bound for but not the greatest lower bound since is also a lower bound. On the other hand no positive number can be a lower bound since given any such there is an such that . Hence must in fact be the greatest lower bound.
Let . Show that and .
Clearly every element of S is less than or equal to 1, so 1 is an upper bound of S. For the same reason, there is no upper bound less than 1 and 1 is the least upper bound.
Clearly every element of S is positive, so 0 is a lower bound of S, and thus if is a lower bound it cannot be the greatest lower bound so we must have that .
Looks OK.Originally Posted by AlexP
Now why don't you go ahead and prove that . You will need to use the Archimedean property of the real numbers.
Clearly 0 is a lower bound of S since for all . Since by the Archimedean property every has a such that , we have that , so for any number we have a lesser element, , so that cannot be a lower bound and 0 is the greatest lower bound.
Could you tell me which of these are worthwhile to try and that I should understand before moving on? Didn't take me long to type all that, and I figure I might as well make my efforts more directed if I can. Thank you, your help is much appreciated.
1. Let S be a nonempty subset of that is bounded below. Prove that .
2. If a set contains one of its upper bounds, show that this upper bound is the supremum of S.
3. Let be nonempty. Show that is an upper bound of S if and only if the conditions and imply that .
4. Let be nonempty. Show that if , then for every number the number is not an upper bound of S, but the number is an upper bound of S.
5. Show that if A and B are bounded subsets of , then is a bounded set. Show that .
6. Let S be a bounded set in and let be a nonempty subset of S. Show that .
7. Let and suppose that belongs to S. If , show that .
8. Show that a nonempty finite set contains its supremum.
I would prove any of these for which you don't immediately see an obvious proof. If you can look at an assertion and see why it is true immediately, then go on to the next one.Originally Posted by AlexP
Let S be a nonempty subset of that is bounded below. Prove that .
Let and let s be any element of S. Since , so that x is an upper bound of and in fact equals since for any we have . Now since and , clearly .
I think you have the idea, but too many minus signs always confuse me and you have at least one sign error  should beOriginally Posted by AlexP
Here is how I would do it :
Since we have that , for given any if then .
I don't know if I will end up skipping any, since even if I do come up with an obvious proof, I still want verification that it's correct.
If a set contains one of its upper bounds, show that this upper bound is the supremum of S.
Let be an upper bound of the set. All that remains to show is that it is the least upper bound. Clearly for any that is small enough, , but by the Archimedean property for every we can find a , so that . So for an arbitrary , we can always find a greater element, so that is not the greatest element and thus cannot be an upper bound, making the least upper bound and thus .
Not only is this not clear, it is not true. Suppose that S is the rational numbers in Then you could take to be for some But is not in S.Originally Posted by AlexP
How about this.Originally Posted by AlexP
Suppose that is an upper bound for . Suppose that is an upper bound for then since and therefore is the least upper bound for .
That actually was my proof, but my brain is so sporadic at times that I think I forgot that I had a proof, came up with a new (and incorrect) one, and forgot about the original one.
I've done a bunch where it's been pretty selfevident that the proofs were correct. Here's one I'd like some feedback on.
Let and be bounded nonempty subsets of , and let . Prove that .
Let and . Since for all and for all , then for all , and is an upper bound of . Also, since and are both least upper bounds, there is clearly no lower upper bound in and .
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Also, I'm not sure about this problem. The notation isn't clear to me.
Let . Define by .
For each , find ; then find .
I'm just confused. In the definition of , is fixed?
This one doesn't quite work. You did show that x+y is an upper bound. But you need to show that it is a least upper bound. The way to do that is to show that any smaller number is not an upper bound. Essentially you need to show that the operation of addition is continuous.Originally Posted by AlexP
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No. x is fixed and you let y vary to find the sup of the h(x,y) which is f(x).Originally Posted by AlexP
I saw this proof to show that addition is continuous...
We have that , so that arbitrarily small changes in and result in arbitrarily small changes in and thus . I'm having a hard time working this particular problem into it though.
think about it. You will get there.Originally Posted by AlexP
ugh...haven't had time to think. I am having a good time with these field courses though. ok, I need to get somewhere with this...I'll take a stab at it.
Let be the smallest value such that and be the smallest value such that . We then have that . Let and , then we have that , so that for all , , and is the least upper bound.
No such or need exist.Originally Posted by AlexP
When you say that is it because of the case where the set contains its supremum? I did mean to specify (since I know they're typically ).
Or how's this?
We'll use instead of and instead of and let them be arbitrary elements of and , respectively. Then plug them into the same inequality, and we have that the change in will be the change in plus the change in , so that for all , and must be the least upper bound.
That is one way to find a counterexample. Another would be to have A being a sequence converging monoticall upward to its limit, which would be the lub.Originally Posted by AlexP
????Originally Posted by AlexP
In other words...
Let and be arbitrary elements of and , respectively. Then we have that , so that arbitrarily small changes in and in result in arbitrarily small changes in , and we have that for all , , where and are the suprema of and , respectively.
Originally Posted by AlexP
and are sets. What is a change in or ?
felt like trying this one.
If , show that there exists an such that .
By the Archimedean property we have that for any there exists an such that (and thus ). So for we can find (since can be made arbitrarily small). Since for all (shown by induction), we have that and , so for all there exists an such that .
If and are closed intervals in , show that if and only if and .
Let . Then (since ) but , so that , contradicting the assumption that .
Suppose . Then there exists , . Since is a closed interval it contains all . Since , or , so because either or since because is a closed interval. This contradicts that and .
If is nonempty, show that is bounded if and only if there exists a closed, bounded interval such that .
If is closed, it is bounded by all , , so is as well by the assumption that .
If is bounded with arbitrary upper bound and lower bound , we can find for some and for some , so that since contains all by definition.
Prove that if and only if .
First note that . Let be given. If then there exists such that for all , . Since and are equal, they have the same , and so if does not exist for one then it does not exist for the other, and vice versa.
looks OK, if a bit unwieldy.Originally Posted by AlexP
The gist of it is your first sentence.
How would you simplify it?
Here's another...
Show that if for all and , then .
Let be given. Since there exists such that for all , . Since we have which implies , so that .
I would stop after the first sentence and refer to the definition of a limit.Originally Posted by AlexP
This one doesn't work. The reason is that your argument is circular.Originally Posted by AlexP
What you need to show is that if x is a small number then x^2 is also small, and you implicitly assume that in your argument when you use in the usual role of .
This is another one of those that is a bit sutble because it is so obvious, and the key is that if
A bit too subtle for me to get, I guess. I have not been thinking about it nonstop for the past ten days because I'm back in school, but with what thought I have given it, it still eludes me.
Keep thinking about it. It is important that you figure this one out for yourself.Originally Posted by AlexP
Show that if for all and , then .
Suppose that , . Since for some we have for all , then for we have so that either , or . For , , so that . Thus if , then , which is a contradiction.
Keep trying.Originally Posted by AlexP
This argument is wrong for several reasons, which it will do you good to figure out.
Show that if for all and , then .
First note that for , . By the definition of a limit, there exists a such that for all we have , . Since , for we also have , so that .
Nope.Originally Posted by AlexP
For
so your estimate does not work.
I have not given up or been taking a break by any means, but I've been working from the homework assignments given on a course website I found, and solutions are included, so I've just been working from that. In this case, however, the solution given does not match mine, yet I have been unable to see what is wrong with mine, if anything.
Show that if is such that where , then .
Suppose and . Note that . It is clear that tends to infinity since for any we can find an such that . But then also tends to infinity since equals some nonzero real number, contradicting that , . Therefore we must have .
Does that work, or if not, what am I missing?
Close, but no cigar.Originally Posted by AlexP
The problem is that because does not exist since is not a real number.
But you can use a trick .
Got it.
Check your PM inbox.
Another case where mine doesn't match the given solution but I still feel that it works.
Suppose that is continuous on and that for every rational number . Prove that for all .
Suppose is continuous on and for and there exists such that . Let be a sequence of rational numbers that converges to . Then converges to but , so is not continuous at , but this is a contradiction. Therefore if for all and is continuous, we must have .
That works.Originally Posted by AlexP
But you have a more direct proof in your argument. To wit:
Suppose is continuous on and for ]. Let and let be a sequence of rational numbers that converges to . Then converges to and by continuity . QED
Yes, that's the proof that was given in the solutions and I do see how it is more direct and nicer.
The solution to this one was not given for some reason.
If and are continuous on , let . If and , show that .
Suppose . Then . By continuity, and . Since , and there exists such that since there exists such that for all , with given, contradicting that . Therefore we must have .
That ndoesn't quite work. You can make it work if you takeOriginally Posted by AlexP
You seem to jump quickly toroof by contradiction. I suggest that, as a rule of thumb, you first try to find a direct proof.
Here is a simple direct proof. Let . Then is continuous since and are continuous. Since for all , and is continuous, it follows that .QED
The fg approach did occur to me but I didn't get anywhere with it. I do admit that I jump to contradiction pretty quickly. It seems it's just easier to figure out what I need to do that way, or something... I will try more for the direct proofs though.
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