bounds

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b!tcoin

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Oct 7, 2019
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hi,
im confused, is this set bounded? and if so, which are the upper and lower bounds.
(0,2) union [3,4]
i think that lower bounds are 0 and -1 but i dont have any idea about the upper ones.
i'd like some help,thanks
 
b!tcoin said:
im confused, is this set bounded? and if so, which are the upper and lower bounds.
(0,2) union [3,4]
Click to expand...
If \(\displaystyle \mathcal{A}=(0,2)\cup [3,4]\) then \(\displaystyle \operatorname{lub}(\mathcal{A})=4~\&~\operatorname{glb}(\mathcal{A})=0\).
Take note that \(\displaystyle 4\in\mathcal{A}\text{ while }0\notin\mathcal{A}\).
 
b!tcoin said:
hi,
im confused, is this set bounded? and if so, which are the upper and lower bounds.
(0,2) union [3,4]
i think that lower bounds are 0 and -1 but i dont have any idea about the upper ones.
i'd like some help,thanks
Click to expand...
Please state the exact wording of the problem you are working on. Details matter!

The fact is that any number less than or equal to 0 is a lower bound; pka has told you about the greatest lower bound. To show that a set is bounded, you only need to find the former. But we don't know yet what you really need to do.
 
b!tcoin said:
hi,
im confused, is this set bounded? and if so, which are the upper and lower bounds.
(0,2) union [3,4]
i think that lower bounds are 0 and -1 but i dont have any idea about the upper ones.
i'd like some help,thanks
Click to expand...
Do you know what the union of the two sets equal?
 
pka said:
If \(\displaystyle \mathcal{A}=(0,2)\cup [3,4]\) then \(\displaystyle \operatorname{lub}(\mathcal{A})=4~\&~\operatorname{glb}(\mathcal{A})=0\).
Take note that \(\displaystyle 4\in\mathcal{A}\text{ while }0\notin\mathcal{A}\).
Click to expand...
Dr.Peterson said:
Please state the exact wording of the problem you are working on. Details matter!
Click to expand...
To bitcoin, I agree that it would be helpful if you would post more about this question
But I was naive enough to hope that my first reply might prompt such a reply from you.,
If I were doing this in a lecture setting, I would insist that you answer:
1) prove that \(\displaystyle 0.1\) is not a lower bound for \(\displaystyle \mathcal{A}.\)
2) prove that \(\displaystyle 3.9\) is not an upper bound for \(\displaystyle \mathcal{A}.\)

Now Bitcoin, if you cannot answer both of those questions, it proves that you are over your head in this material,
That means you need a sitdown session with an instructor.
 
Decide for the following set if is bounded from above and if is bounded from below.I Should also give 2 upper and lower bounds in case they exist.
i can find the upper bound for a set like {0,5,-11,3,12} but this union of sets is a bit more complexed for me.
 
(0,2)∪[3,4] the lower bounds are for this set (NOT GREATEST LOWER BOUNDS=INFIMUM)
0 and -1.
now im stuck on finding the upper ones,because the closed interval confuses me.
 
ok i just came to the conclusion that upper bounds ( 2 of them ) are 4 and 5. hope my answer's correct
 
It appears that you have now stated the actual problem, which explains why you gave two numbers. It's really important to do this, which saves a lot of time figuring out what you are doing.

The given set is (0,2)∪[3,4] .

Any numbers less than or equal to 0 will be lower bounds, so your answer of 0 and -1 is valid; -54893 and -0.5 would also be valid.

Any numbers greater than or equal to 4 will be upper bounds, so your answer of 4 and 5 is valid. 100 and 12345 would also be valid.
 
Thank you for the reply :)
i have one more doubt
regarding Infimum and supremum of this set,i just found that Inf S=0 and Sup S=4
am i correct?
best,
 
Last edited:
Tell me your definition of supremum and infimum, and why you think -1 is the infimum and there is no supremum.
 
b!tcoin said:
regarding Infimum and supremum of this set,i just found that Inf S=0 and Sup S=4
am i correct?
Click to expand...
\(\displaystyle \bf\text{YES, }\sup(S)=4~\&~\inf(S)=0 \)
The supremum is the Least Upper Bound of a set.
The infimum is the Greatest Lower Bound of a set.
 
hello again guys
i just found that this set {1 − 1/2^n : n ∈ N} has no supremum at all.and it has an INFimum of 1/2. can somebody explain me why it has no supremum and why is 1/2 the INF? i have to submit this in a few hours so any help would be greatly appreciated.
Thanks ?
 
What is your source? What was their reasoning? (You don't believe a source with no explanation, do you?)

And, what theorems have you learned about this topic? Have you tried applying them? (Or maybe just the definitions we've already discussed.)
 
an upper bound b is a least upper bound (or supremum), if b ≤ b’ for any other upper bound b’ — we denote it by sup S
an lower bound a is a greatest lower bound (or infimum), if a ≥ a’ for any other lower bound a’ — we denote it by inf S
so because lower bounds of that set were 0 and 1/2 i thought from the infimum definition that a>=a' -> 1/2>=0 for any other bound a' which is 0,and denoted the infimum of the set as 1/2.
does the same thing happen with the supremum?
 
b!tcoin said:
an upper bound b is a least upper bound (or supremum), if b ≤ b’ for any other upper bound b’ — we denote it by sup S an lower bound a is a greatest lower bound (or infimum), if a ≥ a’ for any other lower bound a’ — we denote it by inf S
so because lower bounds of that set were 0 and 1/2 i thought from the infimum definition that a>=a' -> 1/2>=0 for any other bound a' which is 0,and denoted the infimum of the set as 1/2. does the same thing happen with the supremum?
Click to expand...
There are several papers written on quantification issuers with the way you put these definitions.
The arguments are arcane. So I choose to skip them.
Let us just say: \(\displaystyle b=\sup(A)\) the supremum (least upper bound) of the set \(\displaystyle A\) provided that if \(\displaystyle c\) is also an upper of set \(\displaystyle A\) then it must follow that \(\displaystyle \bf{b\le c}\). BTW the same comment (with the obvious modifications) apply for the infimum.
Please note the this relaxed statement does not imply set membership of any kind.
Just yesterday at another site the was some real confusion about the completeness property and set membership.

 
b!tcoin said:
an upper bound b is a least upper bound (or supremum), if b ≤ b’ for any other upper bound b’ — we denote it by sup S
an lower bound a is a greatest lower bound (or infimum), if a ≥ a’ for any other lower bound a’ — we denote it by inf S
so because lower bounds of that set were 0 and 1/2 i thought from the infimum definition that a>=a' -> 1/2>=0 for any other bound a' which is 0,and denoted the infimum of the set as 1/2.
does the same thing happen with the supremum?
Click to expand...

The set, as I understand it, is {1/2, 3/4, 7/8, 15/16, ...}.

There are infinitely many lower bounds, but 0 and 1/2 as you say are among them; and clearly none can be greater than 1/2, so 1/2 is the infimum. But it sounds like you still misunderstand what it means to be a lower bound.

What can you say about the supremum? What upper bounds do you see?

But you haven't answered my questions. Where did you "find" that there is no supremum, and why?
 
upper bounds of the set are : 1 and 2
my friend told me that there is no supremum
b= 2 is a upper bound, and b'=1 is another upper bound.
from the def of the supremum, b<=b' so 2<=1 which is not true, and so this set has no supremum.
thats my thought .
btw my friend told me that there is no supremum here and im struggling to find out why.
 
b!tcoin said:
upper bounds of the set are : 1 and 2
my friend told me that there is no supremum
b= 2 is a upper bound, and b'=1 is another upper bound.
from the def of the supremum, b<=b' so 2<=1 which is not true, and so this set has no supremum.
thats my thought .
btw my friend told me that there is no supremum here and im struggling to find out why.
Click to expand...
Please DO NOT DO THAT! What set are you talking about? We don't know.
Post the exact question. Unless your friend is a PhD mathematician, you will not get better help than here.
So please let us in on the questions.
 
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