How to find supremum and infimum of $(2014,2016]$?
Let $A=(2014,2016]$. Which of the following statements is/are true?
(a) $sup A=2016$ and $inf A=2014$
(b) $sup A=2014$ and $inf A=2016$
(c) $sup A=2016$ and $inf A$ does not exist
(d) $sup A$ does not exist and $inf A=2014$
(e) Both $sup A$ and $inf A$ do not exist.
My try
Since there is no possible least number inf does not exist and $sup(A)=2016$.
Is my argument correct?
real-analysis
edited Dec 9 at 4:29
user21820
38.7k543153
asked Dec 8 at 17:50
emil
417410
add a comment |
Let $A=(2014,2016]$. Which of the following statements is/are true?
(a) $sup A=2016$ and $inf A=2014$
(b) $sup A=2014$ and $inf A=2016$
(c) $sup A=2016$ and $inf A$ does not exist
(d) $sup A$ does not exist and $inf A=2014$
(e) Both $sup A$ and $inf A$ do not exist.
My try
Since there is no possible least number inf does not exist and $sup(A)=2016$.
Is my argument correct?
real-analysis
edited Dec 9 at 4:29
user21820
38.7k543153
asked Dec 8 at 17:50
emil
417410
4
You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
– quanticbolt
Dec 8 at 17:52
add a comment |
Let $A=(2014,2016]$. Which of the following statements is/are true?
(a) $sup A=2016$ and $inf A=2014$
(b) $sup A=2014$ and $inf A=2016$
(c) $sup A=2016$ and $inf A$ does not exist
(d) $sup A$ does not exist and $inf A=2014$
(e) Both $sup A$ and $inf A$ do not exist.
My try
Since there is no possible least number inf does not exist and $sup(A)=2016$.
Is my argument correct?
real-analysis
edited Dec 9 at 4:29
user21820
38.7k543153
asked Dec 8 at 17:50
emil
417410
Let $A=(2014,2016]$. Which of the following statements is/are true?
(a) $sup A=2016$ and $inf A=2014$
(b) $sup A=2014$ and $inf A=2016$
(c) $sup A=2016$ and $inf A$ does not exist
(d) $sup A$ does not exist and $inf A=2014$
(e) Both $sup A$ and $inf A$ do not exist.
My try
Since there is no possible least number inf does not exist and $sup(A)=2016$.
Is my argument correct?
real-analysis
real-analysis
edited Dec 9 at 4:29
user21820
38.7k543153
asked Dec 8 at 17:50
emil
417410
edited Dec 9 at 4:29
user21820
38.7k543153
asked Dec 8 at 17:50
emil
417410
edited Dec 9 at 4:29
user21820
38.7k543153
edited Dec 9 at 4:29
user21820
38.7k543153
edited Dec 9 at 4:29
user21820
38.7k543153
38.7k543153
asked Dec 8 at 17:50
emil
417410
asked Dec 8 at 17:50
emil
417410
asked Dec 8 at 17:50
emil
417410
417410
4
You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
– quanticbolt
Dec 8 at 17:52
add a comment |
4
You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
– quanticbolt
Dec 8 at 17:52
4
4
You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
– quanticbolt
Dec 8 at 17:52
You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
– quanticbolt
Dec 8 at 17:52
add a comment |
3 Answers
3
active
oldest
votes
You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.
answered Dec 8 at 17:53
Mark
5,955415
This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
– emil
Dec 8 at 18:02
4
Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
– Mark
Dec 8 at 18:04
I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
– Shaun
Dec 8 at 18:20
add a comment |
"Since there is no possible least number inf does not exist"
You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.
Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.
answered Dec 8 at 17:56
Shaun
8,582113580
add a comment |
For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.
answered Dec 8 at 17:53
Parcly Taxel
41.2k137199
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.
answered Dec 8 at 17:53
Mark
5,955415
This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
– emil
Dec 8 at 18:02
4
Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
– Mark
Dec 8 at 18:04
I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
– Shaun
Dec 8 at 18:20
add a comment |
You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.
answered Dec 8 at 17:53
Mark
5,955415
This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
– emil
Dec 8 at 18:02
4
Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
– Mark
Dec 8 at 18:04
I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
– Shaun
Dec 8 at 18:20
add a comment |
You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.
answered Dec 8 at 17:53
Mark
5,955415
You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.
answered Dec 8 at 17:53
Mark
5,955415
answered Dec 8 at 17:53
Mark
5,955415
answered Dec 8 at 17:53
Mark
5,955415
answered Dec 8 at 17:53
Mark
5,955415
5,955415
This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
– emil
Dec 8 at 18:02
4
Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
– Mark
Dec 8 at 18:04
I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
– Shaun
Dec 8 at 18:20
add a comment |
This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
– emil
Dec 8 at 18:02
4
Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
– Mark
Dec 8 at 18:04
I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
– Shaun
Dec 8 at 18:20
This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
– emil
Dec 8 at 18:02
This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
– emil
Dec 8 at 18:02
4
4
Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
– Mark
Dec 8 at 18:04
Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
– Mark
Dec 8 at 18:04
I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
– Shaun
Dec 8 at 18:20
I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
– Shaun
Dec 8 at 18:20
add a comment |
"Since there is no possible least number inf does not exist"
You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.
Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.
answered Dec 8 at 17:56
Shaun
8,582113580
add a comment |
"Since there is no possible least number inf does not exist"
You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.
Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.
answered Dec 8 at 17:56
Shaun
8,582113580
add a comment |
"Since there is no possible least number inf does not exist"
You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.
Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.
answered Dec 8 at 17:56
Shaun
8,582113580
"Since there is no possible least number inf does not exist"
You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.
Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.
answered Dec 8 at 17:56
Shaun
8,582113580
answered Dec 8 at 17:56
Shaun
8,582113580
answered Dec 8 at 17:56
Shaun
8,582113580
answered Dec 8 at 17:56
Shaun
8,582113580
8,582113580
add a comment |
add a comment |
For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.
answered Dec 8 at 17:53
Parcly Taxel
41.2k137199
add a comment |
For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.
answered Dec 8 at 17:53
Parcly Taxel
41.2k137199
add a comment |
For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.
answered Dec 8 at 17:53
Parcly Taxel
41.2k137199
For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.
answered Dec 8 at 17:53
Parcly Taxel
41.2k137199
answered Dec 8 at 17:53
Parcly Taxel
41.2k137199
answered Dec 8 at 17:53
Parcly Taxel
41.2k137199
answered Dec 8 at 17:53
Parcly Taxel
41.2k137199
41.2k137199
add a comment |
add a comment |
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You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
– quanticbolt
Dec 8 at 17:52