Mixing
Writing proofs to show that a set is open, not closed and infinite
$begingroup$
I have this problem with my homework
- Let $E$ be a non-empty bounded set of real numbers and put $alpha = sup E ,$
and $beta = inf E$ . Assume that $alpha notin E$ and $beta notin E .$ Which of the following statements
is true and which is false. In each case justify your answer.
(a) $E$ is an open set.
(b) $E$ is not a closed set.
(c) $E$ is an infinite set.
(d) $( alpha , beta ) subset E$
It's about general topology and inf and sup. We have to state whether each statement is true or false and justify by giving nice proofs. My assumptions are:
1)False
2)True
3)True
4)False
Please, I need help in writing appropriate proofs. I'm new to this course and still don't know how to write proofs. Any help is appreciated. I only have problems proving part d), if anyone can help me. Thank you!
real-analysis general-topology supremum-and-infimum
asked Feb 1 at 16:09
PBCPBC
14
$endgroup$
add a comment |
$begingroup$
I have this problem with my homework
- Let $E$ be a non-empty bounded set of real numbers and put $alpha = sup E ,$
and $beta = inf E$ . Assume that $alpha notin E$ and $beta notin E .$ Which of the following statements
is true and which is false. In each case justify your answer.
(a) $E$ is an open set.
(b) $E$ is not a closed set.
(c) $E$ is an infinite set.
(d) $( alpha , beta ) subset E$
It's about general topology and inf and sup. We have to state whether each statement is true or false and justify by giving nice proofs. My assumptions are:
1)False
2)True
3)True
4)False
Please, I need help in writing appropriate proofs. I'm new to this course and still don't know how to write proofs. Any help is appreciated. I only have problems proving part d), if anyone can help me. Thank you!
real-analysis general-topology supremum-and-infimum
asked Feb 1 at 16:09
PBCPBC
14
$endgroup$
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Feb 1 at 16:10
$begingroup$
@YuiTo Cheng: Transcribing an image into text is good; transcribing it into $rmLaTeX$, not as good.
$endgroup$
– Asaf Karagila♦
Feb 1 at 16:26
add a comment |
$begingroup$
I have this problem with my homework
- Let $E$ be a non-empty bounded set of real numbers and put $alpha = sup E ,$
and $beta = inf E$ . Assume that $alpha notin E$ and $beta notin E .$ Which of the following statements
is true and which is false. In each case justify your answer.
(a) $E$ is an open set.
(b) $E$ is not a closed set.
(c) $E$ is an infinite set.
(d) $( alpha , beta ) subset E$
It's about general topology and inf and sup. We have to state whether each statement is true or false and justify by giving nice proofs. My assumptions are:
1)False
2)True
3)True
4)False
Please, I need help in writing appropriate proofs. I'm new to this course and still don't know how to write proofs. Any help is appreciated. I only have problems proving part d), if anyone can help me. Thank you!
real-analysis general-topology supremum-and-infimum
asked Feb 1 at 16:09
PBCPBC
14
$endgroup$
I have this problem with my homework
- Let $E$ be a non-empty bounded set of real numbers and put $alpha = sup E ,$
and $beta = inf E$ . Assume that $alpha notin E$ and $beta notin E .$ Which of the following statements
is true and which is false. In each case justify your answer.
(a) $E$ is an open set.
(b) $E$ is not a closed set.
(c) $E$ is an infinite set.
(d) $( alpha , beta ) subset E$
It's about general topology and inf and sup. We have to state whether each statement is true or false and justify by giving nice proofs. My assumptions are:
1)False
2)True
3)True
4)False
Please, I need help in writing appropriate proofs. I'm new to this course and still don't know how to write proofs. Any help is appreciated. I only have problems proving part d), if anyone can help me. Thank you!
real-analysis general-topology supremum-and-infimum
real-analysis general-topology supremum-and-infimum
asked Feb 1 at 16:09
PBCPBC
14
asked Feb 1 at 16:09
PBCPBC
14
edited Feb 1 at 20:06
PBC
asked Feb 1 at 16:09
PBCPBC
14
asked Feb 1 at 16:09
PBCPBC
14
asked Feb 1 at 16:09
PBCPBC
14
14
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Feb 1 at 16:10
$begingroup$
@YuiTo Cheng: Transcribing an image into text is good; transcribing it into $rmLaTeX$, not as good.
$endgroup$
– Asaf Karagila♦
Feb 1 at 16:26
add a comment |
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Feb 1 at 16:10
$begingroup$
@YuiTo Cheng: Transcribing an image into text is good; transcribing it into $rmLaTeX$, not as good.
$endgroup$
– Asaf Karagila♦
Feb 1 at 16:26
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Feb 1 at 16:10
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Feb 1 at 16:10
$begingroup$
@YuiTo Cheng: Transcribing an image into text is good; transcribing it into $rmLaTeX$, not as good.
$endgroup$
– Asaf Karagila♦
Feb 1 at 16:26
$begingroup$
@YuiTo Cheng: Transcribing an image into text is good; transcribing it into $rmLaTeX$, not as good.
$endgroup$
– Asaf Karagila♦
Feb 1 at 16:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You should do your homework yourself.
But you are right about your decision which is true and false.
If the statement is false, you can give just a simple counterexample.
Hints:
b) Assume $E$ is closed. The limit of each convergent sequence in E is in E. Now, use the property of either infimum or supremum to get a contradition.
c) Assume $E$ is finite. What is the supremum and infimum of a finite set?
answered Feb 1 at 16:18
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
$begingroup$
Thanks for the advice! concerning part c, if E is finite then the set E must contain its supremum and infimum, right? but in my case, I'm given that they do not belong to E. Is this a nice counter example?
$endgroup$
– PBC
Feb 1 at 16:34
$begingroup$
Yes, that argument is correct and sufficient.
$endgroup$
– Mundron Schmidt
Feb 1 at 17:08
add a comment |
$begingroup$
The convention is to show that a statement is false by providing a counter-example and to show that it's true by giving a general proof.
$(a)$ Consider the counter-example $E=(0,1]cup[2,3).alpha=3,beta=0$ but $E$ is not open.
$(b)$ A closed set contains all its limit points. Show that $alpha$ is a limit point of $E$ by noting that $N_epsilon(alpha)cap E-{alpha}nephi$. Since $alphanotin E,E$ is not closed.
$(c)$ If $E$ were to be finite, it would be closed. We have just shown that $E$ is not closed.
$(d)$ The same counter-example as in $(a)$ works here too.
answered Feb 1 at 16:33
Shubham JohriShubham Johri
5,558818
$endgroup$
$begingroup$
Thank you so much, you helped me a lot! Though how did you know that Nϵ(α)∩E−{α}≠ϕ ? Can you please explain this to me.
$endgroup$
– PBC
Feb 1 at 16:41
$begingroup$
@souadbouchahine It was meant to be a clue for you to build on. Anywho, you just have to use the basic properties of supremum. Since $alpha-epsilon$ is not an upper bound of $E,exists xin E|alpha>x>alpha-epsilon$. Thus, $x$ lies in the intersection of $E$ and deleted neighbourhood of $alpha$.
$endgroup$
– Shubham Johri
Feb 1 at 16:46
$begingroup$
Okay, thank you for making this clear!
$endgroup$
– PBC
Feb 1 at 16:52
$begingroup$
But how can the counter example of part a) work in part c) too? if α = 3 and β=0 then (α, β) = (3,0). Is that possible?
$endgroup$
– PBC
Feb 1 at 16:55
$begingroup$
@souadbouchahine For the last option to make sense, it should be $(beta,alpha)$, not the other way round.
$endgroup$
– Shubham Johri
Feb 1 at 20:21
|
show 7 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You should do your homework yourself.
But you are right about your decision which is true and false.
If the statement is false, you can give just a simple counterexample.
Hints:
b) Assume $E$ is closed. The limit of each convergent sequence in E is in E. Now, use the property of either infimum or supremum to get a contradition.
c) Assume $E$ is finite. What is the supremum and infimum of a finite set?
answered Feb 1 at 16:18
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
$begingroup$
Thanks for the advice! concerning part c, if E is finite then the set E must contain its supremum and infimum, right? but in my case, I'm given that they do not belong to E. Is this a nice counter example?
$endgroup$
– PBC
Feb 1 at 16:34
$begingroup$
Yes, that argument is correct and sufficient.
$endgroup$
– Mundron Schmidt
Feb 1 at 17:08
add a comment |
$begingroup$
You should do your homework yourself.
But you are right about your decision which is true and false.
If the statement is false, you can give just a simple counterexample.
Hints:
b) Assume $E$ is closed. The limit of each convergent sequence in E is in E. Now, use the property of either infimum or supremum to get a contradition.
c) Assume $E$ is finite. What is the supremum and infimum of a finite set?
answered Feb 1 at 16:18
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
$begingroup$
Thanks for the advice! concerning part c, if E is finite then the set E must contain its supremum and infimum, right? but in my case, I'm given that they do not belong to E. Is this a nice counter example?
$endgroup$
– PBC
Feb 1 at 16:34
$begingroup$
Yes, that argument is correct and sufficient.
$endgroup$
– Mundron Schmidt
Feb 1 at 17:08
add a comment |
$begingroup$
You should do your homework yourself.
But you are right about your decision which is true and false.
If the statement is false, you can give just a simple counterexample.
Hints:
b) Assume $E$ is closed. The limit of each convergent sequence in E is in E. Now, use the property of either infimum or supremum to get a contradition.
c) Assume $E$ is finite. What is the supremum and infimum of a finite set?
answered Feb 1 at 16:18
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
You should do your homework yourself.
But you are right about your decision which is true and false.
If the statement is false, you can give just a simple counterexample.
Hints:
b) Assume $E$ is closed. The limit of each convergent sequence in E is in E. Now, use the property of either infimum or supremum to get a contradition.
c) Assume $E$ is finite. What is the supremum and infimum of a finite set?
answered Feb 1 at 16:18
Mundron SchmidtMundron Schmidt
7,4942729
answered Feb 1 at 16:18
Mundron SchmidtMundron Schmidt
7,4942729
answered Feb 1 at 16:18
Mundron SchmidtMundron Schmidt
7,4942729
answered Feb 1 at 16:18
Mundron SchmidtMundron Schmidt
7,4942729
7,4942729
$begingroup$
Thanks for the advice! concerning part c, if E is finite then the set E must contain its supremum and infimum, right? but in my case, I'm given that they do not belong to E. Is this a nice counter example?
$endgroup$
– PBC
Feb 1 at 16:34
$begingroup$
Yes, that argument is correct and sufficient.
$endgroup$
– Mundron Schmidt
Feb 1 at 17:08
add a comment |
$begingroup$
Thanks for the advice! concerning part c, if E is finite then the set E must contain its supremum and infimum, right? but in my case, I'm given that they do not belong to E. Is this a nice counter example?
$endgroup$
– PBC
Feb 1 at 16:34
$begingroup$
Yes, that argument is correct and sufficient.
$endgroup$
– Mundron Schmidt
Feb 1 at 17:08
$begingroup$
Thanks for the advice! concerning part c, if E is finite then the set E must contain its supremum and infimum, right? but in my case, I'm given that they do not belong to E. Is this a nice counter example?
$endgroup$
– PBC
Feb 1 at 16:34
$begingroup$
Thanks for the advice! concerning part c, if E is finite then the set E must contain its supremum and infimum, right? but in my case, I'm given that they do not belong to E. Is this a nice counter example?
$endgroup$
– PBC
Feb 1 at 16:34
$begingroup$
Yes, that argument is correct and sufficient.
$endgroup$
– Mundron Schmidt
Feb 1 at 17:08
$begingroup$
Yes, that argument is correct and sufficient.
$endgroup$
– Mundron Schmidt
Feb 1 at 17:08
add a comment |
$begingroup$
The convention is to show that a statement is false by providing a counter-example and to show that it's true by giving a general proof.
$(a)$ Consider the counter-example $E=(0,1]cup[2,3).alpha=3,beta=0$ but $E$ is not open.
$(b)$ A closed set contains all its limit points. Show that $alpha$ is a limit point of $E$ by noting that $N_epsilon(alpha)cap E-{alpha}nephi$. Since $alphanotin E,E$ is not closed.
$(c)$ If $E$ were to be finite, it would be closed. We have just shown that $E$ is not closed.
$(d)$ The same counter-example as in $(a)$ works here too.
answered Feb 1 at 16:33
Shubham JohriShubham Johri
5,558818
$endgroup$
$begingroup$
Thank you so much, you helped me a lot! Though how did you know that Nϵ(α)∩E−{α}≠ϕ ? Can you please explain this to me.
$endgroup$
– PBC
Feb 1 at 16:41
$begingroup$
@souadbouchahine It was meant to be a clue for you to build on. Anywho, you just have to use the basic properties of supremum. Since $alpha-epsilon$ is not an upper bound of $E,exists xin E|alpha>x>alpha-epsilon$. Thus, $x$ lies in the intersection of $E$ and deleted neighbourhood of $alpha$.
$endgroup$
– Shubham Johri
Feb 1 at 16:46
$begingroup$
Okay, thank you for making this clear!
$endgroup$
– PBC
Feb 1 at 16:52
$begingroup$
But how can the counter example of part a) work in part c) too? if α = 3 and β=0 then (α, β) = (3,0). Is that possible?
$endgroup$
– PBC
Feb 1 at 16:55
$begingroup$
@souadbouchahine For the last option to make sense, it should be $(beta,alpha)$, not the other way round.
$endgroup$
– Shubham Johri
Feb 1 at 20:21
|
show 7 more comments
$begingroup$
The convention is to show that a statement is false by providing a counter-example and to show that it's true by giving a general proof.
$(a)$ Consider the counter-example $E=(0,1]cup[2,3).alpha=3,beta=0$ but $E$ is not open.
$(b)$ A closed set contains all its limit points. Show that $alpha$ is a limit point of $E$ by noting that $N_epsilon(alpha)cap E-{alpha}nephi$. Since $alphanotin E,E$ is not closed.
$(c)$ If $E$ were to be finite, it would be closed. We have just shown that $E$ is not closed.
$(d)$ The same counter-example as in $(a)$ works here too.
answered Feb 1 at 16:33
Shubham JohriShubham Johri
5,558818
$endgroup$
$begingroup$
Thank you so much, you helped me a lot! Though how did you know that Nϵ(α)∩E−{α}≠ϕ ? Can you please explain this to me.
$endgroup$
– PBC
Feb 1 at 16:41
$begingroup$
@souadbouchahine It was meant to be a clue for you to build on. Anywho, you just have to use the basic properties of supremum. Since $alpha-epsilon$ is not an upper bound of $E,exists xin E|alpha>x>alpha-epsilon$. Thus, $x$ lies in the intersection of $E$ and deleted neighbourhood of $alpha$.
$endgroup$
– Shubham Johri
Feb 1 at 16:46
$begingroup$
Okay, thank you for making this clear!
$endgroup$
– PBC
Feb 1 at 16:52
$begingroup$
But how can the counter example of part a) work in part c) too? if α = 3 and β=0 then (α, β) = (3,0). Is that possible?
$endgroup$
– PBC
Feb 1 at 16:55
$begingroup$
@souadbouchahine For the last option to make sense, it should be $(beta,alpha)$, not the other way round.
$endgroup$
– Shubham Johri
Feb 1 at 20:21
|
show 7 more comments
$begingroup$
The convention is to show that a statement is false by providing a counter-example and to show that it's true by giving a general proof.
$(a)$ Consider the counter-example $E=(0,1]cup[2,3).alpha=3,beta=0$ but $E$ is not open.
$(b)$ A closed set contains all its limit points. Show that $alpha$ is a limit point of $E$ by noting that $N_epsilon(alpha)cap E-{alpha}nephi$. Since $alphanotin E,E$ is not closed.
$(c)$ If $E$ were to be finite, it would be closed. We have just shown that $E$ is not closed.
$(d)$ The same counter-example as in $(a)$ works here too.
answered Feb 1 at 16:33
Shubham JohriShubham Johri
5,558818
$endgroup$
The convention is to show that a statement is false by providing a counter-example and to show that it's true by giving a general proof.
$(a)$ Consider the counter-example $E=(0,1]cup[2,3).alpha=3,beta=0$ but $E$ is not open.
$(b)$ A closed set contains all its limit points. Show that $alpha$ is a limit point of $E$ by noting that $N_epsilon(alpha)cap E-{alpha}nephi$. Since $alphanotin E,E$ is not closed.
$(c)$ If $E$ were to be finite, it would be closed. We have just shown that $E$ is not closed.
$(d)$ The same counter-example as in $(a)$ works here too.
answered Feb 1 at 16:33
Shubham JohriShubham Johri
5,558818
edited Feb 1 at 16:38
answered Feb 1 at 16:33
Shubham JohriShubham Johri
5,558818
answered Feb 1 at 16:33
Shubham JohriShubham Johri
5,558818
answered Feb 1 at 16:33
Shubham JohriShubham Johri
5,558818
5,558818
$begingroup$
Thank you so much, you helped me a lot! Though how did you know that Nϵ(α)∩E−{α}≠ϕ ? Can you please explain this to me.
$endgroup$
– PBC
Feb 1 at 16:41
$begingroup$
@souadbouchahine It was meant to be a clue for you to build on. Anywho, you just have to use the basic properties of supremum. Since $alpha-epsilon$ is not an upper bound of $E,exists xin E|alpha>x>alpha-epsilon$. Thus, $x$ lies in the intersection of $E$ and deleted neighbourhood of $alpha$.
$endgroup$
– Shubham Johri
Feb 1 at 16:46
$begingroup$
Okay, thank you for making this clear!
$endgroup$
– PBC
Feb 1 at 16:52
$begingroup$
But how can the counter example of part a) work in part c) too? if α = 3 and β=0 then (α, β) = (3,0). Is that possible?
$endgroup$
– PBC
Feb 1 at 16:55
$begingroup$
@souadbouchahine For the last option to make sense, it should be $(beta,alpha)$, not the other way round.
$endgroup$
– Shubham Johri
Feb 1 at 20:21
|
show 7 more comments
$begingroup$
Thank you so much, you helped me a lot! Though how did you know that Nϵ(α)∩E−{α}≠ϕ ? Can you please explain this to me.
$endgroup$
– PBC
Feb 1 at 16:41
$begingroup$
@souadbouchahine It was meant to be a clue for you to build on. Anywho, you just have to use the basic properties of supremum. Since $alpha-epsilon$ is not an upper bound of $E,exists xin E|alpha>x>alpha-epsilon$. Thus, $x$ lies in the intersection of $E$ and deleted neighbourhood of $alpha$.
$endgroup$
– Shubham Johri
Feb 1 at 16:46
$begingroup$
Okay, thank you for making this clear!
$endgroup$
– PBC
Feb 1 at 16:52
$begingroup$
But how can the counter example of part a) work in part c) too? if α = 3 and β=0 then (α, β) = (3,0). Is that possible?
$endgroup$
– PBC
Feb 1 at 16:55
$begingroup$
@souadbouchahine For the last option to make sense, it should be $(beta,alpha)$, not the other way round.
$endgroup$
– Shubham Johri
Feb 1 at 20:21
$begingroup$
Thank you so much, you helped me a lot! Though how did you know that Nϵ(α)∩E−{α}≠ϕ ? Can you please explain this to me.
$endgroup$
– PBC
Feb 1 at 16:41
$begingroup$
Thank you so much, you helped me a lot! Though how did you know that Nϵ(α)∩E−{α}≠ϕ ? Can you please explain this to me.
$endgroup$
– PBC
Feb 1 at 16:41
$begingroup$
@souadbouchahine It was meant to be a clue for you to build on. Anywho, you just have to use the basic properties of supremum. Since $alpha-epsilon$ is not an upper bound of $E,exists xin E|alpha>x>alpha-epsilon$. Thus, $x$ lies in the intersection of $E$ and deleted neighbourhood of $alpha$.
$endgroup$
– Shubham Johri
Feb 1 at 16:46
$begingroup$
@souadbouchahine It was meant to be a clue for you to build on. Anywho, you just have to use the basic properties of supremum. Since $alpha-epsilon$ is not an upper bound of $E,exists xin E|alpha>x>alpha-epsilon$. Thus, $x$ lies in the intersection of $E$ and deleted neighbourhood of $alpha$.
$endgroup$
– Shubham Johri
Feb 1 at 16:46
$begingroup$
Okay, thank you for making this clear!
$endgroup$
– PBC
Feb 1 at 16:52
$begingroup$
Okay, thank you for making this clear!
$endgroup$
– PBC
Feb 1 at 16:52
$begingroup$
But how can the counter example of part a) work in part c) too? if α = 3 and β=0 then (α, β) = (3,0). Is that possible?
$endgroup$
– PBC
Feb 1 at 16:55
$begingroup$
But how can the counter example of part a) work in part c) too? if α = 3 and β=0 then (α, β) = (3,0). Is that possible?
$endgroup$
– PBC
Feb 1 at 16:55
$begingroup$
@souadbouchahine For the last option to make sense, it should be $(beta,alpha)$, not the other way round.
$endgroup$
– Shubham Johri
Feb 1 at 20:21
$begingroup$
@souadbouchahine For the last option to make sense, it should be $(beta,alpha)$, not the other way round.
$endgroup$
– Shubham Johri
Feb 1 at 20:21
|
show 7 more comments
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$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Feb 1 at 16:10
$begingroup$
@YuiTo Cheng: Transcribing an image into text is good; transcribing it into $rmLaTeX$, not as good.
$endgroup$
– Asaf Karagila♦
Feb 1 at 16:26