$f: mathbb R to mathbb R$ be a continuous function, $A ={y in mathbb R: y = lim_{n to infty} f(x_{n}),$ for a...
$f: mathbb R to mathbb R$ be a continuous function and $A$ is a proper subset of $mathbb R$ such that $A ={y in mathbb R: y = lim_{n to infty} f(x_{n}),$ for a sequence $x_{n} to + infty}$. Then the set $A$ is necessarily $-$
$a$ compact set
$b$ closed set
$c$ Singleton set
$d$ none of these
My attempt :
If I take the function $f(x) = sin x$ and $x_{n} = n pi$ and another sequence $y_{n}$ for same function $sin x$ that converges to $1$ or $-1$. Then, I'll have at least two element in $A$ so it is not necessarily a connected set. But this set is compact as well as closed.
I don't know how to choose between other three options.
Can I find a set $A$ such that it's not bounded or it's not closed?
real-analysis
edited Jan 4 at 13:37
mathcounterexamples.net
25.3k21953
asked Jan 4 at 13:28
MathsaddictMathsaddict
2858
add a comment |
$f: mathbb R to mathbb R$ be a continuous function and $A$ is a proper subset of $mathbb R$ such that $A ={y in mathbb R: y = lim_{n to infty} f(x_{n}),$ for a sequence $x_{n} to + infty}$. Then the set $A$ is necessarily $-$
$a$ compact set
$b$ closed set
$c$ Singleton set
$d$ none of these
My attempt :
If I take the function $f(x) = sin x$ and $x_{n} = n pi$ and another sequence $y_{n}$ for same function $sin x$ that converges to $1$ or $-1$. Then, I'll have at least two element in $A$ so it is not necessarily a connected set. But this set is compact as well as closed.
I don't know how to choose between other three options.
Can I find a set $A$ such that it's not bounded or it's not closed?
real-analysis
edited Jan 4 at 13:37
mathcounterexamples.net
25.3k21953
asked Jan 4 at 13:28
MathsaddictMathsaddict
2858
1
The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
– Paul K
Jan 4 at 13:32
It's given in question, "for a sequence" I didn't paid attention to it. Also, for a sequence I could find only one element in A.
– Mathsaddict
Jan 4 at 13:37
@Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
– Mathsaddict
Jan 4 at 13:41
add a comment |
$f: mathbb R to mathbb R$ be a continuous function and $A$ is a proper subset of $mathbb R$ such that $A ={y in mathbb R: y = lim_{n to infty} f(x_{n}),$ for a sequence $x_{n} to + infty}$. Then the set $A$ is necessarily $-$
$a$ compact set
$b$ closed set
$c$ Singleton set
$d$ none of these
My attempt :
If I take the function $f(x) = sin x$ and $x_{n} = n pi$ and another sequence $y_{n}$ for same function $sin x$ that converges to $1$ or $-1$. Then, I'll have at least two element in $A$ so it is not necessarily a connected set. But this set is compact as well as closed.
I don't know how to choose between other three options.
Can I find a set $A$ such that it's not bounded or it's not closed?
real-analysis
edited Jan 4 at 13:37
mathcounterexamples.net
25.3k21953
asked Jan 4 at 13:28
MathsaddictMathsaddict
2858
$f: mathbb R to mathbb R$ be a continuous function and $A$ is a proper subset of $mathbb R$ such that $A ={y in mathbb R: y = lim_{n to infty} f(x_{n}),$ for a sequence $x_{n} to + infty}$. Then the set $A$ is necessarily $-$
$a$ compact set
$b$ closed set
$c$ Singleton set
$d$ none of these
My attempt :
If I take the function $f(x) = sin x$ and $x_{n} = n pi$ and another sequence $y_{n}$ for same function $sin x$ that converges to $1$ or $-1$. Then, I'll have at least two element in $A$ so it is not necessarily a connected set. But this set is compact as well as closed.
I don't know how to choose between other three options.
Can I find a set $A$ such that it's not bounded or it's not closed?
real-analysis
real-analysis
edited Jan 4 at 13:37
mathcounterexamples.net
25.3k21953
asked Jan 4 at 13:28
MathsaddictMathsaddict
2858
edited Jan 4 at 13:37
mathcounterexamples.net
25.3k21953
asked Jan 4 at 13:28
MathsaddictMathsaddict
2858
edited Jan 4 at 13:37
mathcounterexamples.net
25.3k21953
edited Jan 4 at 13:37
mathcounterexamples.net
25.3k21953
edited Jan 4 at 13:37
mathcounterexamples.net
25.3k21953
25.3k21953
asked Jan 4 at 13:28
MathsaddictMathsaddict
2858
asked Jan 4 at 13:28
MathsaddictMathsaddict
2858
asked Jan 4 at 13:28
MathsaddictMathsaddict
2858
2858
1
The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
– Paul K
Jan 4 at 13:32
It's given in question, "for a sequence" I didn't paid attention to it. Also, for a sequence I could find only one element in A.
– Mathsaddict
Jan 4 at 13:37
@Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
– Mathsaddict
Jan 4 at 13:41
add a comment |
1
The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
– Paul K
Jan 4 at 13:32
It's given in question, "for a sequence" I didn't paid attention to it. Also, for a sequence I could find only one element in A.
– Mathsaddict
Jan 4 at 13:37
@Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
– Mathsaddict
Jan 4 at 13:41
1
1
The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
– Paul K
Jan 4 at 13:32
The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
– Paul K
Jan 4 at 13:32
It's given in question, "for a sequence" I didn't paid attention to it. Also, for a sequence I could find only one element in A.
– Mathsaddict
Jan 4 at 13:37
It's given in question, "for a sequence" I didn't paid attention to it. Also, for a sequence I could find only one element in A.
– Mathsaddict
Jan 4 at 13:37
@Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
– Mathsaddict
Jan 4 at 13:41
@Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
– Mathsaddict
Jan 4 at 13:41
add a comment |
1 Answer
1
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votes
a. $A$ may not be a compact set
$f(x)=
begin{cases}
0 & x le 0\
x left(sin x +1 right)& x>0
end{cases}$
You have $A = [0,infty)$
c. $A$ may not be a singleton set
$f(x) = sin x$
$A = [-1,1]$.
b. $A$ is a closed set
If $(y_n)$ is a sequence of $A$ converging to $y$, use a diagonal argument to build a sequence $(x_n)$ such that $limlimits_{n to infty} x_n = infty$ and $limlimits_{n to infty} f(x_n) = y$. Proving that $y in A$ and that $A$ is closed.
FINALLY, THE RIGHT ANSWER IS b.
answered Jan 4 at 13:40
mathcounterexamples.netmathcounterexamples.net
25.3k21953
So I can choose a continuous function and sequence, if I take the set of all limit points of $f(x_{n})$ then I will get $A$, since there always exists a subsequence converging to a limit point. Then I have the counterexamples you mentioned. Now, since the set of limit points is a closed set, so d is correct. Am I right?
– Mathsaddict
Jan 5 at 3:09
No, as b. is correct, d. is not the right answer.
– mathcounterexamples.net
Jan 5 at 7:46
Yes , I meant 'd'. It was typing error. But what I got from your answer is correct?
– Mathsaddict
Jan 5 at 14:27
Yes, that is correct.
– mathcounterexamples.net
Jan 5 at 15:23
add a comment |
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1 Answer
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1 Answer
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votes
a. $A$ may not be a compact set
$f(x)=
begin{cases}
0 & x le 0\
x left(sin x +1 right)& x>0
end{cases}$
You have $A = [0,infty)$
c. $A$ may not be a singleton set
$f(x) = sin x$
$A = [-1,1]$.
b. $A$ is a closed set
If $(y_n)$ is a sequence of $A$ converging to $y$, use a diagonal argument to build a sequence $(x_n)$ such that $limlimits_{n to infty} x_n = infty$ and $limlimits_{n to infty} f(x_n) = y$. Proving that $y in A$ and that $A$ is closed.
FINALLY, THE RIGHT ANSWER IS b.
answered Jan 4 at 13:40
mathcounterexamples.netmathcounterexamples.net
25.3k21953
So I can choose a continuous function and sequence, if I take the set of all limit points of $f(x_{n})$ then I will get $A$, since there always exists a subsequence converging to a limit point. Then I have the counterexamples you mentioned. Now, since the set of limit points is a closed set, so d is correct. Am I right?
– Mathsaddict
Jan 5 at 3:09
No, as b. is correct, d. is not the right answer.
– mathcounterexamples.net
Jan 5 at 7:46
Yes , I meant 'd'. It was typing error. But what I got from your answer is correct?
– Mathsaddict
Jan 5 at 14:27
Yes, that is correct.
– mathcounterexamples.net
Jan 5 at 15:23
add a comment |
a. $A$ may not be a compact set
$f(x)=
begin{cases}
0 & x le 0\
x left(sin x +1 right)& x>0
end{cases}$
You have $A = [0,infty)$
c. $A$ may not be a singleton set
$f(x) = sin x$
$A = [-1,1]$.
b. $A$ is a closed set
If $(y_n)$ is a sequence of $A$ converging to $y$, use a diagonal argument to build a sequence $(x_n)$ such that $limlimits_{n to infty} x_n = infty$ and $limlimits_{n to infty} f(x_n) = y$. Proving that $y in A$ and that $A$ is closed.
FINALLY, THE RIGHT ANSWER IS b.
answered Jan 4 at 13:40
mathcounterexamples.netmathcounterexamples.net
25.3k21953
So I can choose a continuous function and sequence, if I take the set of all limit points of $f(x_{n})$ then I will get $A$, since there always exists a subsequence converging to a limit point. Then I have the counterexamples you mentioned. Now, since the set of limit points is a closed set, so d is correct. Am I right?
– Mathsaddict
Jan 5 at 3:09
No, as b. is correct, d. is not the right answer.
– mathcounterexamples.net
Jan 5 at 7:46
Yes , I meant 'd'. It was typing error. But what I got from your answer is correct?
– Mathsaddict
Jan 5 at 14:27
Yes, that is correct.
– mathcounterexamples.net
Jan 5 at 15:23
add a comment |
a. $A$ may not be a compact set
$f(x)=
begin{cases}
0 & x le 0\
x left(sin x +1 right)& x>0
end{cases}$
You have $A = [0,infty)$
c. $A$ may not be a singleton set
$f(x) = sin x$
$A = [-1,1]$.
b. $A$ is a closed set
If $(y_n)$ is a sequence of $A$ converging to $y$, use a diagonal argument to build a sequence $(x_n)$ such that $limlimits_{n to infty} x_n = infty$ and $limlimits_{n to infty} f(x_n) = y$. Proving that $y in A$ and that $A$ is closed.
FINALLY, THE RIGHT ANSWER IS b.
answered Jan 4 at 13:40
mathcounterexamples.netmathcounterexamples.net
25.3k21953
a. $A$ may not be a compact set
$f(x)=
begin{cases}
0 & x le 0\
x left(sin x +1 right)& x>0
end{cases}$
You have $A = [0,infty)$
c. $A$ may not be a singleton set
$f(x) = sin x$
$A = [-1,1]$.
b. $A$ is a closed set
If $(y_n)$ is a sequence of $A$ converging to $y$, use a diagonal argument to build a sequence $(x_n)$ such that $limlimits_{n to infty} x_n = infty$ and $limlimits_{n to infty} f(x_n) = y$. Proving that $y in A$ and that $A$ is closed.
FINALLY, THE RIGHT ANSWER IS b.
answered Jan 4 at 13:40
mathcounterexamples.netmathcounterexamples.net
25.3k21953
edited Jan 4 at 14:16
answered Jan 4 at 13:40
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Jan 4 at 13:40
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Jan 4 at 13:40
mathcounterexamples.netmathcounterexamples.net
25.3k21953
25.3k21953
So I can choose a continuous function and sequence, if I take the set of all limit points of $f(x_{n})$ then I will get $A$, since there always exists a subsequence converging to a limit point. Then I have the counterexamples you mentioned. Now, since the set of limit points is a closed set, so d is correct. Am I right?
– Mathsaddict
Jan 5 at 3:09
No, as b. is correct, d. is not the right answer.
– mathcounterexamples.net
Jan 5 at 7:46
Yes , I meant 'd'. It was typing error. But what I got from your answer is correct?
– Mathsaddict
Jan 5 at 14:27
Yes, that is correct.
– mathcounterexamples.net
Jan 5 at 15:23
add a comment |
So I can choose a continuous function and sequence, if I take the set of all limit points of $f(x_{n})$ then I will get $A$, since there always exists a subsequence converging to a limit point. Then I have the counterexamples you mentioned. Now, since the set of limit points is a closed set, so d is correct. Am I right?
– Mathsaddict
Jan 5 at 3:09
No, as b. is correct, d. is not the right answer.
– mathcounterexamples.net
Jan 5 at 7:46
Yes , I meant 'd'. It was typing error. But what I got from your answer is correct?
– Mathsaddict
Jan 5 at 14:27
Yes, that is correct.
– mathcounterexamples.net
Jan 5 at 15:23
So I can choose a continuous function and sequence, if I take the set of all limit points of $f(x_{n})$ then I will get $A$, since there always exists a subsequence converging to a limit point. Then I have the counterexamples you mentioned. Now, since the set of limit points is a closed set, so d is correct. Am I right?
– Mathsaddict
Jan 5 at 3:09
So I can choose a continuous function and sequence, if I take the set of all limit points of $f(x_{n})$ then I will get $A$, since there always exists a subsequence converging to a limit point. Then I have the counterexamples you mentioned. Now, since the set of limit points is a closed set, so d is correct. Am I right?
– Mathsaddict
Jan 5 at 3:09
No, as b. is correct, d. is not the right answer.
– mathcounterexamples.net
Jan 5 at 7:46
No, as b. is correct, d. is not the right answer.
– mathcounterexamples.net
Jan 5 at 7:46
Yes , I meant 'd'. It was typing error. But what I got from your answer is correct?
– Mathsaddict
Jan 5 at 14:27
Yes , I meant 'd'. It was typing error. But what I got from your answer is correct?
– Mathsaddict
Jan 5 at 14:27
Yes, that is correct.
– mathcounterexamples.net
Jan 5 at 15:23
Yes, that is correct.
– mathcounterexamples.net
Jan 5 at 15:23
add a comment |
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1
The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
– Paul K
Jan 4 at 13:32
It's given in question, "for a sequence" I didn't paid attention to it. Also, for a sequence I could find only one element in A.
– Mathsaddict
Jan 4 at 13:37
@Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
– Mathsaddict
Jan 4 at 13:41