$f: mathbb R to mathbb R$ be a continuous function, $A ={y in mathbb R: y = lim_{n to infty} f(x_{n}),$ for a...












3














$f: mathbb R to mathbb R$ 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







share|cite|improve this question




















  • 1




    The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
    – Paul K
    17 hours ago










  • 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
    17 hours ago










  • @Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
    – Mathsaddict
    17 hours ago


















3














$f: mathbb R to mathbb R$ 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







share|cite|improve this question




















  • 1




    The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
    – Paul K
    17 hours ago










  • 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
    17 hours ago










  • @Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
    – Mathsaddict
    17 hours ago
















3












3








3







$f: mathbb R to mathbb R$ 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







share|cite|improve this question















$f: mathbb R to mathbb R$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 17 hours ago









mathcounterexamples.net

24.9k21853




24.9k21853










asked 17 hours ago









Mathsaddict

2608




2608








  • 1




    The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
    – Paul K
    17 hours ago










  • 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
    17 hours ago










  • @Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
    – Mathsaddict
    17 hours ago
















  • 1




    The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
    – Paul K
    17 hours ago










  • 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
    17 hours ago










  • @Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
    – Mathsaddict
    17 hours ago










1




1




The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
– Paul K
17 hours ago




The set for $sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit.
– Paul K
17 hours ago












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
17 hours ago




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
17 hours ago












@Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
– Mathsaddict
17 hours ago






@Paul K I took sin x as function, and $npi$ as sequence, then $f(x_{n})$ converges to 0
– Mathsaddict
17 hours ago












1 Answer
1






active

oldest

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1














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.






share|cite|improve this answer























  • 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
    4 hours ago













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1 Answer
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1 Answer
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active

oldest

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active

oldest

votes









1














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.






share|cite|improve this answer























  • 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
    4 hours ago


















1














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.






share|cite|improve this answer























  • 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
    4 hours ago
















1












1








1






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 16 hours ago

























answered 17 hours ago









mathcounterexamples.net

24.9k21853




24.9k21853












  • 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
    4 hours ago




















  • 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
    4 hours ago


















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
4 hours ago






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
4 hours ago




















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