Mixing
Is there a function $f: mathbb{R} to mathbb{R}$ such that $limlimits_{xto p}f(x)=infty$ for every $p in...
$begingroup$
I was wondering if there is a function $f: mathbb{R} to mathbb{R}$ such that its limit at every point is infinite.
I guess not, because what would its graph look like, but then again, I don't know how to prove it.
limits
edited Jan 8 at 8:41
Did
247k23223460
asked Jan 8 at 8:09
user4201961user4201961
730411
$endgroup$
closed as off-topic by Holo, amWhy, Saad, user21820, José Carlos Santos Jan 12 at 11:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, amWhy, Saad, user21820, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 3 more comments
$begingroup$
I was wondering if there is a function $f: mathbb{R} to mathbb{R}$ such that its limit at every point is infinite.
I guess not, because what would its graph look like, but then again, I don't know how to prove it.
limits
edited Jan 8 at 8:41
Did
247k23223460
asked Jan 8 at 8:09
user4201961user4201961
730411
$endgroup$
closed as off-topic by Holo, amWhy, Saad, user21820, José Carlos Santos Jan 12 at 11:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, amWhy, Saad, user21820, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
$endgroup$
– Eevee Trainer
Jan 8 at 8:31
$begingroup$
My intuition tells me that such function doesn't exist. I like to be proven wrong.
$endgroup$
– Math_QED
Jan 8 at 8:40
1
$begingroup$
@EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
$endgroup$
– Anders Kaseorg
Jan 8 at 8:44
2
$begingroup$
No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
$endgroup$
– Song
Jan 8 at 8:44
$begingroup$
Hmmm, I see. Thanks for the insight @AndersKaseorg
$endgroup$
– Eevee Trainer
Jan 8 at 8:45
|
show 3 more comments
$begingroup$
I was wondering if there is a function $f: mathbb{R} to mathbb{R}$ such that its limit at every point is infinite.
I guess not, because what would its graph look like, but then again, I don't know how to prove it.
limits
edited Jan 8 at 8:41
Did
247k23223460
asked Jan 8 at 8:09
user4201961user4201961
730411
$endgroup$
I was wondering if there is a function $f: mathbb{R} to mathbb{R}$ such that its limit at every point is infinite.
I guess not, because what would its graph look like, but then again, I don't know how to prove it.
limits
limits
edited Jan 8 at 8:41
Did
247k23223460
asked Jan 8 at 8:09
user4201961user4201961
730411
edited Jan 8 at 8:41
Did
247k23223460
asked Jan 8 at 8:09
user4201961user4201961
730411
edited Jan 8 at 8:41
Did
247k23223460
edited Jan 8 at 8:41
Did
247k23223460
edited Jan 8 at 8:41
Did
247k23223460
247k23223460
asked Jan 8 at 8:09
user4201961user4201961
730411
asked Jan 8 at 8:09
user4201961user4201961
730411
asked Jan 8 at 8:09
user4201961user4201961
730411
730411
closed as off-topic by Holo, amWhy, Saad, user21820, José Carlos Santos Jan 12 at 11:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, amWhy, Saad, user21820, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Holo, amWhy, Saad, user21820, José Carlos Santos Jan 12 at 11:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, amWhy, Saad, user21820, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
$endgroup$
– Eevee Trainer
Jan 8 at 8:31
$begingroup$
My intuition tells me that such function doesn't exist. I like to be proven wrong.
$endgroup$
– Math_QED
Jan 8 at 8:40
1
$begingroup$
@EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
$endgroup$
– Anders Kaseorg
Jan 8 at 8:44
2
$begingroup$
No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
$endgroup$
– Song
Jan 8 at 8:44
$begingroup$
Hmmm, I see. Thanks for the insight @AndersKaseorg
$endgroup$
– Eevee Trainer
Jan 8 at 8:45
|
show 3 more comments
$begingroup$
While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
$endgroup$
– Eevee Trainer
Jan 8 at 8:31
$begingroup$
My intuition tells me that such function doesn't exist. I like to be proven wrong.
$endgroup$
– Math_QED
Jan 8 at 8:40
1
$begingroup$
@EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
$endgroup$
– Anders Kaseorg
Jan 8 at 8:44
2
$begingroup$
No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
$endgroup$
– Song
Jan 8 at 8:44
$begingroup$
Hmmm, I see. Thanks for the insight @AndersKaseorg
$endgroup$
– Eevee Trainer
Jan 8 at 8:45
$begingroup$
While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
$endgroup$
– Eevee Trainer
Jan 8 at 8:31
$begingroup$
While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
$endgroup$
– Eevee Trainer
Jan 8 at 8:31
$begingroup$
My intuition tells me that such function doesn't exist. I like to be proven wrong.
$endgroup$
– Math_QED
Jan 8 at 8:40
$begingroup$
My intuition tells me that such function doesn't exist. I like to be proven wrong.
$endgroup$
– Math_QED
Jan 8 at 8:40
1
1
$begingroup$
@EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
$endgroup$
– Anders Kaseorg
Jan 8 at 8:44
$begingroup$
@EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
$endgroup$
– Anders Kaseorg
Jan 8 at 8:44
2
2
$begingroup$
No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
$endgroup$
– Song
Jan 8 at 8:44
$begingroup$
No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
$endgroup$
– Song
Jan 8 at 8:44
$begingroup$
Hmmm, I see. Thanks for the insight @AndersKaseorg
$endgroup$
– Eevee Trainer
Jan 8 at 8:45
$begingroup$
Hmmm, I see. Thanks for the insight @AndersKaseorg
$endgroup$
– Eevee Trainer
Jan 8 at 8:45
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
If $f$ is such a function and $M>0$ then
$$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
But
$$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
$$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.
edited Jan 8 at 9:46
SvanN
2,0421422
answered Jan 8 at 8:30
DarmadDarmad
538112
$endgroup$
add a comment |
$begingroup$
Let us show that such a function $f$ does not exist.
Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.
Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.
Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.
answered Jan 8 at 9:32
DidDid
247k23223460
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $f$ is such a function and $M>0$ then
$$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
But
$$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
$$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.
edited Jan 8 at 9:46
SvanN
2,0421422
answered Jan 8 at 8:30
DarmadDarmad
538112
$endgroup$
add a comment |
$begingroup$
If $f$ is such a function and $M>0$ then
$$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
But
$$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
$$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.
edited Jan 8 at 9:46
SvanN
2,0421422
answered Jan 8 at 8:30
DarmadDarmad
538112
$endgroup$
add a comment |
$begingroup$
If $f$ is such a function and $M>0$ then
$$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
But
$$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
$$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.
edited Jan 8 at 9:46
SvanN
2,0421422
answered Jan 8 at 8:30
DarmadDarmad
538112
$endgroup$
If $f$ is such a function and $M>0$ then
$$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
But
$$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
$$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.
edited Jan 8 at 9:46
SvanN
2,0421422
answered Jan 8 at 8:30
DarmadDarmad
538112
edited Jan 8 at 9:46
SvanN
2,0421422
edited Jan 8 at 9:46
SvanN
2,0421422
edited Jan 8 at 9:46
SvanN
2,0421422
2,0421422
answered Jan 8 at 8:30
DarmadDarmad
538112
answered Jan 8 at 8:30
DarmadDarmad
538112
answered Jan 8 at 8:30
DarmadDarmad
538112
538112
add a comment |
add a comment |
$begingroup$
Let us show that such a function $f$ does not exist.
Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.
Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.
Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.
answered Jan 8 at 9:32
DidDid
247k23223460
$endgroup$
add a comment |
$begingroup$
Let us show that such a function $f$ does not exist.
Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.
Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.
Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.
answered Jan 8 at 9:32
DidDid
247k23223460
$endgroup$
add a comment |
$begingroup$
Let us show that such a function $f$ does not exist.
Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.
Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.
Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.
answered Jan 8 at 9:32
DidDid
247k23223460
$endgroup$
Let us show that such a function $f$ does not exist.
Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.
Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.
Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.
answered Jan 8 at 9:32
DidDid
247k23223460
edited Jan 12 at 9:56
answered Jan 8 at 9:32
DidDid
247k23223460
answered Jan 8 at 9:32
DidDid
247k23223460
answered Jan 8 at 9:32
DidDid
247k23223460
247k23223460
add a comment |
add a comment |
$begingroup$
While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
$endgroup$
– Eevee Trainer
Jan 8 at 8:31
$begingroup$
My intuition tells me that such function doesn't exist. I like to be proven wrong.
$endgroup$
– Math_QED
Jan 8 at 8:40
1
$begingroup$
@EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
$endgroup$
– Anders Kaseorg
Jan 8 at 8:44
2
$begingroup$
No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
$endgroup$
– Song
Jan 8 at 8:44
$begingroup$
Hmmm, I see. Thanks for the insight @AndersKaseorg
$endgroup$
– Eevee Trainer
Jan 8 at 8:45