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Does every real function have this weak continuity property?
Clash Royale CLAN TAG#URR8PPP
up vote
49
down vote
favorite
In my research I came across the following question :
Is it true that for every real function $f:mathbbRtomathbbR$, there exists a real sequence $(x_n)_n$, taking infinitely many values, converging to some real number $c$, such that the sequence $(f(x_n))_n$ converges to $f(c)$ ?
set-theory real-analysis sequences-and-series
edited Aug 25 at 9:36
Micah
1885
asked Aug 24 at 10:17
Dattier
612314
add a comment |Â
up vote
49
down vote
favorite
In my research I came across the following question :
Is it true that for every real function $f:mathbbRtomathbbR$, there exists a real sequence $(x_n)_n$, taking infinitely many values, converging to some real number $c$, such that the sequence $(f(x_n))_n$ converges to $f(c)$ ?
set-theory real-analysis sequences-and-series
edited Aug 25 at 9:36
Micah
1885
asked Aug 24 at 10:17
Dattier
612314
1
The question looks interesting, because $f$ is not assumed to have any property (such as continuity). Perhaps prove first this. There is $a$ so that for every $epsilon > 0$, the set $<epsilon $ is uncountable.
– Gerald Edgar
Aug 24 at 10:31
1
The difficulty is to have, $f(x_n)$ converged to $f(a)$ and $x_n$ converged to $a$.
– Dattier
Aug 24 at 10:49
@GeraldEdgar : for your question, you can show that with compacity argument
– Dattier
Aug 24 at 10:52
11
I had to read it like four times before I convinced myself that this is not an immediately false triviality.
– Qfwfq
Aug 24 at 19:36
add a comment |Â
up vote
49
down vote
favorite
up vote
49
down vote
favorite
In my research I came across the following question :
Is it true that for every real function $f:mathbbRtomathbbR$, there exists a real sequence $(x_n)_n$, taking infinitely many values, converging to some real number $c$, such that the sequence $(f(x_n))_n$ converges to $f(c)$ ?
set-theory real-analysis sequences-and-series
edited Aug 25 at 9:36
Micah
1885
asked Aug 24 at 10:17
Dattier
612314
In my research I came across the following question :
Is it true that for every real function $f:mathbbRtomathbbR$, there exists a real sequence $(x_n)_n$, taking infinitely many values, converging to some real number $c$, such that the sequence $(f(x_n))_n$ converges to $f(c)$ ?
set-theory real-analysis sequences-and-series
edited Aug 25 at 9:36
Micah
1885
asked Aug 24 at 10:17
Dattier
612314
edited Aug 25 at 9:36
Micah
1885
edited Aug 25 at 9:36
Micah
1885
edited Aug 25 at 9:36
Micah
1885
1885
asked Aug 24 at 10:17
Dattier
612314
asked Aug 24 at 10:17
Dattier
612314
asked Aug 24 at 10:17
Dattier
612314
612314
1
The question looks interesting, because $f$ is not assumed to have any property (such as continuity). Perhaps prove first this. There is $a$ so that for every $epsilon > 0$, the set $<epsilon $ is uncountable.
– Gerald Edgar
Aug 24 at 10:31
1
The difficulty is to have, $f(x_n)$ converged to $f(a)$ and $x_n$ converged to $a$.
– Dattier
Aug 24 at 10:49
@GeraldEdgar : for your question, you can show that with compacity argument
– Dattier
Aug 24 at 10:52
11
I had to read it like four times before I convinced myself that this is not an immediately false triviality.
– Qfwfq
Aug 24 at 19:36
add a comment |Â
1
The question looks interesting, because $f$ is not assumed to have any property (such as continuity). Perhaps prove first this. There is $a$ so that for every $epsilon > 0$, the set $<epsilon $ is uncountable.
– Gerald Edgar
Aug 24 at 10:31
1
The difficulty is to have, $f(x_n)$ converged to $f(a)$ and $x_n$ converged to $a$.
– Dattier
Aug 24 at 10:49
@GeraldEdgar : for your question, you can show that with compacity argument
– Dattier
Aug 24 at 10:52
11
I had to read it like four times before I convinced myself that this is not an immediately false triviality.
– Qfwfq
Aug 24 at 19:36
1
1
The question looks interesting, because $f$ is not assumed to have any property (such as continuity). Perhaps prove first this. There is $a$ so that for every $epsilon > 0$, the set $<epsilon $ is uncountable.
– Gerald Edgar
Aug 24 at 10:31
The question looks interesting, because $f$ is not assumed to have any property (such as continuity). Perhaps prove first this. There is $a$ so that for every $epsilon > 0$, the set $<epsilon $ is uncountable.
– Gerald Edgar
Aug 24 at 10:31
1
1
The difficulty is to have, $f(x_n)$ converged to $f(a)$ and $x_n$ converged to $a$.
– Dattier
Aug 24 at 10:49
The difficulty is to have, $f(x_n)$ converged to $f(a)$ and $x_n$ converged to $a$.
– Dattier
Aug 24 at 10:49
@GeraldEdgar : for your question, you can show that with compacity argument
– Dattier
Aug 24 at 10:52
@GeraldEdgar : for your question, you can show that with compacity argument
– Dattier
Aug 24 at 10:52
11
11
I had to read it like four times before I convinced myself that this is not an immediately false triviality.
– Qfwfq
Aug 24 at 19:36
I had to read it like four times before I convinced myself that this is not an immediately false triviality.
– Qfwfq
Aug 24 at 19:36
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
59
down vote
accepted
Suppose $f$ were a counterexample. Then, for any $c$, we could find a little interval $(a,b)$ containing $c$ and we could find some $varepsilon>0$ such that all points $xin(a,b)$ except $c$ have $|f(x)-f(c)|>varepsilon$. (Otherwise, by taking smaller and smaller intervals and $varepsilon$'s, we could produce a sequence converging to $c$ with $f$-image converging to $f(c)$.) By shrinking things a little, we can arrange for $a$, $b$, and $varepsilon$ to always be rational. So uncountably many $c$'s must have the same $a$, $b$, and $varepsilon$. Fix such $a$, $b$, and $varepsilon$. So you've got uncountably many $c$'s, all lying in $(a,b)$ and (therefore) all having their $f(c)$'s separated by a distance of at least $varepsilon$ --- a contradiction because there isn't that much room in $mathbb R$.
answered Aug 24 at 11:55
Andreas Blass
55.5k7130212
31
A very quick way to say essentially the same idea: Since $mathbb R^2$ is second countable, not every point in the graph of $f$ can be isolated.
– Monroe Eskew
Aug 24 at 18:21
1
what about the case : $(x_n)_n$ to $c$ and $fracf(x_n)-f(c)x_n-c$ converged ? @MonroeEskew
– Dattier
Aug 26 at 11:29
1
mathoverflow.net/questions/309158/…
– Dattier
Aug 26 at 12:42
add a comment |Â
up vote
48
down vote
This has already been answered satisfactorily, but let me mention that it also follows from the much stronger (and noteworthy) fact that for every $fcolonmathbbRtomathbbR$ there exists a dense $DsubseteqmathbbR$ such that $f|_D$ is continuous (Blumberg's theorem): see here for references. Given that this is true, take $cin D$ and $x_n$ injective converging toward $c$ (which exists since $D$ is dense), and we have $f(x_n)$ converging to $f(c)$ by continuity of $f|_D$.
answered Aug 24 at 12:55
Gro-Tsen
9,05323285
3
@Dattier Blumberg's theorem is not that $f$ is continuous on $D$, it is that $f|_D$ is continuous. The two are quite different. (For some reason, every person who hears about Blumberg's theorem for the first time seems to fall for the same confusion.)
– Gro-Tsen
Aug 24 at 13:07
4
The topology on $D$ is the induced topology. (And as for an example of $fcolon DtomathbbZ$ surjective continuous, consider $f(x)=lfloor x/sqrt2rfloor$ on $D=mathbbQ$.)
– Gro-Tsen
Aug 24 at 13:11
3
The floor function is continuous on $mathbb R setminus mathbb N$.
– Monroe Eskew
Aug 24 at 17:05
1
@Gro-Tsen: The function $xmapsto lfloor sqrt 2xrfloor$ defined on $mathbb Q$ is discontinuous at 0 for what it’s worth.
– Anthony Quas
Aug 24 at 18:04
1
@AnthonyQuas Oh, yeah, right. So remove $0$ from the domain or replace $lfloor x/sqrt2rfloor$ by $lfloor x-sqrt2rfloor$.
– Gro-Tsen
Aug 24 at 19:15
 |Â
show 3 more comments
up vote
29
down vote
It has "been known" since 1908 that for any such function, this holds for all but countably many real numbers $c$, even when we additionally require all the sequences to approach $c$ from the same side.
In May 1908 William Henry Young presented several results for general functions from $mathbb R$ to $mathbb R,$ including a result implying that, given any such function, all but countably many real numbers $c$ have the property you're asking about. These results (for more about them, see my answer here) may have been joint work with his wife, Grace Chisholm Young, and the results were published in the 1908 paper cited below.
Young showed that for co-countably many real numbers $c$ we have
$$f(c) in C^-(f,c) ;; textand ;; f(c) in C^+(f,c)$$
Definition: Given a function $f: mathbb R rightarrow mathbb R$ and $c in mathbb R$, we let $C^-(f,c)$ be the set of all extended real numbers $y$ (i.e. $y$ can be $-infty$ or $+infty$) for which there exists a sequence $leftx_kright$ such that for each $k$ we have $x_k < c,$ and we have $x_k rightarrow c$ and $f(x_k) rightarrow y.$ In other words, $C^-(f,c)$ is the set of all numbers (including $-infty$ and $+infty$) that can be obtained as a limit of $f$-values when using some sequence converging to $c$ from the left. The right version, $C^+(f,c),$ is defined analogously.
Incidentally, the requirement in this definition that each $x_k < c$ (and also each $x_k > c)$ allows you to find such sequences converging to $c$ that have infinitely many values.
William Henry Young, Sulle due funzioni a più valori costituite dai limiti d'una variabile reale a destra e a sinistra di ciascun punto [On the two functions of multiple values that are determined by the left and right limits of a real variable at each point], Atti della Accademia Reale dei Lincei. Rendiconti. Classe di Scienze fisiche, Matematiche e Naturali (5) 17 #9 (1st semestre) (1908), 582-587. [Paper given at session dated 3 May 1908.]
edited Aug 24 at 20:17
Qfwfq
9,683876162
answered Aug 24 at 17:50
Dave L Renfro
1,5271510
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
59
down vote
accepted
Suppose $f$ were a counterexample. Then, for any $c$, we could find a little interval $(a,b)$ containing $c$ and we could find some $varepsilon>0$ such that all points $xin(a,b)$ except $c$ have $|f(x)-f(c)|>varepsilon$. (Otherwise, by taking smaller and smaller intervals and $varepsilon$'s, we could produce a sequence converging to $c$ with $f$-image converging to $f(c)$.) By shrinking things a little, we can arrange for $a$, $b$, and $varepsilon$ to always be rational. So uncountably many $c$'s must have the same $a$, $b$, and $varepsilon$. Fix such $a$, $b$, and $varepsilon$. So you've got uncountably many $c$'s, all lying in $(a,b)$ and (therefore) all having their $f(c)$'s separated by a distance of at least $varepsilon$ --- a contradiction because there isn't that much room in $mathbb R$.
answered Aug 24 at 11:55
Andreas Blass
55.5k7130212
31
A very quick way to say essentially the same idea: Since $mathbb R^2$ is second countable, not every point in the graph of $f$ can be isolated.
– Monroe Eskew
Aug 24 at 18:21
1
what about the case : $(x_n)_n$ to $c$ and $fracf(x_n)-f(c)x_n-c$ converged ? @MonroeEskew
– Dattier
Aug 26 at 11:29
1
mathoverflow.net/questions/309158/…
– Dattier
Aug 26 at 12:42
add a comment |Â
up vote
59
down vote
accepted
Suppose $f$ were a counterexample. Then, for any $c$, we could find a little interval $(a,b)$ containing $c$ and we could find some $varepsilon>0$ such that all points $xin(a,b)$ except $c$ have $|f(x)-f(c)|>varepsilon$. (Otherwise, by taking smaller and smaller intervals and $varepsilon$'s, we could produce a sequence converging to $c$ with $f$-image converging to $f(c)$.) By shrinking things a little, we can arrange for $a$, $b$, and $varepsilon$ to always be rational. So uncountably many $c$'s must have the same $a$, $b$, and $varepsilon$. Fix such $a$, $b$, and $varepsilon$. So you've got uncountably many $c$'s, all lying in $(a,b)$ and (therefore) all having their $f(c)$'s separated by a distance of at least $varepsilon$ --- a contradiction because there isn't that much room in $mathbb R$.
answered Aug 24 at 11:55
Andreas Blass
55.5k7130212
31
A very quick way to say essentially the same idea: Since $mathbb R^2$ is second countable, not every point in the graph of $f$ can be isolated.
– Monroe Eskew
Aug 24 at 18:21
1
what about the case : $(x_n)_n$ to $c$ and $fracf(x_n)-f(c)x_n-c$ converged ? @MonroeEskew
– Dattier
Aug 26 at 11:29
1
mathoverflow.net/questions/309158/…
– Dattier
Aug 26 at 12:42
add a comment |Â
up vote
59
down vote
accepted
up vote
59
down vote
accepted
Suppose $f$ were a counterexample. Then, for any $c$, we could find a little interval $(a,b)$ containing $c$ and we could find some $varepsilon>0$ such that all points $xin(a,b)$ except $c$ have $|f(x)-f(c)|>varepsilon$. (Otherwise, by taking smaller and smaller intervals and $varepsilon$'s, we could produce a sequence converging to $c$ with $f$-image converging to $f(c)$.) By shrinking things a little, we can arrange for $a$, $b$, and $varepsilon$ to always be rational. So uncountably many $c$'s must have the same $a$, $b$, and $varepsilon$. Fix such $a$, $b$, and $varepsilon$. So you've got uncountably many $c$'s, all lying in $(a,b)$ and (therefore) all having their $f(c)$'s separated by a distance of at least $varepsilon$ --- a contradiction because there isn't that much room in $mathbb R$.
answered Aug 24 at 11:55
Andreas Blass
55.5k7130212
Suppose $f$ were a counterexample. Then, for any $c$, we could find a little interval $(a,b)$ containing $c$ and we could find some $varepsilon>0$ such that all points $xin(a,b)$ except $c$ have $|f(x)-f(c)|>varepsilon$. (Otherwise, by taking smaller and smaller intervals and $varepsilon$'s, we could produce a sequence converging to $c$ with $f$-image converging to $f(c)$.) By shrinking things a little, we can arrange for $a$, $b$, and $varepsilon$ to always be rational. So uncountably many $c$'s must have the same $a$, $b$, and $varepsilon$. Fix such $a$, $b$, and $varepsilon$. So you've got uncountably many $c$'s, all lying in $(a,b)$ and (therefore) all having their $f(c)$'s separated by a distance of at least $varepsilon$ --- a contradiction because there isn't that much room in $mathbb R$.
answered Aug 24 at 11:55
Andreas Blass
55.5k7130212
answered Aug 24 at 11:55
Andreas Blass
55.5k7130212
answered Aug 24 at 11:55
Andreas Blass
55.5k7130212
answered Aug 24 at 11:55
Andreas Blass
55.5k7130212
55.5k7130212
31
A very quick way to say essentially the same idea: Since $mathbb R^2$ is second countable, not every point in the graph of $f$ can be isolated.
– Monroe Eskew
Aug 24 at 18:21
1
what about the case : $(x_n)_n$ to $c$ and $fracf(x_n)-f(c)x_n-c$ converged ? @MonroeEskew
– Dattier
Aug 26 at 11:29
1
mathoverflow.net/questions/309158/…
– Dattier
Aug 26 at 12:42
add a comment |Â
31
A very quick way to say essentially the same idea: Since $mathbb R^2$ is second countable, not every point in the graph of $f$ can be isolated.
– Monroe Eskew
Aug 24 at 18:21
1
what about the case : $(x_n)_n$ to $c$ and $fracf(x_n)-f(c)x_n-c$ converged ? @MonroeEskew
– Dattier
Aug 26 at 11:29
1
mathoverflow.net/questions/309158/…
– Dattier
Aug 26 at 12:42
31
31
A very quick way to say essentially the same idea: Since $mathbb R^2$ is second countable, not every point in the graph of $f$ can be isolated.
– Monroe Eskew
Aug 24 at 18:21
A very quick way to say essentially the same idea: Since $mathbb R^2$ is second countable, not every point in the graph of $f$ can be isolated.
– Monroe Eskew
Aug 24 at 18:21
1
1
what about the case : $(x_n)_n$ to $c$ and $fracf(x_n)-f(c)x_n-c$ converged ? @MonroeEskew
– Dattier
Aug 26 at 11:29
what about the case : $(x_n)_n$ to $c$ and $fracf(x_n)-f(c)x_n-c$ converged ? @MonroeEskew
– Dattier
Aug 26 at 11:29
1
1
mathoverflow.net/questions/309158/…
– Dattier
Aug 26 at 12:42
mathoverflow.net/questions/309158/…
– Dattier
Aug 26 at 12:42
add a comment |Â
up vote
48
down vote
This has already been answered satisfactorily, but let me mention that it also follows from the much stronger (and noteworthy) fact that for every $fcolonmathbbRtomathbbR$ there exists a dense $DsubseteqmathbbR$ such that $f|_D$ is continuous (Blumberg's theorem): see here for references. Given that this is true, take $cin D$ and $x_n$ injective converging toward $c$ (which exists since $D$ is dense), and we have $f(x_n)$ converging to $f(c)$ by continuity of $f|_D$.
answered Aug 24 at 12:55
Gro-Tsen
9,05323285
3
@Dattier Blumberg's theorem is not that $f$ is continuous on $D$, it is that $f|_D$ is continuous. The two are quite different. (For some reason, every person who hears about Blumberg's theorem for the first time seems to fall for the same confusion.)
– Gro-Tsen
Aug 24 at 13:07
4
The topology on $D$ is the induced topology. (And as for an example of $fcolon DtomathbbZ$ surjective continuous, consider $f(x)=lfloor x/sqrt2rfloor$ on $D=mathbbQ$.)
– Gro-Tsen
Aug 24 at 13:11
3
The floor function is continuous on $mathbb R setminus mathbb N$.
– Monroe Eskew
Aug 24 at 17:05
1
@Gro-Tsen: The function $xmapsto lfloor sqrt 2xrfloor$ defined on $mathbb Q$ is discontinuous at 0 for what it’s worth.
– Anthony Quas
Aug 24 at 18:04
1
@AnthonyQuas Oh, yeah, right. So remove $0$ from the domain or replace $lfloor x/sqrt2rfloor$ by $lfloor x-sqrt2rfloor$.
– Gro-Tsen
Aug 24 at 19:15
 |Â
show 3 more comments
up vote
48
down vote
This has already been answered satisfactorily, but let me mention that it also follows from the much stronger (and noteworthy) fact that for every $fcolonmathbbRtomathbbR$ there exists a dense $DsubseteqmathbbR$ such that $f|_D$ is continuous (Blumberg's theorem): see here for references. Given that this is true, take $cin D$ and $x_n$ injective converging toward $c$ (which exists since $D$ is dense), and we have $f(x_n)$ converging to $f(c)$ by continuity of $f|_D$.
answered Aug 24 at 12:55
Gro-Tsen
9,05323285
3
@Dattier Blumberg's theorem is not that $f$ is continuous on $D$, it is that $f|_D$ is continuous. The two are quite different. (For some reason, every person who hears about Blumberg's theorem for the first time seems to fall for the same confusion.)
– Gro-Tsen
Aug 24 at 13:07
4
The topology on $D$ is the induced topology. (And as for an example of $fcolon DtomathbbZ$ surjective continuous, consider $f(x)=lfloor x/sqrt2rfloor$ on $D=mathbbQ$.)
– Gro-Tsen
Aug 24 at 13:11
3
The floor function is continuous on $mathbb R setminus mathbb N$.
– Monroe Eskew
Aug 24 at 17:05
1
@Gro-Tsen: The function $xmapsto lfloor sqrt 2xrfloor$ defined on $mathbb Q$ is discontinuous at 0 for what it’s worth.
– Anthony Quas
Aug 24 at 18:04
1
@AnthonyQuas Oh, yeah, right. So remove $0$ from the domain or replace $lfloor x/sqrt2rfloor$ by $lfloor x-sqrt2rfloor$.
– Gro-Tsen
Aug 24 at 19:15
 |Â
show 3 more comments
up vote
48
down vote
up vote
48
down vote
This has already been answered satisfactorily, but let me mention that it also follows from the much stronger (and noteworthy) fact that for every $fcolonmathbbRtomathbbR$ there exists a dense $DsubseteqmathbbR$ such that $f|_D$ is continuous (Blumberg's theorem): see here for references. Given that this is true, take $cin D$ and $x_n$ injective converging toward $c$ (which exists since $D$ is dense), and we have $f(x_n)$ converging to $f(c)$ by continuity of $f|_D$.
answered Aug 24 at 12:55
Gro-Tsen
9,05323285
This has already been answered satisfactorily, but let me mention that it also follows from the much stronger (and noteworthy) fact that for every $fcolonmathbbRtomathbbR$ there exists a dense $DsubseteqmathbbR$ such that $f|_D$ is continuous (Blumberg's theorem): see here for references. Given that this is true, take $cin D$ and $x_n$ injective converging toward $c$ (which exists since $D$ is dense), and we have $f(x_n)$ converging to $f(c)$ by continuity of $f|_D$.
answered Aug 24 at 12:55
Gro-Tsen
9,05323285
answered Aug 24 at 12:55
Gro-Tsen
9,05323285
answered Aug 24 at 12:55
Gro-Tsen
9,05323285
answered Aug 24 at 12:55
Gro-Tsen
9,05323285
9,05323285
3
@Dattier Blumberg's theorem is not that $f$ is continuous on $D$, it is that $f|_D$ is continuous. The two are quite different. (For some reason, every person who hears about Blumberg's theorem for the first time seems to fall for the same confusion.)
– Gro-Tsen
Aug 24 at 13:07
4
The topology on $D$ is the induced topology. (And as for an example of $fcolon DtomathbbZ$ surjective continuous, consider $f(x)=lfloor x/sqrt2rfloor$ on $D=mathbbQ$.)
– Gro-Tsen
Aug 24 at 13:11
3
The floor function is continuous on $mathbb R setminus mathbb N$.
– Monroe Eskew
Aug 24 at 17:05
1
@Gro-Tsen: The function $xmapsto lfloor sqrt 2xrfloor$ defined on $mathbb Q$ is discontinuous at 0 for what it’s worth.
– Anthony Quas
Aug 24 at 18:04
1
@AnthonyQuas Oh, yeah, right. So remove $0$ from the domain or replace $lfloor x/sqrt2rfloor$ by $lfloor x-sqrt2rfloor$.
– Gro-Tsen
Aug 24 at 19:15
 |Â
show 3 more comments
3
@Dattier Blumberg's theorem is not that $f$ is continuous on $D$, it is that $f|_D$ is continuous. The two are quite different. (For some reason, every person who hears about Blumberg's theorem for the first time seems to fall for the same confusion.)
– Gro-Tsen
Aug 24 at 13:07
4
The topology on $D$ is the induced topology. (And as for an example of $fcolon DtomathbbZ$ surjective continuous, consider $f(x)=lfloor x/sqrt2rfloor$ on $D=mathbbQ$.)
– Gro-Tsen
Aug 24 at 13:11
3
The floor function is continuous on $mathbb R setminus mathbb N$.
– Monroe Eskew
Aug 24 at 17:05
1
@Gro-Tsen: The function $xmapsto lfloor sqrt 2xrfloor$ defined on $mathbb Q$ is discontinuous at 0 for what it’s worth.
– Anthony Quas
Aug 24 at 18:04
1
@AnthonyQuas Oh, yeah, right. So remove $0$ from the domain or replace $lfloor x/sqrt2rfloor$ by $lfloor x-sqrt2rfloor$.
– Gro-Tsen
Aug 24 at 19:15
3
3
@Dattier Blumberg's theorem is not that $f$ is continuous on $D$, it is that $f|_D$ is continuous. The two are quite different. (For some reason, every person who hears about Blumberg's theorem for the first time seems to fall for the same confusion.)
– Gro-Tsen
Aug 24 at 13:07
@Dattier Blumberg's theorem is not that $f$ is continuous on $D$, it is that $f|_D$ is continuous. The two are quite different. (For some reason, every person who hears about Blumberg's theorem for the first time seems to fall for the same confusion.)
– Gro-Tsen
Aug 24 at 13:07
4
4
The topology on $D$ is the induced topology. (And as for an example of $fcolon DtomathbbZ$ surjective continuous, consider $f(x)=lfloor x/sqrt2rfloor$ on $D=mathbbQ$.)
– Gro-Tsen
Aug 24 at 13:11
The topology on $D$ is the induced topology. (And as for an example of $fcolon DtomathbbZ$ surjective continuous, consider $f(x)=lfloor x/sqrt2rfloor$ on $D=mathbbQ$.)
– Gro-Tsen
Aug 24 at 13:11
3
3
The floor function is continuous on $mathbb R setminus mathbb N$.
– Monroe Eskew
Aug 24 at 17:05
The floor function is continuous on $mathbb R setminus mathbb N$.
– Monroe Eskew
Aug 24 at 17:05
1
1
@Gro-Tsen: The function $xmapsto lfloor sqrt 2xrfloor$ defined on $mathbb Q$ is discontinuous at 0 for what it’s worth.
– Anthony Quas
Aug 24 at 18:04
@Gro-Tsen: The function $xmapsto lfloor sqrt 2xrfloor$ defined on $mathbb Q$ is discontinuous at 0 for what it’s worth.
– Anthony Quas
Aug 24 at 18:04
1
1
@AnthonyQuas Oh, yeah, right. So remove $0$ from the domain or replace $lfloor x/sqrt2rfloor$ by $lfloor x-sqrt2rfloor$.
– Gro-Tsen
Aug 24 at 19:15
@AnthonyQuas Oh, yeah, right. So remove $0$ from the domain or replace $lfloor x/sqrt2rfloor$ by $lfloor x-sqrt2rfloor$.
– Gro-Tsen
Aug 24 at 19:15
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up vote
29
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It has "been known" since 1908 that for any such function, this holds for all but countably many real numbers $c$, even when we additionally require all the sequences to approach $c$ from the same side.
In May 1908 William Henry Young presented several results for general functions from $mathbb R$ to $mathbb R,$ including a result implying that, given any such function, all but countably many real numbers $c$ have the property you're asking about. These results (for more about them, see my answer here) may have been joint work with his wife, Grace Chisholm Young, and the results were published in the 1908 paper cited below.
Young showed that for co-countably many real numbers $c$ we have
$$f(c) in C^-(f,c) ;; textand ;; f(c) in C^+(f,c)$$
Definition: Given a function $f: mathbb R rightarrow mathbb R$ and $c in mathbb R$, we let $C^-(f,c)$ be the set of all extended real numbers $y$ (i.e. $y$ can be $-infty$ or $+infty$) for which there exists a sequence $leftx_kright$ such that for each $k$ we have $x_k < c,$ and we have $x_k rightarrow c$ and $f(x_k) rightarrow y.$ In other words, $C^-(f,c)$ is the set of all numbers (including $-infty$ and $+infty$) that can be obtained as a limit of $f$-values when using some sequence converging to $c$ from the left. The right version, $C^+(f,c),$ is defined analogously.
Incidentally, the requirement in this definition that each $x_k < c$ (and also each $x_k > c)$ allows you to find such sequences converging to $c$ that have infinitely many values.
William Henry Young, Sulle due funzioni a più valori costituite dai limiti d'una variabile reale a destra e a sinistra di ciascun punto [On the two functions of multiple values that are determined by the left and right limits of a real variable at each point], Atti della Accademia Reale dei Lincei. Rendiconti. Classe di Scienze fisiche, Matematiche e Naturali (5) 17 #9 (1st semestre) (1908), 582-587. [Paper given at session dated 3 May 1908.]
edited Aug 24 at 20:17
Qfwfq
9,683876162
answered Aug 24 at 17:50
Dave L Renfro
1,5271510
add a comment |Â
up vote
29
down vote
It has "been known" since 1908 that for any such function, this holds for all but countably many real numbers $c$, even when we additionally require all the sequences to approach $c$ from the same side.
In May 1908 William Henry Young presented several results for general functions from $mathbb R$ to $mathbb R,$ including a result implying that, given any such function, all but countably many real numbers $c$ have the property you're asking about. These results (for more about them, see my answer here) may have been joint work with his wife, Grace Chisholm Young, and the results were published in the 1908 paper cited below.
Young showed that for co-countably many real numbers $c$ we have
$$f(c) in C^-(f,c) ;; textand ;; f(c) in C^+(f,c)$$
Definition: Given a function $f: mathbb R rightarrow mathbb R$ and $c in mathbb R$, we let $C^-(f,c)$ be the set of all extended real numbers $y$ (i.e. $y$ can be $-infty$ or $+infty$) for which there exists a sequence $leftx_kright$ such that for each $k$ we have $x_k < c,$ and we have $x_k rightarrow c$ and $f(x_k) rightarrow y.$ In other words, $C^-(f,c)$ is the set of all numbers (including $-infty$ and $+infty$) that can be obtained as a limit of $f$-values when using some sequence converging to $c$ from the left. The right version, $C^+(f,c),$ is defined analogously.
Incidentally, the requirement in this definition that each $x_k < c$ (and also each $x_k > c)$ allows you to find such sequences converging to $c$ that have infinitely many values.
William Henry Young, Sulle due funzioni a più valori costituite dai limiti d'una variabile reale a destra e a sinistra di ciascun punto [On the two functions of multiple values that are determined by the left and right limits of a real variable at each point], Atti della Accademia Reale dei Lincei. Rendiconti. Classe di Scienze fisiche, Matematiche e Naturali (5) 17 #9 (1st semestre) (1908), 582-587. [Paper given at session dated 3 May 1908.]
edited Aug 24 at 20:17
Qfwfq
9,683876162
answered Aug 24 at 17:50
Dave L Renfro
1,5271510
add a comment |Â
up vote
29
down vote
up vote
29
down vote
It has "been known" since 1908 that for any such function, this holds for all but countably many real numbers $c$, even when we additionally require all the sequences to approach $c$ from the same side.
In May 1908 William Henry Young presented several results for general functions from $mathbb R$ to $mathbb R,$ including a result implying that, given any such function, all but countably many real numbers $c$ have the property you're asking about. These results (for more about them, see my answer here) may have been joint work with his wife, Grace Chisholm Young, and the results were published in the 1908 paper cited below.
Young showed that for co-countably many real numbers $c$ we have
$$f(c) in C^-(f,c) ;; textand ;; f(c) in C^+(f,c)$$
Definition: Given a function $f: mathbb R rightarrow mathbb R$ and $c in mathbb R$, we let $C^-(f,c)$ be the set of all extended real numbers $y$ (i.e. $y$ can be $-infty$ or $+infty$) for which there exists a sequence $leftx_kright$ such that for each $k$ we have $x_k < c,$ and we have $x_k rightarrow c$ and $f(x_k) rightarrow y.$ In other words, $C^-(f,c)$ is the set of all numbers (including $-infty$ and $+infty$) that can be obtained as a limit of $f$-values when using some sequence converging to $c$ from the left. The right version, $C^+(f,c),$ is defined analogously.
Incidentally, the requirement in this definition that each $x_k < c$ (and also each $x_k > c)$ allows you to find such sequences converging to $c$ that have infinitely many values.
William Henry Young, Sulle due funzioni a più valori costituite dai limiti d'una variabile reale a destra e a sinistra di ciascun punto [On the two functions of multiple values that are determined by the left and right limits of a real variable at each point], Atti della Accademia Reale dei Lincei. Rendiconti. Classe di Scienze fisiche, Matematiche e Naturali (5) 17 #9 (1st semestre) (1908), 582-587. [Paper given at session dated 3 May 1908.]
edited Aug 24 at 20:17
Qfwfq
9,683876162
answered Aug 24 at 17:50
Dave L Renfro
1,5271510
It has "been known" since 1908 that for any such function, this holds for all but countably many real numbers $c$, even when we additionally require all the sequences to approach $c$ from the same side.
In May 1908 William Henry Young presented several results for general functions from $mathbb R$ to $mathbb R,$ including a result implying that, given any such function, all but countably many real numbers $c$ have the property you're asking about. These results (for more about them, see my answer here) may have been joint work with his wife, Grace Chisholm Young, and the results were published in the 1908 paper cited below.
Young showed that for co-countably many real numbers $c$ we have
$$f(c) in C^-(f,c) ;; textand ;; f(c) in C^+(f,c)$$
Definition: Given a function $f: mathbb R rightarrow mathbb R$ and $c in mathbb R$, we let $C^-(f,c)$ be the set of all extended real numbers $y$ (i.e. $y$ can be $-infty$ or $+infty$) for which there exists a sequence $leftx_kright$ such that for each $k$ we have $x_k < c,$ and we have $x_k rightarrow c$ and $f(x_k) rightarrow y.$ In other words, $C^-(f,c)$ is the set of all numbers (including $-infty$ and $+infty$) that can be obtained as a limit of $f$-values when using some sequence converging to $c$ from the left. The right version, $C^+(f,c),$ is defined analogously.
Incidentally, the requirement in this definition that each $x_k < c$ (and also each $x_k > c)$ allows you to find such sequences converging to $c$ that have infinitely many values.
William Henry Young, Sulle due funzioni a più valori costituite dai limiti d'una variabile reale a destra e a sinistra di ciascun punto [On the two functions of multiple values that are determined by the left and right limits of a real variable at each point], Atti della Accademia Reale dei Lincei. Rendiconti. Classe di Scienze fisiche, Matematiche e Naturali (5) 17 #9 (1st semestre) (1908), 582-587. [Paper given at session dated 3 May 1908.]
edited Aug 24 at 20:17
Qfwfq
9,683876162
answered Aug 24 at 17:50
Dave L Renfro
1,5271510
edited Aug 24 at 20:17
Qfwfq
9,683876162
edited Aug 24 at 20:17
Qfwfq
9,683876162
edited Aug 24 at 20:17
Qfwfq
9,683876162
9,683876162
answered Aug 24 at 17:50
Dave L Renfro
1,5271510
answered Aug 24 at 17:50
Dave L Renfro
1,5271510
answered Aug 24 at 17:50
Dave L Renfro
1,5271510
1,5271510
add a comment |Â
add a comment |Â
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1
The question looks interesting, because $f$ is not assumed to have any property (such as continuity). Perhaps prove first this. There is $a$ so that for every $epsilon > 0$, the set $<epsilon $ is uncountable.
– Gerald Edgar
Aug 24 at 10:31
1
The difficulty is to have, $f(x_n)$ converged to $f(a)$ and $x_n$ converged to $a$.
– Dattier
Aug 24 at 10:49
@GeraldEdgar : for your question, you can show that with compacity argument
– Dattier
Aug 24 at 10:52
11
I had to read it like four times before I convinced myself that this is not an immediately false triviality.
– Qfwfq
Aug 24 at 19:36