A problem in real analysis of a topological nature












5












$begingroup$


Let $f: R to R$ be a function such that the closure of its graph contains as a subset the graph of a uniformly continuous function. Does there exist a dense subset $S$ of $R$ such that the restricted function $f|S: S to R$ is uniformly continuous?






real-analysis







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$endgroup$








  • 1




    $begingroup$
    If $f|_S$ is uniformly continuous then it can be natuarlly extended to a uniformly continuous function on the closure of $S$. Hence $f$ itself has to be uniformly continuous. If you weaken the assumption to "there is a sequence of sets $S_n$ such that their union is dense in $R$ and f restricted to $S_n$ is uniformly continuous on $S_n$" then it's true for measurable functions by Lusin's Theorem.
    $endgroup$
    – Martin Kell
    Dec 26 '18 at 16:27






  • 1




    $begingroup$
    @MartinKell: It's true that $f|_S$ will extend to a uniformly continuous function (call it $g$), but $f$ need not agree with $g$ on $S^c$, so we cannot conclude that $f$ is uniformly continuous.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 16:44
















5












$begingroup$


Let $f: R to R$ be a function such that the closure of its graph contains as a subset the graph of a uniformly continuous function. Does there exist a dense subset $S$ of $R$ such that the restricted function $f|S: S to R$ is uniformly continuous?






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If $f|_S$ is uniformly continuous then it can be natuarlly extended to a uniformly continuous function on the closure of $S$. Hence $f$ itself has to be uniformly continuous. If you weaken the assumption to "there is a sequence of sets $S_n$ such that their union is dense in $R$ and f restricted to $S_n$ is uniformly continuous on $S_n$" then it's true for measurable functions by Lusin's Theorem.
    $endgroup$
    – Martin Kell
    Dec 26 '18 at 16:27






  • 1




    $begingroup$
    @MartinKell: It's true that $f|_S$ will extend to a uniformly continuous function (call it $g$), but $f$ need not agree with $g$ on $S^c$, so we cannot conclude that $f$ is uniformly continuous.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 16:44














5












5








5





$begingroup$


Let $f: R to R$ be a function such that the closure of its graph contains as a subset the graph of a uniformly continuous function. Does there exist a dense subset $S$ of $R$ such that the restricted function $f|S: S to R$ is uniformly continuous?






real-analysis







share|cite|improve this question









$endgroup$




Let $f: R to R$ be a function such that the closure of its graph contains as a subset the graph of a uniformly continuous function. Does there exist a dense subset $S$ of $R$ such that the restricted function $f|S: S to R$ is uniformly continuous?






real-analysis




real-analysis






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asked Dec 26 '18 at 14:58









James BaxterJames Baxter

32313




32313








  • 1




    $begingroup$
    If $f|_S$ is uniformly continuous then it can be natuarlly extended to a uniformly continuous function on the closure of $S$. Hence $f$ itself has to be uniformly continuous. If you weaken the assumption to "there is a sequence of sets $S_n$ such that their union is dense in $R$ and f restricted to $S_n$ is uniformly continuous on $S_n$" then it's true for measurable functions by Lusin's Theorem.
    $endgroup$
    – Martin Kell
    Dec 26 '18 at 16:27






  • 1




    $begingroup$
    @MartinKell: It's true that $f|_S$ will extend to a uniformly continuous function (call it $g$), but $f$ need not agree with $g$ on $S^c$, so we cannot conclude that $f$ is uniformly continuous.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 16:44














  • 1




    $begingroup$
    If $f|_S$ is uniformly continuous then it can be natuarlly extended to a uniformly continuous function on the closure of $S$. Hence $f$ itself has to be uniformly continuous. If you weaken the assumption to "there is a sequence of sets $S_n$ such that their union is dense in $R$ and f restricted to $S_n$ is uniformly continuous on $S_n$" then it's true for measurable functions by Lusin's Theorem.
    $endgroup$
    – Martin Kell
    Dec 26 '18 at 16:27






  • 1




    $begingroup$
    @MartinKell: It's true that $f|_S$ will extend to a uniformly continuous function (call it $g$), but $f$ need not agree with $g$ on $S^c$, so we cannot conclude that $f$ is uniformly continuous.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 16:44








1




1




$begingroup$
If $f|_S$ is uniformly continuous then it can be natuarlly extended to a uniformly continuous function on the closure of $S$. Hence $f$ itself has to be uniformly continuous. If you weaken the assumption to "there is a sequence of sets $S_n$ such that their union is dense in $R$ and f restricted to $S_n$ is uniformly continuous on $S_n$" then it's true for measurable functions by Lusin's Theorem.
$endgroup$
– Martin Kell
Dec 26 '18 at 16:27




$begingroup$
If $f|_S$ is uniformly continuous then it can be natuarlly extended to a uniformly continuous function on the closure of $S$. Hence $f$ itself has to be uniformly continuous. If you weaken the assumption to "there is a sequence of sets $S_n$ such that their union is dense in $R$ and f restricted to $S_n$ is uniformly continuous on $S_n$" then it's true for measurable functions by Lusin's Theorem.
$endgroup$
– Martin Kell
Dec 26 '18 at 16:27




1




1




$begingroup$
@MartinKell: It's true that $f|_S$ will extend to a uniformly continuous function (call it $g$), but $f$ need not agree with $g$ on $S^c$, so we cannot conclude that $f$ is uniformly continuous.
$endgroup$
– Nate Eldredge
Dec 26 '18 at 16:44




$begingroup$
@MartinKell: It's true that $f|_S$ will extend to a uniformly continuous function (call it $g$), but $f$ need not agree with $g$ on $S^c$, so we cannot conclude that $f$ is uniformly continuous.
$endgroup$
– Nate Eldredge
Dec 26 '18 at 16:44










1 Answer
1






active

oldest

votes


















7












$begingroup$

Consider the following modification of the Dirichlet "popcorn" function:
$$f(x) = begin{cases} 1/q, & text{$x in mathbb{Q}$, $x=p/q$ in lowest terms} \
-1, & x notin mathbb{Q},, x < 0 \
-2, & x notin mathbb{Q}, , x > 0.end{cases}$$
Since every real number can be approximated by rationals with arbitrarily large denominator, the closure of the graph of $f$ contains the $x$-axis, which is the uniformly continuous function $0$.



Let $S subset mathbb{R}$ be dense. If $f|_S$ is uniformly continuous, then it extends to a unique uniformly continuous function $g$ on all of $mathbb{R}$, and we have $f=g$ on $S$.



If $S$ contains a negative irrational number $x$, then $g(x) = f(x) = -1$. Let $y$ be any positive number in $S$. If $y$ is rational, we have $g(y) = f(y) = 0$. Then by the continuity of $g$, there would have to be some $z in S$ with $f(z) = g(z) in (-3/4, -1/4)$ which is impossible. If $y$ is irrational, we get a similar contradiction since $g(y) = f(y) = -2$. So $S$ does not contain any negative irrational. Similarly, $S$ does not contain any positive irrational.



So we must have $S subset mathbb{Q}$. But this is similarly impossible. The rationals in $S$ cannot all have the same denominator (in lowest terms), so let $x_1 = p_1/q_1, x_2 = p_2/q_2 in S$, where $q_1 < q_2$. Then by the continuity of $g$ there must be some $y in S$ with $f(y) = g(y) in (frac{1}{q_1}, frac{1}{q_1+1})$, but $f$ never takes on any such value.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f|_S$ for any dense $S$. Thus $Ssubseteqmathbb Rsetminusmathbb Q$, but then we have problems around $0$.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:08








  • 2




    $begingroup$
    @Wojowu: It's not quite that simple, I think. For instance, the function $f=1_mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 17:16












  • $begingroup$
    You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:24










  • $begingroup$
    @მამუკაჯიბლაძე: Yes, thank you. Fixed.
    $endgroup$
    – Nate Eldredge
    Dec 27 '18 at 15:49











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1 Answer
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7












$begingroup$

Consider the following modification of the Dirichlet "popcorn" function:
$$f(x) = begin{cases} 1/q, & text{$x in mathbb{Q}$, $x=p/q$ in lowest terms} \
-1, & x notin mathbb{Q},, x < 0 \
-2, & x notin mathbb{Q}, , x > 0.end{cases}$$
Since every real number can be approximated by rationals with arbitrarily large denominator, the closure of the graph of $f$ contains the $x$-axis, which is the uniformly continuous function $0$.



Let $S subset mathbb{R}$ be dense. If $f|_S$ is uniformly continuous, then it extends to a unique uniformly continuous function $g$ on all of $mathbb{R}$, and we have $f=g$ on $S$.



If $S$ contains a negative irrational number $x$, then $g(x) = f(x) = -1$. Let $y$ be any positive number in $S$. If $y$ is rational, we have $g(y) = f(y) = 0$. Then by the continuity of $g$, there would have to be some $z in S$ with $f(z) = g(z) in (-3/4, -1/4)$ which is impossible. If $y$ is irrational, we get a similar contradiction since $g(y) = f(y) = -2$. So $S$ does not contain any negative irrational. Similarly, $S$ does not contain any positive irrational.



So we must have $S subset mathbb{Q}$. But this is similarly impossible. The rationals in $S$ cannot all have the same denominator (in lowest terms), so let $x_1 = p_1/q_1, x_2 = p_2/q_2 in S$, where $q_1 < q_2$. Then by the continuity of $g$ there must be some $y in S$ with $f(y) = g(y) in (frac{1}{q_1}, frac{1}{q_1+1})$, but $f$ never takes on any such value.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f|_S$ for any dense $S$. Thus $Ssubseteqmathbb Rsetminusmathbb Q$, but then we have problems around $0$.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:08








  • 2




    $begingroup$
    @Wojowu: It's not quite that simple, I think. For instance, the function $f=1_mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 17:16












  • $begingroup$
    You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:24










  • $begingroup$
    @მამუკაჯიბლაძე: Yes, thank you. Fixed.
    $endgroup$
    – Nate Eldredge
    Dec 27 '18 at 15:49
















7












$begingroup$

Consider the following modification of the Dirichlet "popcorn" function:
$$f(x) = begin{cases} 1/q, & text{$x in mathbb{Q}$, $x=p/q$ in lowest terms} \
-1, & x notin mathbb{Q},, x < 0 \
-2, & x notin mathbb{Q}, , x > 0.end{cases}$$
Since every real number can be approximated by rationals with arbitrarily large denominator, the closure of the graph of $f$ contains the $x$-axis, which is the uniformly continuous function $0$.



Let $S subset mathbb{R}$ be dense. If $f|_S$ is uniformly continuous, then it extends to a unique uniformly continuous function $g$ on all of $mathbb{R}$, and we have $f=g$ on $S$.



If $S$ contains a negative irrational number $x$, then $g(x) = f(x) = -1$. Let $y$ be any positive number in $S$. If $y$ is rational, we have $g(y) = f(y) = 0$. Then by the continuity of $g$, there would have to be some $z in S$ with $f(z) = g(z) in (-3/4, -1/4)$ which is impossible. If $y$ is irrational, we get a similar contradiction since $g(y) = f(y) = -2$. So $S$ does not contain any negative irrational. Similarly, $S$ does not contain any positive irrational.



So we must have $S subset mathbb{Q}$. But this is similarly impossible. The rationals in $S$ cannot all have the same denominator (in lowest terms), so let $x_1 = p_1/q_1, x_2 = p_2/q_2 in S$, where $q_1 < q_2$. Then by the continuity of $g$ there must be some $y in S$ with $f(y) = g(y) in (frac{1}{q_1}, frac{1}{q_1+1})$, but $f$ never takes on any such value.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f|_S$ for any dense $S$. Thus $Ssubseteqmathbb Rsetminusmathbb Q$, but then we have problems around $0$.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:08








  • 2




    $begingroup$
    @Wojowu: It's not quite that simple, I think. For instance, the function $f=1_mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 17:16












  • $begingroup$
    You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:24










  • $begingroup$
    @მამუკაჯიბლაძე: Yes, thank you. Fixed.
    $endgroup$
    – Nate Eldredge
    Dec 27 '18 at 15:49














7












7








7





$begingroup$

Consider the following modification of the Dirichlet "popcorn" function:
$$f(x) = begin{cases} 1/q, & text{$x in mathbb{Q}$, $x=p/q$ in lowest terms} \
-1, & x notin mathbb{Q},, x < 0 \
-2, & x notin mathbb{Q}, , x > 0.end{cases}$$
Since every real number can be approximated by rationals with arbitrarily large denominator, the closure of the graph of $f$ contains the $x$-axis, which is the uniformly continuous function $0$.



Let $S subset mathbb{R}$ be dense. If $f|_S$ is uniformly continuous, then it extends to a unique uniformly continuous function $g$ on all of $mathbb{R}$, and we have $f=g$ on $S$.



If $S$ contains a negative irrational number $x$, then $g(x) = f(x) = -1$. Let $y$ be any positive number in $S$. If $y$ is rational, we have $g(y) = f(y) = 0$. Then by the continuity of $g$, there would have to be some $z in S$ with $f(z) = g(z) in (-3/4, -1/4)$ which is impossible. If $y$ is irrational, we get a similar contradiction since $g(y) = f(y) = -2$. So $S$ does not contain any negative irrational. Similarly, $S$ does not contain any positive irrational.



So we must have $S subset mathbb{Q}$. But this is similarly impossible. The rationals in $S$ cannot all have the same denominator (in lowest terms), so let $x_1 = p_1/q_1, x_2 = p_2/q_2 in S$, where $q_1 < q_2$. Then by the continuity of $g$ there must be some $y in S$ with $f(y) = g(y) in (frac{1}{q_1}, frac{1}{q_1+1})$, but $f$ never takes on any such value.






share|cite|improve this answer











$endgroup$



Consider the following modification of the Dirichlet "popcorn" function:
$$f(x) = begin{cases} 1/q, & text{$x in mathbb{Q}$, $x=p/q$ in lowest terms} \
-1, & x notin mathbb{Q},, x < 0 \
-2, & x notin mathbb{Q}, , x > 0.end{cases}$$
Since every real number can be approximated by rationals with arbitrarily large denominator, the closure of the graph of $f$ contains the $x$-axis, which is the uniformly continuous function $0$.



Let $S subset mathbb{R}$ be dense. If $f|_S$ is uniformly continuous, then it extends to a unique uniformly continuous function $g$ on all of $mathbb{R}$, and we have $f=g$ on $S$.



If $S$ contains a negative irrational number $x$, then $g(x) = f(x) = -1$. Let $y$ be any positive number in $S$. If $y$ is rational, we have $g(y) = f(y) = 0$. Then by the continuity of $g$, there would have to be some $z in S$ with $f(z) = g(z) in (-3/4, -1/4)$ which is impossible. If $y$ is irrational, we get a similar contradiction since $g(y) = f(y) = -2$. So $S$ does not contain any negative irrational. Similarly, $S$ does not contain any positive irrational.



So we must have $S subset mathbb{Q}$. But this is similarly impossible. The rationals in $S$ cannot all have the same denominator (in lowest terms), so let $x_1 = p_1/q_1, x_2 = p_2/q_2 in S$, where $q_1 < q_2$. Then by the continuity of $g$ there must be some $y in S$ with $f(y) = g(y) in (frac{1}{q_1}, frac{1}{q_1+1})$, but $f$ never takes on any such value.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 27 '18 at 15:49

























answered Dec 26 '18 at 17:01









Nate EldredgeNate Eldredge

19.9k368114




19.9k368114












  • $begingroup$
    Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f|_S$ for any dense $S$. Thus $Ssubseteqmathbb Rsetminusmathbb Q$, but then we have problems around $0$.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:08








  • 2




    $begingroup$
    @Wojowu: It's not quite that simple, I think. For instance, the function $f=1_mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 17:16












  • $begingroup$
    You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:24










  • $begingroup$
    @მამუკაჯიბლაძე: Yes, thank you. Fixed.
    $endgroup$
    – Nate Eldredge
    Dec 27 '18 at 15:49


















  • $begingroup$
    Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f|_S$ for any dense $S$. Thus $Ssubseteqmathbb Rsetminusmathbb Q$, but then we have problems around $0$.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:08








  • 2




    $begingroup$
    @Wojowu: It's not quite that simple, I think. For instance, the function $f=1_mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational.
    $endgroup$
    – Nate Eldredge
    Dec 26 '18 at 17:16












  • $begingroup$
    You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph.
    $endgroup$
    – Wojowu
    Dec 26 '18 at 17:24










  • $begingroup$
    @მამუკაჯიბლაძე: Yes, thank you. Fixed.
    $endgroup$
    – Nate Eldredge
    Dec 27 '18 at 15:49
















$begingroup$
Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f|_S$ for any dense $S$. Thus $Ssubseteqmathbb Rsetminusmathbb Q$, but then we have problems around $0$.
$endgroup$
– Wojowu
Dec 26 '18 at 17:08






$begingroup$
Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f|_S$ for any dense $S$. Thus $Ssubseteqmathbb Rsetminusmathbb Q$, but then we have problems around $0$.
$endgroup$
– Wojowu
Dec 26 '18 at 17:08






2




2




$begingroup$
@Wojowu: It's not quite that simple, I think. For instance, the function $f=1_mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational.
$endgroup$
– Nate Eldredge
Dec 26 '18 at 17:16






$begingroup$
@Wojowu: It's not quite that simple, I think. For instance, the function $f=1_mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational.
$endgroup$
– Nate Eldredge
Dec 26 '18 at 17:16














$begingroup$
You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph.
$endgroup$
– Wojowu
Dec 26 '18 at 17:24




$begingroup$
You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph.
$endgroup$
– Wojowu
Dec 26 '18 at 17:24












$begingroup$
@მამუკაჯიბლაძე: Yes, thank you. Fixed.
$endgroup$
– Nate Eldredge
Dec 27 '18 at 15:49




$begingroup$
@მამუკაჯიბლაძე: Yes, thank you. Fixed.
$endgroup$
– Nate Eldredge
Dec 27 '18 at 15:49


















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