A problem in real analysis of a topological nature












3














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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  • 1




    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.
    – Martin Kell
    6 hours ago






  • 1




    @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.
    – Nate Eldredge
    6 hours ago










  • @Nate: true, the conclusion was too quick. However it shows that a dense subset where f is uniformly continuous is rather restrictive. A counterexample is just sin(1/x) with arbitrary value at zero: 1st for all but the zero value the function is continuous so it’s graph closed at (x,f(x)), 2nd for no value at 0 is the function continuous. Especially it’s not uniformly continuous. Btw. usng non-compactness is $mathbb{R}$ one may contruct even continuous counterexamples.
    – Martin Kell
    6 hours ago
















3














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


















  • 1




    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.
    – Martin Kell
    6 hours ago






  • 1




    @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.
    – Nate Eldredge
    6 hours ago










  • @Nate: true, the conclusion was too quick. However it shows that a dense subset where f is uniformly continuous is rather restrictive. A counterexample is just sin(1/x) with arbitrary value at zero: 1st for all but the zero value the function is continuous so it’s graph closed at (x,f(x)), 2nd for no value at 0 is the function continuous. Especially it’s not uniformly continuous. Btw. usng non-compactness is $mathbb{R}$ one may contruct even continuous counterexamples.
    – Martin Kell
    6 hours ago














3












3








3







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













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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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 8 hours ago









James Baxter

1668




1668








  • 1




    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.
    – Martin Kell
    6 hours ago






  • 1




    @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.
    – Nate Eldredge
    6 hours ago










  • @Nate: true, the conclusion was too quick. However it shows that a dense subset where f is uniformly continuous is rather restrictive. A counterexample is just sin(1/x) with arbitrary value at zero: 1st for all but the zero value the function is continuous so it’s graph closed at (x,f(x)), 2nd for no value at 0 is the function continuous. Especially it’s not uniformly continuous. Btw. usng non-compactness is $mathbb{R}$ one may contruct even continuous counterexamples.
    – Martin Kell
    6 hours ago














  • 1




    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.
    – Martin Kell
    6 hours ago






  • 1




    @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.
    – Nate Eldredge
    6 hours ago










  • @Nate: true, the conclusion was too quick. However it shows that a dense subset where f is uniformly continuous is rather restrictive. A counterexample is just sin(1/x) with arbitrary value at zero: 1st for all but the zero value the function is continuous so it’s graph closed at (x,f(x)), 2nd for no value at 0 is the function continuous. Especially it’s not uniformly continuous. Btw. usng non-compactness is $mathbb{R}$ one may contruct even continuous counterexamples.
    – Martin Kell
    6 hours ago








1




1




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.
– Martin Kell
6 hours ago




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.
– Martin Kell
6 hours ago




1




1




@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.
– Nate Eldredge
6 hours ago




@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.
– Nate Eldredge
6 hours ago












@Nate: true, the conclusion was too quick. However it shows that a dense subset where f is uniformly continuous is rather restrictive. A counterexample is just sin(1/x) with arbitrary value at zero: 1st for all but the zero value the function is continuous so it’s graph closed at (x,f(x)), 2nd for no value at 0 is the function continuous. Especially it’s not uniformly continuous. Btw. usng non-compactness is $mathbb{R}$ one may contruct even continuous counterexamples.
– Martin Kell
6 hours ago




@Nate: true, the conclusion was too quick. However it shows that a dense subset where f is uniformly continuous is rather restrictive. A counterexample is just sin(1/x) with arbitrary value at zero: 1st for all but the zero value the function is continuous so it’s graph closed at (x,f(x)), 2nd for no value at 0 is the function continuous. Especially it’s not uniformly continuous. Btw. usng non-compactness is $mathbb{R}$ one may contruct even continuous counterexamples.
– Martin Kell
6 hours ago










1 Answer
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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 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





















  • 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 $0$.
    – Wojowu
    5 hours ago












  • @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.
    – Nate Eldredge
    5 hours ago












  • 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.
    – Wojowu
    5 hours ago











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3














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 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





















  • 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 $0$.
    – Wojowu
    5 hours ago












  • @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.
    – Nate Eldredge
    5 hours ago












  • 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.
    – Wojowu
    5 hours ago
















3














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 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





















  • 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 $0$.
    – Wojowu
    5 hours ago












  • @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.
    – Nate Eldredge
    5 hours ago












  • 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.
    – Wojowu
    5 hours ago














3












3








3






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 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












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 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










answered 6 hours ago









Nate Eldredge

19.6k365109




19.6k365109












  • 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 $0$.
    – Wojowu
    5 hours ago












  • @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.
    – Nate Eldredge
    5 hours ago












  • 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.
    – Wojowu
    5 hours ago


















  • 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 $0$.
    – Wojowu
    5 hours ago












  • @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.
    – Nate Eldredge
    5 hours ago












  • 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.
    – Wojowu
    5 hours ago
















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 $0$.
– Wojowu
5 hours ago






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 $0$.
– Wojowu
5 hours ago














@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.
– Nate Eldredge
5 hours ago






@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.
– Nate Eldredge
5 hours ago














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.
– Wojowu
5 hours ago




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.
– Wojowu
5 hours ago


















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Trompette piccolo

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Ne doit pas être confondu avec trompette de poche. Trompette piccolo en si bémol, avec des branches d'embouchure différentes pour le si bémol (la plus courte) ou la (la plus longue) Trompette piccolo La trompette piccolo est la plus petite de la famille des trompettes. Elle joue une octave plus haut que la trompette standard en si bémol. La plupart des trompettes piccolo sont conçues pour jouer dans les deux tonalités de si bémol ou la, en utilisant une branche d'embouchure différente pour chaque tonalité. Le tube de la trompette piccolo en si bémol a une longueur moitié de celle de la trompette normale en si bémol. Des trompettes piccolo en sol, fa, ut sont également fabriquées, mais elles sont rares. La trompette soprano en ré, aussi connu comme la « trompette Bach », a été inventée vers 1890 par le facteur belge Victor-Charles Mahillon pour jouer les parties hautes de la trompette dans la musique de Bach et Haendel. La trompette piccolo moderne permet aux
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Slow SSRS Report in dynamic grouping and multiple parameters

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0 I have developed a report which has 5 dynamic row groups from where users can choose in parameter selection. The report has 12 parameters so that user can get required information as per their parameter choice. Each parameter has multi value selection option with numerous lists of values (example: Customer parameter has 35,000 values and Item parameter has 2,500 values). Now, the report takes huge time to load parameter values as well as slow in preview (although, server resource is adequate). I am looking for a solution to improve the parameter selection performance without reducing the number of parameter and dynamic grouping sql-server reporting-services ssrs-2012 share | improve this question
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Simon Yates (cyclisme)

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Pour les articles homonymes, voir Simon Yates et Yates. Simon Yates.mw-parser-output .entete.cyclisme{background-image:url("//upload.wikimedia.org/wikipedia/commons/thumb/8/86/Cycling_%28road%29_pictogram.svg/45px-Cycling_%28road%29_pictogram.svg.png")} Simon Yates lors du Tour d'Alberta 2014 Informations Naissance 7 août 1992 (26 ans) Bury Nationalité Britannique Équipe actuelle Mitchelton-Scott Spécialités Grimpeur, cyclisme sur piste Équipes amateurs 2013 100% Me Équipes professionnelles 01.2014-06.2016 [ n 1 ] Orica-GreenEDGE 07.2016-12.2016 [ n 2 ] Orica-BikeExchange 2017 Orica-Scott 2018- Mitchelton-Scott Principales victoires 1 classement mondial UCI World Tour 2018 1 titre mondial Champion du monde de course aux points (2013) 1 grand tour Tour d'Espagne 2018 2 classements annexes de grands tours Classement du meilleur jeune Tour de France 2017 Classement du combiné Tour d'
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