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Prove continuity on a function at every irrational point and discontinuity at every rational point.
Clash Royale CLAN TAG#URR8PPP
up vote
12
down vote
favorite
Consider the function:
$f(x)= begincases
1/n quad &textif $x= m/n$ in simplest form \
0 quad &textif $x in mathbbRsetminusmathbbQ$
endcases $
Prove that the function is continuous at every irrational point and also that the function is not continuous at every rational point. Also, we can say that the function is continuous at some point $k$ if $displaystylelim_x to k f(x)=f(k)$.
I was thinking of doing an epsilon delta proof backwards using the fact that $mathbbQ$ is dense in $mathbbR$ for rational points and irrational points. Any ways on how to expand on this are welcome.
real-analysis
asked Oct 4 '12 at 11:54
Casquibaldo
3721720
add a comment |Â
up vote
12
down vote
favorite
Consider the function:
$f(x)= begincases
1/n quad &textif $x= m/n$ in simplest form \
0 quad &textif $x in mathbbRsetminusmathbbQ$
endcases $
Prove that the function is continuous at every irrational point and also that the function is not continuous at every rational point. Also, we can say that the function is continuous at some point $k$ if $displaystylelim_x to k f(x)=f(k)$.
I was thinking of doing an epsilon delta proof backwards using the fact that $mathbbQ$ is dense in $mathbbR$ for rational points and irrational points. Any ways on how to expand on this are welcome.
real-analysis
asked Oct 4 '12 at 11:54
Casquibaldo
3721720
2
The way to do this is to show that in arbitrarily small neighborhoods of an irrational point $x$, rationals approximating $x$ must have large denominators in reduced form.
– Seth
Oct 4 '12 at 12:11
1
Hint: $|frac ab-frac xy|=|fracay-bxby|ge frac1by$ if $frac abnefrac xy$
– Hagen von Eitzen
Oct 4 '12 at 12:31
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
Consider the function:
$f(x)= begincases
1/n quad &textif $x= m/n$ in simplest form \
0 quad &textif $x in mathbbRsetminusmathbbQ$
endcases $
Prove that the function is continuous at every irrational point and also that the function is not continuous at every rational point. Also, we can say that the function is continuous at some point $k$ if $displaystylelim_x to k f(x)=f(k)$.
I was thinking of doing an epsilon delta proof backwards using the fact that $mathbbQ$ is dense in $mathbbR$ for rational points and irrational points. Any ways on how to expand on this are welcome.
real-analysis
asked Oct 4 '12 at 11:54
Casquibaldo
3721720
Consider the function:
$f(x)= begincases
1/n quad &textif $x= m/n$ in simplest form \
0 quad &textif $x in mathbbRsetminusmathbbQ$
endcases $
Prove that the function is continuous at every irrational point and also that the function is not continuous at every rational point. Also, we can say that the function is continuous at some point $k$ if $displaystylelim_x to k f(x)=f(k)$.
I was thinking of doing an epsilon delta proof backwards using the fact that $mathbbQ$ is dense in $mathbbR$ for rational points and irrational points. Any ways on how to expand on this are welcome.
real-analysis
asked Oct 4 '12 at 11:54
Casquibaldo
3721720
asked Oct 4 '12 at 11:54
Casquibaldo
3721720
asked Oct 4 '12 at 11:54
Casquibaldo
3721720
asked Oct 4 '12 at 11:54
Casquibaldo
3721720
3721720
2
The way to do this is to show that in arbitrarily small neighborhoods of an irrational point $x$, rationals approximating $x$ must have large denominators in reduced form.
– Seth
Oct 4 '12 at 12:11
1
Hint: $|frac ab-frac xy|=|fracay-bxby|ge frac1by$ if $frac abnefrac xy$
– Hagen von Eitzen
Oct 4 '12 at 12:31
add a comment |Â
2
The way to do this is to show that in arbitrarily small neighborhoods of an irrational point $x$, rationals approximating $x$ must have large denominators in reduced form.
– Seth
Oct 4 '12 at 12:11
1
Hint: $|frac ab-frac xy|=|fracay-bxby|ge frac1by$ if $frac abnefrac xy$
– Hagen von Eitzen
Oct 4 '12 at 12:31
2
2
The way to do this is to show that in arbitrarily small neighborhoods of an irrational point $x$, rationals approximating $x$ must have large denominators in reduced form.
– Seth
Oct 4 '12 at 12:11
The way to do this is to show that in arbitrarily small neighborhoods of an irrational point $x$, rationals approximating $x$ must have large denominators in reduced form.
– Seth
Oct 4 '12 at 12:11
1
1
Hint: $|frac ab-frac xy|=|fracay-bxby|ge frac1by$ if $frac abnefrac xy$
– Hagen von Eitzen
Oct 4 '12 at 12:31
Hint: $|frac ab-frac xy|=|fracay-bxby|ge frac1by$ if $frac abnefrac xy$
– Hagen von Eitzen
Oct 4 '12 at 12:31
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
You cannot do this using simple rules about limits, like de l?Hôpital's rule or the continuity of "analytic expressions", etc. You have to go back to the actual definition of continuity when solving this problem. Here are a few hints:
In the first place you want to prove different things about a given $xinmathbb R$, depending on whether $x$ is rational or irrational.
If $x$ is rational then $f(x)$ is a certain positive number. It is claimed that $f$ is not continuous at $x$. Assume to the contrary that $f$ is continuous at $x$. Then there is a neighborhood $U:=U_delta(x)$ such that $f(t)>f(x)over2$ for all $tin U$. Hm?
If $x$ is irrational then $f(x)=0$. It is claimed that $f$ is continuous at $x$. This means that for any $epsilon>0$ we should be able to produce a neighborhood $U:=U_delta(x)$ such that $f(t)<epsilon$ for all $tin U$. Which points $tinmathbb R$ could violate the condition $f(t)<epsilon $?
answered Oct 4 '12 at 12:52
Christian Blatter
164k7107306
Are they going to be the rationals?
– Casquibaldo
Oct 4 '12 at 13:27
@Casquibaldo: If you mean by "they" the $t$ in the last line of my answer: Yes, but only selected rationals.
– Christian Blatter
Oct 4 '12 at 13:31
add a comment |Â
up vote
5
down vote
This function is known as Thomae's function and is an excellent example of a function that is continuous at uncountably infinite number of points while discontinuous at countably infinite number of points. Note that the reverse result is not possible, i.e. one cannot have a function that is discontinuous on an uncountably infinite dense set and continuous on a countably infinite dense set.
As for the proof , the discontinuity part follows from the fact that irrationals are dense in $ R $ as you rightly thought. The continuity part is a little bit more involved and it would help to employ the Archimedean principle in tandem with the $ epsilon$-$delta $ definition.
Crudely, look at $ x=sqrt3 $ , see that using the non-terminating decimal expansion i.e. $ sqrt 3=1.732148ldots $ , one can forge a sequence of rationals $ 1 , 17/10 ,173/100 ,ldots $
that converges to $ x $, while at the same time your $ f(x_n) $ is $ 1/10,1/100,1/1000,ldots $ which converges to $ 0 $ which is nothing but $ f(x) $. This is merely an illustration of the continuity and not a rigorous proof.
edited Oct 30 '16 at 5:43
Michael Hardy
204k23186462
answered Oct 4 '12 at 16:12
Vishesh
1,4661126
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
You cannot do this using simple rules about limits, like de l?Hôpital's rule or the continuity of "analytic expressions", etc. You have to go back to the actual definition of continuity when solving this problem. Here are a few hints:
In the first place you want to prove different things about a given $xinmathbb R$, depending on whether $x$ is rational or irrational.
If $x$ is rational then $f(x)$ is a certain positive number. It is claimed that $f$ is not continuous at $x$. Assume to the contrary that $f$ is continuous at $x$. Then there is a neighborhood $U:=U_delta(x)$ such that $f(t)>f(x)over2$ for all $tin U$. Hm?
If $x$ is irrational then $f(x)=0$. It is claimed that $f$ is continuous at $x$. This means that for any $epsilon>0$ we should be able to produce a neighborhood $U:=U_delta(x)$ such that $f(t)<epsilon$ for all $tin U$. Which points $tinmathbb R$ could violate the condition $f(t)<epsilon $?
answered Oct 4 '12 at 12:52
Christian Blatter
164k7107306
Are they going to be the rationals?
– Casquibaldo
Oct 4 '12 at 13:27
@Casquibaldo: If you mean by "they" the $t$ in the last line of my answer: Yes, but only selected rationals.
– Christian Blatter
Oct 4 '12 at 13:31
add a comment |Â
up vote
7
down vote
You cannot do this using simple rules about limits, like de l?Hôpital's rule or the continuity of "analytic expressions", etc. You have to go back to the actual definition of continuity when solving this problem. Here are a few hints:
In the first place you want to prove different things about a given $xinmathbb R$, depending on whether $x$ is rational or irrational.
If $x$ is rational then $f(x)$ is a certain positive number. It is claimed that $f$ is not continuous at $x$. Assume to the contrary that $f$ is continuous at $x$. Then there is a neighborhood $U:=U_delta(x)$ such that $f(t)>f(x)over2$ for all $tin U$. Hm?
If $x$ is irrational then $f(x)=0$. It is claimed that $f$ is continuous at $x$. This means that for any $epsilon>0$ we should be able to produce a neighborhood $U:=U_delta(x)$ such that $f(t)<epsilon$ for all $tin U$. Which points $tinmathbb R$ could violate the condition $f(t)<epsilon $?
answered Oct 4 '12 at 12:52
Christian Blatter
164k7107306
Are they going to be the rationals?
– Casquibaldo
Oct 4 '12 at 13:27
@Casquibaldo: If you mean by "they" the $t$ in the last line of my answer: Yes, but only selected rationals.
– Christian Blatter
Oct 4 '12 at 13:31
add a comment |Â
up vote
7
down vote
up vote
7
down vote
You cannot do this using simple rules about limits, like de l?Hôpital's rule or the continuity of "analytic expressions", etc. You have to go back to the actual definition of continuity when solving this problem. Here are a few hints:
In the first place you want to prove different things about a given $xinmathbb R$, depending on whether $x$ is rational or irrational.
If $x$ is rational then $f(x)$ is a certain positive number. It is claimed that $f$ is not continuous at $x$. Assume to the contrary that $f$ is continuous at $x$. Then there is a neighborhood $U:=U_delta(x)$ such that $f(t)>f(x)over2$ for all $tin U$. Hm?
If $x$ is irrational then $f(x)=0$. It is claimed that $f$ is continuous at $x$. This means that for any $epsilon>0$ we should be able to produce a neighborhood $U:=U_delta(x)$ such that $f(t)<epsilon$ for all $tin U$. Which points $tinmathbb R$ could violate the condition $f(t)<epsilon $?
answered Oct 4 '12 at 12:52
Christian Blatter
164k7107306
You cannot do this using simple rules about limits, like de l?Hôpital's rule or the continuity of "analytic expressions", etc. You have to go back to the actual definition of continuity when solving this problem. Here are a few hints:
In the first place you want to prove different things about a given $xinmathbb R$, depending on whether $x$ is rational or irrational.
If $x$ is rational then $f(x)$ is a certain positive number. It is claimed that $f$ is not continuous at $x$. Assume to the contrary that $f$ is continuous at $x$. Then there is a neighborhood $U:=U_delta(x)$ such that $f(t)>f(x)over2$ for all $tin U$. Hm?
If $x$ is irrational then $f(x)=0$. It is claimed that $f$ is continuous at $x$. This means that for any $epsilon>0$ we should be able to produce a neighborhood $U:=U_delta(x)$ such that $f(t)<epsilon$ for all $tin U$. Which points $tinmathbb R$ could violate the condition $f(t)<epsilon $?
answered Oct 4 '12 at 12:52
Christian Blatter
164k7107306
answered Oct 4 '12 at 12:52
Christian Blatter
164k7107306
answered Oct 4 '12 at 12:52
Christian Blatter
164k7107306
answered Oct 4 '12 at 12:52
Christian Blatter
164k7107306
164k7107306
Are they going to be the rationals?
– Casquibaldo
Oct 4 '12 at 13:27
@Casquibaldo: If you mean by "they" the $t$ in the last line of my answer: Yes, but only selected rationals.
– Christian Blatter
Oct 4 '12 at 13:31
add a comment |Â
Are they going to be the rationals?
– Casquibaldo
Oct 4 '12 at 13:27
@Casquibaldo: If you mean by "they" the $t$ in the last line of my answer: Yes, but only selected rationals.
– Christian Blatter
Oct 4 '12 at 13:31
Are they going to be the rationals?
– Casquibaldo
Oct 4 '12 at 13:27
Are they going to be the rationals?
– Casquibaldo
Oct 4 '12 at 13:27
@Casquibaldo: If you mean by "they" the $t$ in the last line of my answer: Yes, but only selected rationals.
– Christian Blatter
Oct 4 '12 at 13:31
@Casquibaldo: If you mean by "they" the $t$ in the last line of my answer: Yes, but only selected rationals.
– Christian Blatter
Oct 4 '12 at 13:31
add a comment |Â
up vote
5
down vote
This function is known as Thomae's function and is an excellent example of a function that is continuous at uncountably infinite number of points while discontinuous at countably infinite number of points. Note that the reverse result is not possible, i.e. one cannot have a function that is discontinuous on an uncountably infinite dense set and continuous on a countably infinite dense set.
As for the proof , the discontinuity part follows from the fact that irrationals are dense in $ R $ as you rightly thought. The continuity part is a little bit more involved and it would help to employ the Archimedean principle in tandem with the $ epsilon$-$delta $ definition.
Crudely, look at $ x=sqrt3 $ , see that using the non-terminating decimal expansion i.e. $ sqrt 3=1.732148ldots $ , one can forge a sequence of rationals $ 1 , 17/10 ,173/100 ,ldots $
that converges to $ x $, while at the same time your $ f(x_n) $ is $ 1/10,1/100,1/1000,ldots $ which converges to $ 0 $ which is nothing but $ f(x) $. This is merely an illustration of the continuity and not a rigorous proof.
edited Oct 30 '16 at 5:43
Michael Hardy
204k23186462
answered Oct 4 '12 at 16:12
Vishesh
1,4661126
add a comment |Â
up vote
5
down vote
This function is known as Thomae's function and is an excellent example of a function that is continuous at uncountably infinite number of points while discontinuous at countably infinite number of points. Note that the reverse result is not possible, i.e. one cannot have a function that is discontinuous on an uncountably infinite dense set and continuous on a countably infinite dense set.
As for the proof , the discontinuity part follows from the fact that irrationals are dense in $ R $ as you rightly thought. The continuity part is a little bit more involved and it would help to employ the Archimedean principle in tandem with the $ epsilon$-$delta $ definition.
Crudely, look at $ x=sqrt3 $ , see that using the non-terminating decimal expansion i.e. $ sqrt 3=1.732148ldots $ , one can forge a sequence of rationals $ 1 , 17/10 ,173/100 ,ldots $
that converges to $ x $, while at the same time your $ f(x_n) $ is $ 1/10,1/100,1/1000,ldots $ which converges to $ 0 $ which is nothing but $ f(x) $. This is merely an illustration of the continuity and not a rigorous proof.
edited Oct 30 '16 at 5:43
Michael Hardy
204k23186462
answered Oct 4 '12 at 16:12
Vishesh
1,4661126
add a comment |Â
up vote
5
down vote
up vote
5
down vote
This function is known as Thomae's function and is an excellent example of a function that is continuous at uncountably infinite number of points while discontinuous at countably infinite number of points. Note that the reverse result is not possible, i.e. one cannot have a function that is discontinuous on an uncountably infinite dense set and continuous on a countably infinite dense set.
As for the proof , the discontinuity part follows from the fact that irrationals are dense in $ R $ as you rightly thought. The continuity part is a little bit more involved and it would help to employ the Archimedean principle in tandem with the $ epsilon$-$delta $ definition.
Crudely, look at $ x=sqrt3 $ , see that using the non-terminating decimal expansion i.e. $ sqrt 3=1.732148ldots $ , one can forge a sequence of rationals $ 1 , 17/10 ,173/100 ,ldots $
that converges to $ x $, while at the same time your $ f(x_n) $ is $ 1/10,1/100,1/1000,ldots $ which converges to $ 0 $ which is nothing but $ f(x) $. This is merely an illustration of the continuity and not a rigorous proof.
edited Oct 30 '16 at 5:43
Michael Hardy
204k23186462
answered Oct 4 '12 at 16:12
Vishesh
1,4661126
This function is known as Thomae's function and is an excellent example of a function that is continuous at uncountably infinite number of points while discontinuous at countably infinite number of points. Note that the reverse result is not possible, i.e. one cannot have a function that is discontinuous on an uncountably infinite dense set and continuous on a countably infinite dense set.
As for the proof , the discontinuity part follows from the fact that irrationals are dense in $ R $ as you rightly thought. The continuity part is a little bit more involved and it would help to employ the Archimedean principle in tandem with the $ epsilon$-$delta $ definition.
Crudely, look at $ x=sqrt3 $ , see that using the non-terminating decimal expansion i.e. $ sqrt 3=1.732148ldots $ , one can forge a sequence of rationals $ 1 , 17/10 ,173/100 ,ldots $
that converges to $ x $, while at the same time your $ f(x_n) $ is $ 1/10,1/100,1/1000,ldots $ which converges to $ 0 $ which is nothing but $ f(x) $. This is merely an illustration of the continuity and not a rigorous proof.
edited Oct 30 '16 at 5:43
Michael Hardy
204k23186462
answered Oct 4 '12 at 16:12
Vishesh
1,4661126
edited Oct 30 '16 at 5:43
Michael Hardy
204k23186462
edited Oct 30 '16 at 5:43
Michael Hardy
204k23186462
edited Oct 30 '16 at 5:43
Michael Hardy
204k23186462
204k23186462
answered Oct 4 '12 at 16:12
Vishesh
1,4661126
answered Oct 4 '12 at 16:12
Vishesh
1,4661126
answered Oct 4 '12 at 16:12
Vishesh
1,4661126
1,4661126
add a comment |Â
add a comment |Â
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2
The way to do this is to show that in arbitrarily small neighborhoods of an irrational point $x$, rationals approximating $x$ must have large denominators in reduced form.
– Seth
Oct 4 '12 at 12:11
1
Hint: $|frac ab-frac xy|=|fracay-bxby|ge frac1by$ if $frac abnefrac xy$
– Hagen von Eitzen
Oct 4 '12 at 12:31