Is extension of continuous function on $Bbb Q$ is continuous on $Bbb R$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Continuous extension of a functionContinuous FunctionIs the following true for $f: Bbb R^3 rightarrow Bbb R$ continuous?Continuous function on $BbbR$ with neither open nor closed imageContinuous extension of uniformly continuous functionsIs function continuous, bounded?$f:Bbb Rto Bbb R$ be a continuous function such that $f(i)=0forall iin Bbb Z$Existence of continuous function $f$ on $Bbb R$ which vanishes exactly on $Asubset Bbb R$Prove that, any continuous and periodic function on $BbbR$ is uniformly continuous on $BbbR$.Absolutely integrable function on $Bbb R$
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Is extension of continuous function on $Bbb Q$ is continuous on $Bbb R$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Continuous extension of a functionContinuous FunctionIs the following true for $f: Bbb R^3 rightarrow Bbb R$ continuous?Continuous function on $BbbR$ with neither open nor closed imageContinuous extension of uniformly continuous functionsIs function continuous, bounded?$f:Bbb Rto Bbb R$ be a continuous function such that $f(i)=0forall iin Bbb Z$Existence of continuous function $f$ on $Bbb R$ which vanishes exactly on $Asubset Bbb R$Prove that, any continuous and periodic function on $BbbR$ is uniformly continuous on $BbbR$.Absolutely integrable function on $Bbb R$
$begingroup$
If $f:Bbb Q to Bbb Q$ is a continuous function then $f$ can be extented to $g: Bbb R to Bbb R$ such that $g$ is continuous
Is it true?
real-analysis
edited Apr 1 at 14:43
Chinnapparaj R
6,60721029
asked Apr 1 at 14:22
Deepali Deepali
111
$endgroup$
add a comment |
$begingroup$
If $f:Bbb Q to Bbb Q$ is a continuous function then $f$ can be extented to $g: Bbb R to Bbb R$ such that $g$ is continuous
Is it true?
real-analysis
edited Apr 1 at 14:43
Chinnapparaj R
6,60721029
asked Apr 1 at 14:22
Deepali Deepali
111
$endgroup$
$begingroup$
In general, $f$ needs to be uniformly continuous on $mathbbQ$, but the more specific condition could be considered "cauchy continuity" that is, cauchy sequences in $mathbbQ$ must map to a Cauchy sequence in $mathbbQ$ under $f$
$endgroup$
– rubikscube09
Apr 1 at 14:33
1
$begingroup$
@rubikscube09 No, $f$ need not be uniformly continuous on $Bbb Q$; the function $f(x)=x^2$ certainly extends to a continuous function on $Bbb R$. A continuous function from $Bbb Q$ to $Bbb R$ extends continuously to $Bbb R$ if and only if its restriction to every bounded subset of $Bbb Q$ is uniformly continuous
$endgroup$
– David C. Ullrich
Apr 1 at 14:46
add a comment |
$begingroup$
If $f:Bbb Q to Bbb Q$ is a continuous function then $f$ can be extented to $g: Bbb R to Bbb R$ such that $g$ is continuous
Is it true?
real-analysis
edited Apr 1 at 14:43
Chinnapparaj R
6,60721029
asked Apr 1 at 14:22
Deepali Deepali
111
$endgroup$
If $f:Bbb Q to Bbb Q$ is a continuous function then $f$ can be extented to $g: Bbb R to Bbb R$ such that $g$ is continuous
Is it true?
real-analysis
real-analysis
edited Apr 1 at 14:43
Chinnapparaj R
6,60721029
asked Apr 1 at 14:22
Deepali Deepali
111
edited Apr 1 at 14:43
Chinnapparaj R
6,60721029
asked Apr 1 at 14:22
Deepali Deepali
111
edited Apr 1 at 14:43
Chinnapparaj R
6,60721029
edited Apr 1 at 14:43
Chinnapparaj R
6,60721029
edited Apr 1 at 14:43
Chinnapparaj R
6,60721029
6,60721029
asked Apr 1 at 14:22
Deepali Deepali
111
asked Apr 1 at 14:22
Deepali Deepali
111
asked Apr 1 at 14:22
Deepali Deepali
111
111
$begingroup$
In general, $f$ needs to be uniformly continuous on $mathbbQ$, but the more specific condition could be considered "cauchy continuity" that is, cauchy sequences in $mathbbQ$ must map to a Cauchy sequence in $mathbbQ$ under $f$
$endgroup$
– rubikscube09
Apr 1 at 14:33
1
$begingroup$
@rubikscube09 No, $f$ need not be uniformly continuous on $Bbb Q$; the function $f(x)=x^2$ certainly extends to a continuous function on $Bbb R$. A continuous function from $Bbb Q$ to $Bbb R$ extends continuously to $Bbb R$ if and only if its restriction to every bounded subset of $Bbb Q$ is uniformly continuous
$endgroup$
– David C. Ullrich
Apr 1 at 14:46
add a comment |
$begingroup$
In general, $f$ needs to be uniformly continuous on $mathbbQ$, but the more specific condition could be considered "cauchy continuity" that is, cauchy sequences in $mathbbQ$ must map to a Cauchy sequence in $mathbbQ$ under $f$
$endgroup$
– rubikscube09
Apr 1 at 14:33
1
$begingroup$
@rubikscube09 No, $f$ need not be uniformly continuous on $Bbb Q$; the function $f(x)=x^2$ certainly extends to a continuous function on $Bbb R$. A continuous function from $Bbb Q$ to $Bbb R$ extends continuously to $Bbb R$ if and only if its restriction to every bounded subset of $Bbb Q$ is uniformly continuous
$endgroup$
– David C. Ullrich
Apr 1 at 14:46
$begingroup$
In general, $f$ needs to be uniformly continuous on $mathbbQ$, but the more specific condition could be considered "cauchy continuity" that is, cauchy sequences in $mathbbQ$ must map to a Cauchy sequence in $mathbbQ$ under $f$
$endgroup$
– rubikscube09
Apr 1 at 14:33
$begingroup$
In general, $f$ needs to be uniformly continuous on $mathbbQ$, but the more specific condition could be considered "cauchy continuity" that is, cauchy sequences in $mathbbQ$ must map to a Cauchy sequence in $mathbbQ$ under $f$
$endgroup$
– rubikscube09
Apr 1 at 14:33
1
1
$begingroup$
@rubikscube09 No, $f$ need not be uniformly continuous on $Bbb Q$; the function $f(x)=x^2$ certainly extends to a continuous function on $Bbb R$. A continuous function from $Bbb Q$ to $Bbb R$ extends continuously to $Bbb R$ if and only if its restriction to every bounded subset of $Bbb Q$ is uniformly continuous
$endgroup$
– David C. Ullrich
Apr 1 at 14:46
$begingroup$
@rubikscube09 No, $f$ need not be uniformly continuous on $Bbb Q$; the function $f(x)=x^2$ certainly extends to a continuous function on $Bbb R$. A continuous function from $Bbb Q$ to $Bbb R$ extends continuously to $Bbb R$ if and only if its restriction to every bounded subset of $Bbb Q$ is uniformly continuous
$endgroup$
– David C. Ullrich
Apr 1 at 14:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: What if$$f(x)=begincases0&text if x<sqrt2\1&text if x>sqrt2?endcases$$
answered Apr 1 at 14:24
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
3
$begingroup$
Not that there's any problem with that example, but I bet $1/(x^2-2)$ would be clearer to some readers, one in particular...
$endgroup$
– David C. Ullrich
Apr 1 at 14:43
$begingroup$
Nice! I would have posted it if I had had that idea.
$endgroup$
– José Carlos Santos
Apr 1 at 14:45
add a comment |
$begingroup$
A function $f_Bbb Q: Bbb Qto Bbb R$ (including funcitons that are $Bbb Qto Bbb Q$) can be extended to a continuous function $f_Bbb R:Bbb Rto Bbb R$ iff it is sequentially continuous. In other words, if for any real number $r$ and sequences $x_n, y_n$ of rational numbers with $x_n, y_nto r$, we have $f(x_n) - f(y_n) to 0$.
answered Apr 1 at 14:28
ArthurArthur
123k7122211
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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active
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votes
$begingroup$
Hint: What if$$f(x)=begincases0&text if x<sqrt2\1&text if x>sqrt2?endcases$$
answered Apr 1 at 14:24
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
3
$begingroup$
Not that there's any problem with that example, but I bet $1/(x^2-2)$ would be clearer to some readers, one in particular...
$endgroup$
– David C. Ullrich
Apr 1 at 14:43
$begingroup$
Nice! I would have posted it if I had had that idea.
$endgroup$
– José Carlos Santos
Apr 1 at 14:45
add a comment |
$begingroup$
Hint: What if$$f(x)=begincases0&text if x<sqrt2\1&text if x>sqrt2?endcases$$
answered Apr 1 at 14:24
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
3
$begingroup$
Not that there's any problem with that example, but I bet $1/(x^2-2)$ would be clearer to some readers, one in particular...
$endgroup$
– David C. Ullrich
Apr 1 at 14:43
$begingroup$
Nice! I would have posted it if I had had that idea.
$endgroup$
– José Carlos Santos
Apr 1 at 14:45
add a comment |
$begingroup$
Hint: What if$$f(x)=begincases0&text if x<sqrt2\1&text if x>sqrt2?endcases$$
answered Apr 1 at 14:24
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
Hint: What if$$f(x)=begincases0&text if x<sqrt2\1&text if x>sqrt2?endcases$$
answered Apr 1 at 14:24
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 1 at 14:24
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 1 at 14:24
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 1 at 14:24
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
3
$begingroup$
Not that there's any problem with that example, but I bet $1/(x^2-2)$ would be clearer to some readers, one in particular...
$endgroup$
– David C. Ullrich
Apr 1 at 14:43
$begingroup$
Nice! I would have posted it if I had had that idea.
$endgroup$
– José Carlos Santos
Apr 1 at 14:45
add a comment |
3
$begingroup$
Not that there's any problem with that example, but I bet $1/(x^2-2)$ would be clearer to some readers, one in particular...
$endgroup$
– David C. Ullrich
Apr 1 at 14:43
$begingroup$
Nice! I would have posted it if I had had that idea.
$endgroup$
– José Carlos Santos
Apr 1 at 14:45
3
3
$begingroup$
Not that there's any problem with that example, but I bet $1/(x^2-2)$ would be clearer to some readers, one in particular...
$endgroup$
– David C. Ullrich
Apr 1 at 14:43
$begingroup$
Not that there's any problem with that example, but I bet $1/(x^2-2)$ would be clearer to some readers, one in particular...
$endgroup$
– David C. Ullrich
Apr 1 at 14:43
$begingroup$
Nice! I would have posted it if I had had that idea.
$endgroup$
– José Carlos Santos
Apr 1 at 14:45
$begingroup$
Nice! I would have posted it if I had had that idea.
$endgroup$
– José Carlos Santos
Apr 1 at 14:45
add a comment |
$begingroup$
A function $f_Bbb Q: Bbb Qto Bbb R$ (including funcitons that are $Bbb Qto Bbb Q$) can be extended to a continuous function $f_Bbb R:Bbb Rto Bbb R$ iff it is sequentially continuous. In other words, if for any real number $r$ and sequences $x_n, y_n$ of rational numbers with $x_n, y_nto r$, we have $f(x_n) - f(y_n) to 0$.
answered Apr 1 at 14:28
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
A function $f_Bbb Q: Bbb Qto Bbb R$ (including funcitons that are $Bbb Qto Bbb Q$) can be extended to a continuous function $f_Bbb R:Bbb Rto Bbb R$ iff it is sequentially continuous. In other words, if for any real number $r$ and sequences $x_n, y_n$ of rational numbers with $x_n, y_nto r$, we have $f(x_n) - f(y_n) to 0$.
answered Apr 1 at 14:28
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
A function $f_Bbb Q: Bbb Qto Bbb R$ (including funcitons that are $Bbb Qto Bbb Q$) can be extended to a continuous function $f_Bbb R:Bbb Rto Bbb R$ iff it is sequentially continuous. In other words, if for any real number $r$ and sequences $x_n, y_n$ of rational numbers with $x_n, y_nto r$, we have $f(x_n) - f(y_n) to 0$.
answered Apr 1 at 14:28
ArthurArthur
123k7122211
$endgroup$
A function $f_Bbb Q: Bbb Qto Bbb R$ (including funcitons that are $Bbb Qto Bbb Q$) can be extended to a continuous function $f_Bbb R:Bbb Rto Bbb R$ iff it is sequentially continuous. In other words, if for any real number $r$ and sequences $x_n, y_n$ of rational numbers with $x_n, y_nto r$, we have $f(x_n) - f(y_n) to 0$.
answered Apr 1 at 14:28
ArthurArthur
123k7122211
answered Apr 1 at 14:28
ArthurArthur
123k7122211
answered Apr 1 at 14:28
ArthurArthur
123k7122211
answered Apr 1 at 14:28
ArthurArthur
123k7122211
123k7122211
add a comment |
add a comment |
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$begingroup$
In general, $f$ needs to be uniformly continuous on $mathbbQ$, but the more specific condition could be considered "cauchy continuity" that is, cauchy sequences in $mathbbQ$ must map to a Cauchy sequence in $mathbbQ$ under $f$
$endgroup$
– rubikscube09
Apr 1 at 14:33
1
$begingroup$
@rubikscube09 No, $f$ need not be uniformly continuous on $Bbb Q$; the function $f(x)=x^2$ certainly extends to a continuous function on $Bbb R$. A continuous function from $Bbb Q$ to $Bbb R$ extends continuously to $Bbb R$ if and only if its restriction to every bounded subset of $Bbb Q$ is uniformly continuous
$endgroup$
– David C. Ullrich
Apr 1 at 14:46