Existence of a pointwise convergent subsequenceWhen does pointwise convergence imply uniform convergence?Are Arzelà–Ascoli theorems results of similar theorems on normed spaces, metric spaces or other spaces?Does Arzelà-Ascoli theorem hold for pointwise convergence?Subsequence Convergencelooking for proof that this uniformly bounded sequence of functions has no pointwise convergent subsequenceHow to show a set is compact in a function space?Having trouble showing existence of pointwise convergent subsequence of a sequence of real-valued pointwise bounded functions on a countable set.Arzela-Ascoli Theorem: Is only pointwise boundedness required?Pointwise a.e. convergence vs. Pointwise convergenceExistence of pointwise convergent subsequence of increasing functions
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Existence of a pointwise convergent subsequence
When does pointwise convergence imply uniform convergence?Are Arzelà–Ascoli theorems results of similar theorems on normed spaces, metric spaces or other spaces?Does Arzelà-Ascoli theorem hold for pointwise convergence?Subsequence Convergencelooking for proof that this uniformly bounded sequence of functions has no pointwise convergent subsequenceHow to show a set is compact in a function space?Having trouble showing existence of pointwise convergent subsequence of a sequence of real-valued pointwise bounded functions on a countable set.Arzela-Ascoli Theorem: Is only pointwise boundedness required?Pointwise a.e. convergence vs. Pointwise convergenceExistence of pointwise convergent subsequence of increasing functions
$begingroup$
Let $X$ be a compact metric space and $f_m:Xto[0,1]$ a continuous function for each $minmathbb N$.
Does there necessarily exist a $f:Xto[0,1]$ (not necessarily continuous) and a subsequence $(f_m_k)$ such that $f_m_k(x)to f(x)$ for each $xin X$ pointwise?
Note that the equicontinuity of $(f_m)_minmathbb N$ is not assumed, so that the Arzelà–Ascoli theorem is of no use here. However, the desired conclusion is also weaker: the supposed limit function $f$ need not be continuous and only pointwise convergence is required.
Any suggestion would be appreciated.
real-analysis general-topology
asked Jun 4 at 20:59
triple_sectriple_sec
16.2k21954
$endgroup$
add a comment |
$begingroup$
Let $X$ be a compact metric space and $f_m:Xto[0,1]$ a continuous function for each $minmathbb N$.
Does there necessarily exist a $f:Xto[0,1]$ (not necessarily continuous) and a subsequence $(f_m_k)$ such that $f_m_k(x)to f(x)$ for each $xin X$ pointwise?
Note that the equicontinuity of $(f_m)_minmathbb N$ is not assumed, so that the Arzelà–Ascoli theorem is of no use here. However, the desired conclusion is also weaker: the supposed limit function $f$ need not be continuous and only pointwise convergence is required.
Any suggestion would be appreciated.
real-analysis general-topology
asked Jun 4 at 20:59
triple_sectriple_sec
16.2k21954
$endgroup$
1
$begingroup$
$X$ has a countable dense subset. Perhaps you could define $f$ there first?
$endgroup$
– copper.hat
Jun 4 at 21:26
$begingroup$
@copper.hat Yes, that’s a good idea for a first step. If $x^1,x^2,ldots$ is dense, then, by Cantor’s diagonal method, there is a subsequence such that $f_m_k(x^j)$ converges as $ktoinfty$ for every $jinmathbb N$. Using this result, the next step would be showing that $f_m_k(x)$ is Cauchy in $[0,1]$ for each $xin X$, but it seems tricky, as convergence at the $x^j$s need not be uniform.
$endgroup$
– triple_sec
Jun 4 at 21:28
$begingroup$
Robert has a nice counterexample below.
$endgroup$
– copper.hat
Jun 4 at 21:32
$begingroup$
If functions $f_n:mathbbRrightarrow [0,1]$ are nondecreasing in x for each n, then (regardless of continuity or behavior with n) your argument of showing convergence over a subsequence on a countable but dense subset can be used to prove convergence everywhere in $mathbbR$.
$endgroup$
– Michael
Jun 5 at 0:07
$begingroup$
The same does not hold for functions $f_n:mathbbR^2rightarrow [0,1]$ that are entrywise nondecreasing as we can have discontinuity across a diagonal and then adapt one of the 1-d counterexamples given below on that diagonal.
$endgroup$
– Michael
Jun 5 at 0:29
add a comment |
$begingroup$
Let $X$ be a compact metric space and $f_m:Xto[0,1]$ a continuous function for each $minmathbb N$.
Does there necessarily exist a $f:Xto[0,1]$ (not necessarily continuous) and a subsequence $(f_m_k)$ such that $f_m_k(x)to f(x)$ for each $xin X$ pointwise?
Note that the equicontinuity of $(f_m)_minmathbb N$ is not assumed, so that the Arzelà–Ascoli theorem is of no use here. However, the desired conclusion is also weaker: the supposed limit function $f$ need not be continuous and only pointwise convergence is required.
Any suggestion would be appreciated.
real-analysis general-topology
asked Jun 4 at 20:59
triple_sectriple_sec
16.2k21954
$endgroup$
Let $X$ be a compact metric space and $f_m:Xto[0,1]$ a continuous function for each $minmathbb N$.
Does there necessarily exist a $f:Xto[0,1]$ (not necessarily continuous) and a subsequence $(f_m_k)$ such that $f_m_k(x)to f(x)$ for each $xin X$ pointwise?
Note that the equicontinuity of $(f_m)_minmathbb N$ is not assumed, so that the Arzelà–Ascoli theorem is of no use here. However, the desired conclusion is also weaker: the supposed limit function $f$ need not be continuous and only pointwise convergence is required.
Any suggestion would be appreciated.
real-analysis general-topology
real-analysis general-topology
asked Jun 4 at 20:59
triple_sectriple_sec
16.2k21954
asked Jun 4 at 20:59
triple_sectriple_sec
16.2k21954
asked Jun 4 at 20:59
triple_sectriple_sec
16.2k21954
asked Jun 4 at 20:59
triple_sectriple_sec
16.2k21954
asked Jun 4 at 20:59
triple_sectriple_sec
16.2k21954
16.2k21954
1
$begingroup$
$X$ has a countable dense subset. Perhaps you could define $f$ there first?
$endgroup$
– copper.hat
Jun 4 at 21:26
$begingroup$
@copper.hat Yes, that’s a good idea for a first step. If $x^1,x^2,ldots$ is dense, then, by Cantor’s diagonal method, there is a subsequence such that $f_m_k(x^j)$ converges as $ktoinfty$ for every $jinmathbb N$. Using this result, the next step would be showing that $f_m_k(x)$ is Cauchy in $[0,1]$ for each $xin X$, but it seems tricky, as convergence at the $x^j$s need not be uniform.
$endgroup$
– triple_sec
Jun 4 at 21:28
$begingroup$
Robert has a nice counterexample below.
$endgroup$
– copper.hat
Jun 4 at 21:32
$begingroup$
If functions $f_n:mathbbRrightarrow [0,1]$ are nondecreasing in x for each n, then (regardless of continuity or behavior with n) your argument of showing convergence over a subsequence on a countable but dense subset can be used to prove convergence everywhere in $mathbbR$.
$endgroup$
– Michael
Jun 5 at 0:07
$begingroup$
The same does not hold for functions $f_n:mathbbR^2rightarrow [0,1]$ that are entrywise nondecreasing as we can have discontinuity across a diagonal and then adapt one of the 1-d counterexamples given below on that diagonal.
$endgroup$
– Michael
Jun 5 at 0:29
add a comment |
1
$begingroup$
$X$ has a countable dense subset. Perhaps you could define $f$ there first?
$endgroup$
– copper.hat
Jun 4 at 21:26
$begingroup$
@copper.hat Yes, that’s a good idea for a first step. If $x^1,x^2,ldots$ is dense, then, by Cantor’s diagonal method, there is a subsequence such that $f_m_k(x^j)$ converges as $ktoinfty$ for every $jinmathbb N$. Using this result, the next step would be showing that $f_m_k(x)$ is Cauchy in $[0,1]$ for each $xin X$, but it seems tricky, as convergence at the $x^j$s need not be uniform.
$endgroup$
– triple_sec
Jun 4 at 21:28
$begingroup$
Robert has a nice counterexample below.
$endgroup$
– copper.hat
Jun 4 at 21:32
$begingroup$
If functions $f_n:mathbbRrightarrow [0,1]$ are nondecreasing in x for each n, then (regardless of continuity or behavior with n) your argument of showing convergence over a subsequence on a countable but dense subset can be used to prove convergence everywhere in $mathbbR$.
$endgroup$
– Michael
Jun 5 at 0:07
$begingroup$
The same does not hold for functions $f_n:mathbbR^2rightarrow [0,1]$ that are entrywise nondecreasing as we can have discontinuity across a diagonal and then adapt one of the 1-d counterexamples given below on that diagonal.
$endgroup$
– Michael
Jun 5 at 0:29
1
1
$begingroup$
$X$ has a countable dense subset. Perhaps you could define $f$ there first?
$endgroup$
– copper.hat
Jun 4 at 21:26
$begingroup$
$X$ has a countable dense subset. Perhaps you could define $f$ there first?
$endgroup$
– copper.hat
Jun 4 at 21:26
$begingroup$
@copper.hat Yes, that’s a good idea for a first step. If $x^1,x^2,ldots$ is dense, then, by Cantor’s diagonal method, there is a subsequence such that $f_m_k(x^j)$ converges as $ktoinfty$ for every $jinmathbb N$. Using this result, the next step would be showing that $f_m_k(x)$ is Cauchy in $[0,1]$ for each $xin X$, but it seems tricky, as convergence at the $x^j$s need not be uniform.
$endgroup$
– triple_sec
Jun 4 at 21:28
$begingroup$
@copper.hat Yes, that’s a good idea for a first step. If $x^1,x^2,ldots$ is dense, then, by Cantor’s diagonal method, there is a subsequence such that $f_m_k(x^j)$ converges as $ktoinfty$ for every $jinmathbb N$. Using this result, the next step would be showing that $f_m_k(x)$ is Cauchy in $[0,1]$ for each $xin X$, but it seems tricky, as convergence at the $x^j$s need not be uniform.
$endgroup$
– triple_sec
Jun 4 at 21:28
$begingroup$
Robert has a nice counterexample below.
$endgroup$
– copper.hat
Jun 4 at 21:32
$begingroup$
Robert has a nice counterexample below.
$endgroup$
– copper.hat
Jun 4 at 21:32
$begingroup$
If functions $f_n:mathbbRrightarrow [0,1]$ are nondecreasing in x for each n, then (regardless of continuity or behavior with n) your argument of showing convergence over a subsequence on a countable but dense subset can be used to prove convergence everywhere in $mathbbR$.
$endgroup$
– Michael
Jun 5 at 0:07
$begingroup$
If functions $f_n:mathbbRrightarrow [0,1]$ are nondecreasing in x for each n, then (regardless of continuity or behavior with n) your argument of showing convergence over a subsequence on a countable but dense subset can be used to prove convergence everywhere in $mathbbR$.
$endgroup$
– Michael
Jun 5 at 0:07
$begingroup$
The same does not hold for functions $f_n:mathbbR^2rightarrow [0,1]$ that are entrywise nondecreasing as we can have discontinuity across a diagonal and then adapt one of the 1-d counterexamples given below on that diagonal.
$endgroup$
– Michael
Jun 5 at 0:29
$begingroup$
The same does not hold for functions $f_n:mathbbR^2rightarrow [0,1]$ that are entrywise nondecreasing as we can have discontinuity across a diagonal and then adapt one of the 1-d counterexamples given below on that diagonal.
$endgroup$
– Michael
Jun 5 at 0:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, there does not. Consider $f_n(x) = cos(4^n pi x)$ on $[0,1]$ (with range in $[-1,1]$ not $[0,1]$, but you can transform it). Then for any subsequence $f_n_k$, there is
some $x in [0,1]$ such that $f_n_k(x) ge cos(pi/4)$ if $k$ is odd and $le -cos(pi/4)$ if $k$ is even. All you need to do is choose the base-$4$ digits of $x$ correctly.
EDIT: Any choice of the first $n$ digits after the "decimal" point leaves $x$ in an interval of length $4^-n$ on which $f_n$ goes either from $1$ to $-1$ or $-1$ to $1$. Choose the $n+1$'th digit to be $0$ or $3$ to make $f_n(x) ge cos(pi/4)$ or $le -cos(pi/4)$ (which is which depends on the previous choices of digits). For example, if you choose the first $3$ digits to be $0,3,0$, that says $x$ is between $0 times 4^-1 + 3 times 4^-2 + 0 times 4^-3 = 3/16$ and $0 times 4^-1 + 3 times 4^-2 + 1 times 4^-3 = 13/64$,
say $x = 3/16 + 4^-3 t$, $0 le t le 1$. On this interval, $f_3(x) = cos(4^3 pi x) = cos(pi t)$ goes from $1$ at $t=0$ to $-1$ at $t=1$. If you want $f_3(x) ge cos(pi/4)$, choose the $4$'th digit to be $0$, if you want $f_3(x) le - cos(pi/4)$ choose the $4$'th digit to be $1$.
answered Jun 4 at 21:31
Robert IsraelRobert Israel
338k23232490
$endgroup$
$begingroup$
What do mean by "choose the base-$4$ digits of $x$ correctly"? Could you illustrate it with an example please? For instance the case where $n_k=k$. How do you construct $x$? Thanks.
$endgroup$
– Dog_69
Jun 4 at 22:45
add a comment |
$begingroup$
A similar example using more machinery hence perhaps less tricky in the details: Let $X=[0,2pi]$, $f_n(t)=cos(nt)$. If $f_n_k(x)to f(x)$ for (almost) every $x$ then Dominated Convergence shows that $||f_n_k-f||_2to0$; this is impossible by orthogonality (for example, $||f_n-f_m||_2^2=2pi$.)
answered Jun 4 at 23:35
David C. UllrichDavid C. Ullrich
63.1k44298
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
No, there does not. Consider $f_n(x) = cos(4^n pi x)$ on $[0,1]$ (with range in $[-1,1]$ not $[0,1]$, but you can transform it). Then for any subsequence $f_n_k$, there is
some $x in [0,1]$ such that $f_n_k(x) ge cos(pi/4)$ if $k$ is odd and $le -cos(pi/4)$ if $k$ is even. All you need to do is choose the base-$4$ digits of $x$ correctly.
EDIT: Any choice of the first $n$ digits after the "decimal" point leaves $x$ in an interval of length $4^-n$ on which $f_n$ goes either from $1$ to $-1$ or $-1$ to $1$. Choose the $n+1$'th digit to be $0$ or $3$ to make $f_n(x) ge cos(pi/4)$ or $le -cos(pi/4)$ (which is which depends on the previous choices of digits). For example, if you choose the first $3$ digits to be $0,3,0$, that says $x$ is between $0 times 4^-1 + 3 times 4^-2 + 0 times 4^-3 = 3/16$ and $0 times 4^-1 + 3 times 4^-2 + 1 times 4^-3 = 13/64$,
say $x = 3/16 + 4^-3 t$, $0 le t le 1$. On this interval, $f_3(x) = cos(4^3 pi x) = cos(pi t)$ goes from $1$ at $t=0$ to $-1$ at $t=1$. If you want $f_3(x) ge cos(pi/4)$, choose the $4$'th digit to be $0$, if you want $f_3(x) le - cos(pi/4)$ choose the $4$'th digit to be $1$.
answered Jun 4 at 21:31
Robert IsraelRobert Israel
338k23232490
$endgroup$
$begingroup$
What do mean by "choose the base-$4$ digits of $x$ correctly"? Could you illustrate it with an example please? For instance the case where $n_k=k$. How do you construct $x$? Thanks.
$endgroup$
– Dog_69
Jun 4 at 22:45
add a comment |
$begingroup$
No, there does not. Consider $f_n(x) = cos(4^n pi x)$ on $[0,1]$ (with range in $[-1,1]$ not $[0,1]$, but you can transform it). Then for any subsequence $f_n_k$, there is
some $x in [0,1]$ such that $f_n_k(x) ge cos(pi/4)$ if $k$ is odd and $le -cos(pi/4)$ if $k$ is even. All you need to do is choose the base-$4$ digits of $x$ correctly.
EDIT: Any choice of the first $n$ digits after the "decimal" point leaves $x$ in an interval of length $4^-n$ on which $f_n$ goes either from $1$ to $-1$ or $-1$ to $1$. Choose the $n+1$'th digit to be $0$ or $3$ to make $f_n(x) ge cos(pi/4)$ or $le -cos(pi/4)$ (which is which depends on the previous choices of digits). For example, if you choose the first $3$ digits to be $0,3,0$, that says $x$ is between $0 times 4^-1 + 3 times 4^-2 + 0 times 4^-3 = 3/16$ and $0 times 4^-1 + 3 times 4^-2 + 1 times 4^-3 = 13/64$,
say $x = 3/16 + 4^-3 t$, $0 le t le 1$. On this interval, $f_3(x) = cos(4^3 pi x) = cos(pi t)$ goes from $1$ at $t=0$ to $-1$ at $t=1$. If you want $f_3(x) ge cos(pi/4)$, choose the $4$'th digit to be $0$, if you want $f_3(x) le - cos(pi/4)$ choose the $4$'th digit to be $1$.
answered Jun 4 at 21:31
Robert IsraelRobert Israel
338k23232490
$endgroup$
$begingroup$
What do mean by "choose the base-$4$ digits of $x$ correctly"? Could you illustrate it with an example please? For instance the case where $n_k=k$. How do you construct $x$? Thanks.
$endgroup$
– Dog_69
Jun 4 at 22:45
add a comment |
$begingroup$
No, there does not. Consider $f_n(x) = cos(4^n pi x)$ on $[0,1]$ (with range in $[-1,1]$ not $[0,1]$, but you can transform it). Then for any subsequence $f_n_k$, there is
some $x in [0,1]$ such that $f_n_k(x) ge cos(pi/4)$ if $k$ is odd and $le -cos(pi/4)$ if $k$ is even. All you need to do is choose the base-$4$ digits of $x$ correctly.
EDIT: Any choice of the first $n$ digits after the "decimal" point leaves $x$ in an interval of length $4^-n$ on which $f_n$ goes either from $1$ to $-1$ or $-1$ to $1$. Choose the $n+1$'th digit to be $0$ or $3$ to make $f_n(x) ge cos(pi/4)$ or $le -cos(pi/4)$ (which is which depends on the previous choices of digits). For example, if you choose the first $3$ digits to be $0,3,0$, that says $x$ is between $0 times 4^-1 + 3 times 4^-2 + 0 times 4^-3 = 3/16$ and $0 times 4^-1 + 3 times 4^-2 + 1 times 4^-3 = 13/64$,
say $x = 3/16 + 4^-3 t$, $0 le t le 1$. On this interval, $f_3(x) = cos(4^3 pi x) = cos(pi t)$ goes from $1$ at $t=0$ to $-1$ at $t=1$. If you want $f_3(x) ge cos(pi/4)$, choose the $4$'th digit to be $0$, if you want $f_3(x) le - cos(pi/4)$ choose the $4$'th digit to be $1$.
answered Jun 4 at 21:31
Robert IsraelRobert Israel
338k23232490
$endgroup$
No, there does not. Consider $f_n(x) = cos(4^n pi x)$ on $[0,1]$ (with range in $[-1,1]$ not $[0,1]$, but you can transform it). Then for any subsequence $f_n_k$, there is
some $x in [0,1]$ such that $f_n_k(x) ge cos(pi/4)$ if $k$ is odd and $le -cos(pi/4)$ if $k$ is even. All you need to do is choose the base-$4$ digits of $x$ correctly.
EDIT: Any choice of the first $n$ digits after the "decimal" point leaves $x$ in an interval of length $4^-n$ on which $f_n$ goes either from $1$ to $-1$ or $-1$ to $1$. Choose the $n+1$'th digit to be $0$ or $3$ to make $f_n(x) ge cos(pi/4)$ or $le -cos(pi/4)$ (which is which depends on the previous choices of digits). For example, if you choose the first $3$ digits to be $0,3,0$, that says $x$ is between $0 times 4^-1 + 3 times 4^-2 + 0 times 4^-3 = 3/16$ and $0 times 4^-1 + 3 times 4^-2 + 1 times 4^-3 = 13/64$,
say $x = 3/16 + 4^-3 t$, $0 le t le 1$. On this interval, $f_3(x) = cos(4^3 pi x) = cos(pi t)$ goes from $1$ at $t=0$ to $-1$ at $t=1$. If you want $f_3(x) ge cos(pi/4)$, choose the $4$'th digit to be $0$, if you want $f_3(x) le - cos(pi/4)$ choose the $4$'th digit to be $1$.
answered Jun 4 at 21:31
Robert IsraelRobert Israel
338k23232490
edited Jun 4 at 23:41
answered Jun 4 at 21:31
Robert IsraelRobert Israel
338k23232490
answered Jun 4 at 21:31
Robert IsraelRobert Israel
338k23232490
answered Jun 4 at 21:31
Robert IsraelRobert Israel
338k23232490
338k23232490
$begingroup$
What do mean by "choose the base-$4$ digits of $x$ correctly"? Could you illustrate it with an example please? For instance the case where $n_k=k$. How do you construct $x$? Thanks.
$endgroup$
– Dog_69
Jun 4 at 22:45
add a comment |
$begingroup$
What do mean by "choose the base-$4$ digits of $x$ correctly"? Could you illustrate it with an example please? For instance the case where $n_k=k$. How do you construct $x$? Thanks.
$endgroup$
– Dog_69
Jun 4 at 22:45
$begingroup$
What do mean by "choose the base-$4$ digits of $x$ correctly"? Could you illustrate it with an example please? For instance the case where $n_k=k$. How do you construct $x$? Thanks.
$endgroup$
– Dog_69
Jun 4 at 22:45
$begingroup$
What do mean by "choose the base-$4$ digits of $x$ correctly"? Could you illustrate it with an example please? For instance the case where $n_k=k$. How do you construct $x$? Thanks.
$endgroup$
– Dog_69
Jun 4 at 22:45
add a comment |
$begingroup$
A similar example using more machinery hence perhaps less tricky in the details: Let $X=[0,2pi]$, $f_n(t)=cos(nt)$. If $f_n_k(x)to f(x)$ for (almost) every $x$ then Dominated Convergence shows that $||f_n_k-f||_2to0$; this is impossible by orthogonality (for example, $||f_n-f_m||_2^2=2pi$.)
answered Jun 4 at 23:35
David C. UllrichDavid C. Ullrich
63.1k44298
$endgroup$
add a comment |
$begingroup$
A similar example using more machinery hence perhaps less tricky in the details: Let $X=[0,2pi]$, $f_n(t)=cos(nt)$. If $f_n_k(x)to f(x)$ for (almost) every $x$ then Dominated Convergence shows that $||f_n_k-f||_2to0$; this is impossible by orthogonality (for example, $||f_n-f_m||_2^2=2pi$.)
answered Jun 4 at 23:35
David C. UllrichDavid C. Ullrich
63.1k44298
$endgroup$
add a comment |
$begingroup$
A similar example using more machinery hence perhaps less tricky in the details: Let $X=[0,2pi]$, $f_n(t)=cos(nt)$. If $f_n_k(x)to f(x)$ for (almost) every $x$ then Dominated Convergence shows that $||f_n_k-f||_2to0$; this is impossible by orthogonality (for example, $||f_n-f_m||_2^2=2pi$.)
answered Jun 4 at 23:35
David C. UllrichDavid C. Ullrich
63.1k44298
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A similar example using more machinery hence perhaps less tricky in the details: Let $X=[0,2pi]$, $f_n(t)=cos(nt)$. If $f_n_k(x)to f(x)$ for (almost) every $x$ then Dominated Convergence shows that $||f_n_k-f||_2to0$; this is impossible by orthogonality (for example, $||f_n-f_m||_2^2=2pi$.)
answered Jun 4 at 23:35
David C. UllrichDavid C. Ullrich
63.1k44298
answered Jun 4 at 23:35
David C. UllrichDavid C. Ullrich
63.1k44298
answered Jun 4 at 23:35
David C. UllrichDavid C. Ullrich
63.1k44298
answered Jun 4 at 23:35
David C. UllrichDavid C. Ullrich
63.1k44298
63.1k44298
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$X$ has a countable dense subset. Perhaps you could define $f$ there first?
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– copper.hat
Jun 4 at 21:26
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@copper.hat Yes, that’s a good idea for a first step. If $x^1,x^2,ldots$ is dense, then, by Cantor’s diagonal method, there is a subsequence such that $f_m_k(x^j)$ converges as $ktoinfty$ for every $jinmathbb N$. Using this result, the next step would be showing that $f_m_k(x)$ is Cauchy in $[0,1]$ for each $xin X$, but it seems tricky, as convergence at the $x^j$s need not be uniform.
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– triple_sec
Jun 4 at 21:28
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Robert has a nice counterexample below.
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– copper.hat
Jun 4 at 21:32
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If functions $f_n:mathbbRrightarrow [0,1]$ are nondecreasing in x for each n, then (regardless of continuity or behavior with n) your argument of showing convergence over a subsequence on a countable but dense subset can be used to prove convergence everywhere in $mathbbR$.
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– Michael
Jun 5 at 0:07
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The same does not hold for functions $f_n:mathbbR^2rightarrow [0,1]$ that are entrywise nondecreasing as we can have discontinuity across a diagonal and then adapt one of the 1-d counterexamples given below on that diagonal.
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– Michael
Jun 5 at 0:29