Mixing
Any continuous mapping on compact set is uniformly continous
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I've got some problems while I'm proving that statement.
I'm trying to show that with using RAA(it is PMA exercise Problem 10, chapter 4).
Here is my idea.
Suppose $f$ is continuous but not uniformly. Then there is some $varepsilon>0$ which satisfies that for any $delta>0$, there is $x,yin X$ such that
$$d(x,y)<deltaland d(f(x),f(y))gevarepsilon.$$
In other words, there are sequences ${p_n}$ and ${q_n}$ in $X$ so that for any $delta>0$ there is some $nge N$ such that
$$d(p_n,q_n)<deltaland d(f(p_n),f(q_n))>varepsilon.$$
Since $X$ is compact metric space, for any sequence in $X$ has a convergent subsequence. Let $p_{n_k}to ain X$ and $q_{n_k}to bin X$. To simplify let $p_{n_k}=a_k$ and $q_{n_k}=b_k$. Then
$$0le d_X(b_k,a)le d_X(b_k,a_k)+d_X(a_k,a).$$
Since both $d_X(b_k,a_k)$ and $d_X(a_k,a)$ converges to $0$ as $kto infty$, $d_X(b_k,a)to0$, i.e., $a=b$.
Since $f$ is continuous,
$$f(a)=limlimits_{xto a}f(x)=limlimits_{ktoinfty}f(a_k)=limlimits_{ktoinfty}f(b_k).$$
That is, for fixed $varepsilon>0$, there are $N_1>0$ and $N_2>0$ such that
$$kge N_1Longrightarrow d_Y(f(a),f(a_k))<frac{varepsilon}{2}$$
and
$$kge N_2Longrightarrow d_Y(f(a),f(b_k))<frac{varepsilon}{2}.$$
Hence, if $kgemax(N_1,N_2)$,
$$d_Y(f(a_k),f(b_k))le d_Y(f(a_k),f(a))+d_Y(f(a),f(b_k))<frac{varepsilon}{2}+frac{varepsilon}{2}=varepsilon.$$
Thus I got the result that $d_Y(f(p_{n_k}),f(q_{n_k}))<varepsilon$, but I think it is not sufficient. Even though $d_Y(f(p_{n_k}),f(q_{n_k}))<varepsilon$ for all $kge max(N_1,N_2)$, it does not guaratee that $d_Y(f(p_n),f(q_n))<varepsilon$ for all $nge N$ for some $N$.
Is there something I am missing? I cannot catch it.
analysis continuity uniform-continuity
asked Jan 27 at 15:32
Caenorhabditis ElegansCaenorhabditis Elegans
161
$endgroup$
add a comment |
$begingroup$
I've got some problems while I'm proving that statement.
I'm trying to show that with using RAA(it is PMA exercise Problem 10, chapter 4).
Here is my idea.
Suppose $f$ is continuous but not uniformly. Then there is some $varepsilon>0$ which satisfies that for any $delta>0$, there is $x,yin X$ such that
$$d(x,y)<deltaland d(f(x),f(y))gevarepsilon.$$
In other words, there are sequences ${p_n}$ and ${q_n}$ in $X$ so that for any $delta>0$ there is some $nge N$ such that
$$d(p_n,q_n)<deltaland d(f(p_n),f(q_n))>varepsilon.$$
Since $X$ is compact metric space, for any sequence in $X$ has a convergent subsequence. Let $p_{n_k}to ain X$ and $q_{n_k}to bin X$. To simplify let $p_{n_k}=a_k$ and $q_{n_k}=b_k$. Then
$$0le d_X(b_k,a)le d_X(b_k,a_k)+d_X(a_k,a).$$
Since both $d_X(b_k,a_k)$ and $d_X(a_k,a)$ converges to $0$ as $kto infty$, $d_X(b_k,a)to0$, i.e., $a=b$.
Since $f$ is continuous,
$$f(a)=limlimits_{xto a}f(x)=limlimits_{ktoinfty}f(a_k)=limlimits_{ktoinfty}f(b_k).$$
That is, for fixed $varepsilon>0$, there are $N_1>0$ and $N_2>0$ such that
$$kge N_1Longrightarrow d_Y(f(a),f(a_k))<frac{varepsilon}{2}$$
and
$$kge N_2Longrightarrow d_Y(f(a),f(b_k))<frac{varepsilon}{2}.$$
Hence, if $kgemax(N_1,N_2)$,
$$d_Y(f(a_k),f(b_k))le d_Y(f(a_k),f(a))+d_Y(f(a),f(b_k))<frac{varepsilon}{2}+frac{varepsilon}{2}=varepsilon.$$
Thus I got the result that $d_Y(f(p_{n_k}),f(q_{n_k}))<varepsilon$, but I think it is not sufficient. Even though $d_Y(f(p_{n_k}),f(q_{n_k}))<varepsilon$ for all $kge max(N_1,N_2)$, it does not guaratee that $d_Y(f(p_n),f(q_n))<varepsilon$ for all $nge N$ for some $N$.
Is there something I am missing? I cannot catch it.
analysis continuity uniform-continuity
asked Jan 27 at 15:32
Caenorhabditis ElegansCaenorhabditis Elegans
161
$endgroup$
$begingroup$
Related: math.stackexchange.com/questions/110573/…
$endgroup$
– Exp ikx
Jan 27 at 15:37
$begingroup$
@Expikx It will be useful, but I need to prove that via RAA, not directly.
$endgroup$
– Caenorhabditis Elegans
Jan 27 at 15:50
add a comment |
$begingroup$
I've got some problems while I'm proving that statement.
I'm trying to show that with using RAA(it is PMA exercise Problem 10, chapter 4).
Here is my idea.
Suppose $f$ is continuous but not uniformly. Then there is some $varepsilon>0$ which satisfies that for any $delta>0$, there is $x,yin X$ such that
$$d(x,y)<deltaland d(f(x),f(y))gevarepsilon.$$
In other words, there are sequences ${p_n}$ and ${q_n}$ in $X$ so that for any $delta>0$ there is some $nge N$ such that
$$d(p_n,q_n)<deltaland d(f(p_n),f(q_n))>varepsilon.$$
Since $X$ is compact metric space, for any sequence in $X$ has a convergent subsequence. Let $p_{n_k}to ain X$ and $q_{n_k}to bin X$. To simplify let $p_{n_k}=a_k$ and $q_{n_k}=b_k$. Then
$$0le d_X(b_k,a)le d_X(b_k,a_k)+d_X(a_k,a).$$
Since both $d_X(b_k,a_k)$ and $d_X(a_k,a)$ converges to $0$ as $kto infty$, $d_X(b_k,a)to0$, i.e., $a=b$.
Since $f$ is continuous,
$$f(a)=limlimits_{xto a}f(x)=limlimits_{ktoinfty}f(a_k)=limlimits_{ktoinfty}f(b_k).$$
That is, for fixed $varepsilon>0$, there are $N_1>0$ and $N_2>0$ such that
$$kge N_1Longrightarrow d_Y(f(a),f(a_k))<frac{varepsilon}{2}$$
and
$$kge N_2Longrightarrow d_Y(f(a),f(b_k))<frac{varepsilon}{2}.$$
Hence, if $kgemax(N_1,N_2)$,
$$d_Y(f(a_k),f(b_k))le d_Y(f(a_k),f(a))+d_Y(f(a),f(b_k))<frac{varepsilon}{2}+frac{varepsilon}{2}=varepsilon.$$
Thus I got the result that $d_Y(f(p_{n_k}),f(q_{n_k}))<varepsilon$, but I think it is not sufficient. Even though $d_Y(f(p_{n_k}),f(q_{n_k}))<varepsilon$ for all $kge max(N_1,N_2)$, it does not guaratee that $d_Y(f(p_n),f(q_n))<varepsilon$ for all $nge N$ for some $N$.
Is there something I am missing? I cannot catch it.
analysis continuity uniform-continuity
asked Jan 27 at 15:32
Caenorhabditis ElegansCaenorhabditis Elegans
161
$endgroup$
I've got some problems while I'm proving that statement.
I'm trying to show that with using RAA(it is PMA exercise Problem 10, chapter 4).
Here is my idea.
Suppose $f$ is continuous but not uniformly. Then there is some $varepsilon>0$ which satisfies that for any $delta>0$, there is $x,yin X$ such that
$$d(x,y)<deltaland d(f(x),f(y))gevarepsilon.$$
In other words, there are sequences ${p_n}$ and ${q_n}$ in $X$ so that for any $delta>0$ there is some $nge N$ such that
$$d(p_n,q_n)<deltaland d(f(p_n),f(q_n))>varepsilon.$$
Since $X$ is compact metric space, for any sequence in $X$ has a convergent subsequence. Let $p_{n_k}to ain X$ and $q_{n_k}to bin X$. To simplify let $p_{n_k}=a_k$ and $q_{n_k}=b_k$. Then
$$0le d_X(b_k,a)le d_X(b_k,a_k)+d_X(a_k,a).$$
Since both $d_X(b_k,a_k)$ and $d_X(a_k,a)$ converges to $0$ as $kto infty$, $d_X(b_k,a)to0$, i.e., $a=b$.
Since $f$ is continuous,
$$f(a)=limlimits_{xto a}f(x)=limlimits_{ktoinfty}f(a_k)=limlimits_{ktoinfty}f(b_k).$$
That is, for fixed $varepsilon>0$, there are $N_1>0$ and $N_2>0$ such that
$$kge N_1Longrightarrow d_Y(f(a),f(a_k))<frac{varepsilon}{2}$$
and
$$kge N_2Longrightarrow d_Y(f(a),f(b_k))<frac{varepsilon}{2}.$$
Hence, if $kgemax(N_1,N_2)$,
$$d_Y(f(a_k),f(b_k))le d_Y(f(a_k),f(a))+d_Y(f(a),f(b_k))<frac{varepsilon}{2}+frac{varepsilon}{2}=varepsilon.$$
Thus I got the result that $d_Y(f(p_{n_k}),f(q_{n_k}))<varepsilon$, but I think it is not sufficient. Even though $d_Y(f(p_{n_k}),f(q_{n_k}))<varepsilon$ for all $kge max(N_1,N_2)$, it does not guaratee that $d_Y(f(p_n),f(q_n))<varepsilon$ for all $nge N$ for some $N$.
Is there something I am missing? I cannot catch it.
analysis continuity uniform-continuity
analysis continuity uniform-continuity
asked Jan 27 at 15:32
Caenorhabditis ElegansCaenorhabditis Elegans
161
asked Jan 27 at 15:32
Caenorhabditis ElegansCaenorhabditis Elegans
161
asked Jan 27 at 15:32
Caenorhabditis ElegansCaenorhabditis Elegans
161
asked Jan 27 at 15:32
Caenorhabditis ElegansCaenorhabditis Elegans
161
asked Jan 27 at 15:32
Caenorhabditis ElegansCaenorhabditis Elegans
161
161
$begingroup$
Related: math.stackexchange.com/questions/110573/…
$endgroup$
– Exp ikx
Jan 27 at 15:37
$begingroup$
@Expikx It will be useful, but I need to prove that via RAA, not directly.
$endgroup$
– Caenorhabditis Elegans
Jan 27 at 15:50
add a comment |
$begingroup$
Related: math.stackexchange.com/questions/110573/…
$endgroup$
– Exp ikx
Jan 27 at 15:37
$begingroup$
@Expikx It will be useful, but I need to prove that via RAA, not directly.
$endgroup$
– Caenorhabditis Elegans
Jan 27 at 15:50
$begingroup$
Related: math.stackexchange.com/questions/110573/…
$endgroup$
– Exp ikx
Jan 27 at 15:37
$begingroup$
Related: math.stackexchange.com/questions/110573/…
$endgroup$
– Exp ikx
Jan 27 at 15:37
$begingroup$
@Expikx It will be useful, but I need to prove that via RAA, not directly.
$endgroup$
– Caenorhabditis Elegans
Jan 27 at 15:50
$begingroup$
@Expikx It will be useful, but I need to prove that via RAA, not directly.
$endgroup$
– Caenorhabditis Elegans
Jan 27 at 15:50
add a comment |
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$begingroup$
Related: math.stackexchange.com/questions/110573/…
$endgroup$
– Exp ikx
Jan 27 at 15:37
$begingroup$
@Expikx It will be useful, but I need to prove that via RAA, not directly.
$endgroup$
– Caenorhabditis Elegans
Jan 27 at 15:50