Mixing
Let $X$ be a compact metric space. Prove that $C[X]$ contains a countable subset that separates points.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
My proof seems way too trivial, so I am really doubting it.
Equip $X$ with any norm $|cdot|_X$ (note that $|cdot|_X : X rightarrow Bbb R$ is continuous).
Since $Bbb Q setminus 0$ is countable, we may enumerate it; i.e. $Bbb Q setminus 0 = q_1, q_2, ldots $.
Then define $f_n: Xto mathbbR$ by $f_n(x) = q_n|x|_X, forall n in Bbb N$.
Each $f_n$ is continuous on X as $|cdot|_X$ is continuous.
Take $x,y in X$ such that $x neq y$. Then:
$$|f_n(x) - f_n(y)| = |q_n| cdot||x|_X - |y|_X| > 0$$ since $|x|_X neq |y|_X$ since $x neq y$ (and since $q_n neq 0, forall n in Bbb N$).
Thus, $forall x,y in X$ for which $x neq y, exists n in Bbb N$ such that $f_n(x) neq f_n(y)$. So $f_n_n in Bbb N$ separates points.
If I am way off the mark, hints only please!
real-analysis metric-spaces
edited Jul 20 at 23:18
mechanodroid
22.2k52041
asked Jul 20 at 22:59
Bojack Horseman
1026
add a comment |Â
up vote
1
down vote
favorite
My proof seems way too trivial, so I am really doubting it.
Equip $X$ with any norm $|cdot|_X$ (note that $|cdot|_X : X rightarrow Bbb R$ is continuous).
Since $Bbb Q setminus 0$ is countable, we may enumerate it; i.e. $Bbb Q setminus 0 = q_1, q_2, ldots $.
Then define $f_n: Xto mathbbR$ by $f_n(x) = q_n|x|_X, forall n in Bbb N$.
Each $f_n$ is continuous on X as $|cdot|_X$ is continuous.
Take $x,y in X$ such that $x neq y$. Then:
$$|f_n(x) - f_n(y)| = |q_n| cdot||x|_X - |y|_X| > 0$$ since $|x|_X neq |y|_X$ since $x neq y$ (and since $q_n neq 0, forall n in Bbb N$).
Thus, $forall x,y in X$ for which $x neq y, exists n in Bbb N$ such that $f_n(x) neq f_n(y)$. So $f_n_n in Bbb N$ separates points.
If I am way off the mark, hints only please!
real-analysis metric-spaces
edited Jul 20 at 23:18
mechanodroid
22.2k52041
asked Jul 20 at 22:59
Bojack Horseman
1026
If $x ne y$ it can still be $|x| = |y|$.
– mechanodroid
Jul 20 at 23:03
If $X$ is not a vector space (or contained within one), then we can't equip it with a norm
– Omnomnomnom
Jul 20 at 23:05
Thank you, I knew I overlooked something silly.
– Bojack Horseman
Jul 20 at 23:06
It's not a deep result so don't worry about the complexity of the proof
– DanielWainfleet
Jul 21 at 4:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My proof seems way too trivial, so I am really doubting it.
Equip $X$ with any norm $|cdot|_X$ (note that $|cdot|_X : X rightarrow Bbb R$ is continuous).
Since $Bbb Q setminus 0$ is countable, we may enumerate it; i.e. $Bbb Q setminus 0 = q_1, q_2, ldots $.
Then define $f_n: Xto mathbbR$ by $f_n(x) = q_n|x|_X, forall n in Bbb N$.
Each $f_n$ is continuous on X as $|cdot|_X$ is continuous.
Take $x,y in X$ such that $x neq y$. Then:
$$|f_n(x) - f_n(y)| = |q_n| cdot||x|_X - |y|_X| > 0$$ since $|x|_X neq |y|_X$ since $x neq y$ (and since $q_n neq 0, forall n in Bbb N$).
Thus, $forall x,y in X$ for which $x neq y, exists n in Bbb N$ such that $f_n(x) neq f_n(y)$. So $f_n_n in Bbb N$ separates points.
If I am way off the mark, hints only please!
real-analysis metric-spaces
edited Jul 20 at 23:18
mechanodroid
22.2k52041
asked Jul 20 at 22:59
Bojack Horseman
1026
My proof seems way too trivial, so I am really doubting it.
Equip $X$ with any norm $|cdot|_X$ (note that $|cdot|_X : X rightarrow Bbb R$ is continuous).
Since $Bbb Q setminus 0$ is countable, we may enumerate it; i.e. $Bbb Q setminus 0 = q_1, q_2, ldots $.
Then define $f_n: Xto mathbbR$ by $f_n(x) = q_n|x|_X, forall n in Bbb N$.
Each $f_n$ is continuous on X as $|cdot|_X$ is continuous.
Take $x,y in X$ such that $x neq y$. Then:
$$|f_n(x) - f_n(y)| = |q_n| cdot||x|_X - |y|_X| > 0$$ since $|x|_X neq |y|_X$ since $x neq y$ (and since $q_n neq 0, forall n in Bbb N$).
Thus, $forall x,y in X$ for which $x neq y, exists n in Bbb N$ such that $f_n(x) neq f_n(y)$. So $f_n_n in Bbb N$ separates points.
If I am way off the mark, hints only please!
real-analysis metric-spaces
edited Jul 20 at 23:18
mechanodroid
22.2k52041
asked Jul 20 at 22:59
Bojack Horseman
1026
edited Jul 20 at 23:18
mechanodroid
22.2k52041
edited Jul 20 at 23:18
mechanodroid
22.2k52041
edited Jul 20 at 23:18
mechanodroid
22.2k52041
22.2k52041
asked Jul 20 at 22:59
Bojack Horseman
1026
asked Jul 20 at 22:59
Bojack Horseman
1026
asked Jul 20 at 22:59
Bojack Horseman
1026
1026
If $x ne y$ it can still be $|x| = |y|$.
– mechanodroid
Jul 20 at 23:03
If $X$ is not a vector space (or contained within one), then we can't equip it with a norm
– Omnomnomnom
Jul 20 at 23:05
Thank you, I knew I overlooked something silly.
– Bojack Horseman
Jul 20 at 23:06
It's not a deep result so don't worry about the complexity of the proof
– DanielWainfleet
Jul 21 at 4:37
add a comment |Â
If $x ne y$ it can still be $|x| = |y|$.
– mechanodroid
Jul 20 at 23:03
If $X$ is not a vector space (or contained within one), then we can't equip it with a norm
– Omnomnomnom
Jul 20 at 23:05
Thank you, I knew I overlooked something silly.
– Bojack Horseman
Jul 20 at 23:06
It's not a deep result so don't worry about the complexity of the proof
– DanielWainfleet
Jul 21 at 4:37
If $x ne y$ it can still be $|x| = |y|$.
– mechanodroid
Jul 20 at 23:03
If $x ne y$ it can still be $|x| = |y|$.
– mechanodroid
Jul 20 at 23:03
If $X$ is not a vector space (or contained within one), then we can't equip it with a norm
– Omnomnomnom
Jul 20 at 23:05
If $X$ is not a vector space (or contained within one), then we can't equip it with a norm
– Omnomnomnom
Jul 20 at 23:05
Thank you, I knew I overlooked something silly.
– Bojack Horseman
Jul 20 at 23:06
Thank you, I knew I overlooked something silly.
– Bojack Horseman
Jul 20 at 23:06
It's not a deep result so don't worry about the complexity of the proof
– DanielWainfleet
Jul 21 at 4:37
It's not a deep result so don't worry about the complexity of the proof
– DanielWainfleet
Jul 21 at 4:37
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
A compact metric space is separable so let $D$ be a countable dense subset of $(X,d)$.
Define $f_q in C(X)$ as $f_q(x) = d(x,q)$ for $q in D$.
If $x, y in X$ with $x ne y$ then pick a sequence $(q_n)_n$ in $D$ which converges to $x$. Then $(q_n)_n$ doesn't converge to $y$ so there exists $varepsilon > 0$ and a subsequence $(q_p(n))_n$ such that $d(y, q_p(n)) ge varepsilon$. There exists $n in mathbbN$ such that $d(x, q_p(n)) < varepsilon$ so $f_q_p(n)(x) < varepsilon$ and $f_q_p(n)(y) ge varepsilon$. Hence $f_q_p(n)(x) ne f_q_p(n)(y)$.
Therefore $f_q_q in D$ is a countable subset of $C(X)$ which separates points.
answered Jul 20 at 23:15
mechanodroid
22.2k52041
1
As usual I was about to post a much more complicated solution............................+1
– DanielWainfleet
Jul 21 at 4:43
add a comment |Â
up vote
1
down vote
It is very wrong to say that $$xneq yimplies|x|neq |y|$$
Note that the way $f_n$ is defined, whether or not $f_n(x)=f_n(y)$ does not actually depend on $n$ so it is unlikely that this family will be useful.
Finally, and that's actually a hint, you haven't used the fact that $X$ is compact.
answered Jul 20 at 23:05
Arnaud Mortier
19k22159
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
A compact metric space is separable so let $D$ be a countable dense subset of $(X,d)$.
Define $f_q in C(X)$ as $f_q(x) = d(x,q)$ for $q in D$.
If $x, y in X$ with $x ne y$ then pick a sequence $(q_n)_n$ in $D$ which converges to $x$. Then $(q_n)_n$ doesn't converge to $y$ so there exists $varepsilon > 0$ and a subsequence $(q_p(n))_n$ such that $d(y, q_p(n)) ge varepsilon$. There exists $n in mathbbN$ such that $d(x, q_p(n)) < varepsilon$ so $f_q_p(n)(x) < varepsilon$ and $f_q_p(n)(y) ge varepsilon$. Hence $f_q_p(n)(x) ne f_q_p(n)(y)$.
Therefore $f_q_q in D$ is a countable subset of $C(X)$ which separates points.
answered Jul 20 at 23:15
mechanodroid
22.2k52041
1
As usual I was about to post a much more complicated solution............................+1
– DanielWainfleet
Jul 21 at 4:43
add a comment |Â
up vote
3
down vote
accepted
A compact metric space is separable so let $D$ be a countable dense subset of $(X,d)$.
Define $f_q in C(X)$ as $f_q(x) = d(x,q)$ for $q in D$.
If $x, y in X$ with $x ne y$ then pick a sequence $(q_n)_n$ in $D$ which converges to $x$. Then $(q_n)_n$ doesn't converge to $y$ so there exists $varepsilon > 0$ and a subsequence $(q_p(n))_n$ such that $d(y, q_p(n)) ge varepsilon$. There exists $n in mathbbN$ such that $d(x, q_p(n)) < varepsilon$ so $f_q_p(n)(x) < varepsilon$ and $f_q_p(n)(y) ge varepsilon$. Hence $f_q_p(n)(x) ne f_q_p(n)(y)$.
Therefore $f_q_q in D$ is a countable subset of $C(X)$ which separates points.
answered Jul 20 at 23:15
mechanodroid
22.2k52041
1
As usual I was about to post a much more complicated solution............................+1
– DanielWainfleet
Jul 21 at 4:43
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
A compact metric space is separable so let $D$ be a countable dense subset of $(X,d)$.
Define $f_q in C(X)$ as $f_q(x) = d(x,q)$ for $q in D$.
If $x, y in X$ with $x ne y$ then pick a sequence $(q_n)_n$ in $D$ which converges to $x$. Then $(q_n)_n$ doesn't converge to $y$ so there exists $varepsilon > 0$ and a subsequence $(q_p(n))_n$ such that $d(y, q_p(n)) ge varepsilon$. There exists $n in mathbbN$ such that $d(x, q_p(n)) < varepsilon$ so $f_q_p(n)(x) < varepsilon$ and $f_q_p(n)(y) ge varepsilon$. Hence $f_q_p(n)(x) ne f_q_p(n)(y)$.
Therefore $f_q_q in D$ is a countable subset of $C(X)$ which separates points.
answered Jul 20 at 23:15
mechanodroid
22.2k52041
A compact metric space is separable so let $D$ be a countable dense subset of $(X,d)$.
Define $f_q in C(X)$ as $f_q(x) = d(x,q)$ for $q in D$.
If $x, y in X$ with $x ne y$ then pick a sequence $(q_n)_n$ in $D$ which converges to $x$. Then $(q_n)_n$ doesn't converge to $y$ so there exists $varepsilon > 0$ and a subsequence $(q_p(n))_n$ such that $d(y, q_p(n)) ge varepsilon$. There exists $n in mathbbN$ such that $d(x, q_p(n)) < varepsilon$ so $f_q_p(n)(x) < varepsilon$ and $f_q_p(n)(y) ge varepsilon$. Hence $f_q_p(n)(x) ne f_q_p(n)(y)$.
Therefore $f_q_q in D$ is a countable subset of $C(X)$ which separates points.
answered Jul 20 at 23:15
mechanodroid
22.2k52041
answered Jul 20 at 23:15
mechanodroid
22.2k52041
answered Jul 20 at 23:15
mechanodroid
22.2k52041
answered Jul 20 at 23:15
mechanodroid
22.2k52041
22.2k52041
1
As usual I was about to post a much more complicated solution............................+1
– DanielWainfleet
Jul 21 at 4:43
add a comment |Â
1
As usual I was about to post a much more complicated solution............................+1
– DanielWainfleet
Jul 21 at 4:43
1
1
As usual I was about to post a much more complicated solution............................+1
– DanielWainfleet
Jul 21 at 4:43
As usual I was about to post a much more complicated solution............................+1
– DanielWainfleet
Jul 21 at 4:43
add a comment |Â
up vote
1
down vote
It is very wrong to say that $$xneq yimplies|x|neq |y|$$
Note that the way $f_n$ is defined, whether or not $f_n(x)=f_n(y)$ does not actually depend on $n$ so it is unlikely that this family will be useful.
Finally, and that's actually a hint, you haven't used the fact that $X$ is compact.
answered Jul 20 at 23:05
Arnaud Mortier
19k22159
add a comment |Â
up vote
1
down vote
It is very wrong to say that $$xneq yimplies|x|neq |y|$$
Note that the way $f_n$ is defined, whether or not $f_n(x)=f_n(y)$ does not actually depend on $n$ so it is unlikely that this family will be useful.
Finally, and that's actually a hint, you haven't used the fact that $X$ is compact.
answered Jul 20 at 23:05
Arnaud Mortier
19k22159
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is very wrong to say that $$xneq yimplies|x|neq |y|$$
Note that the way $f_n$ is defined, whether or not $f_n(x)=f_n(y)$ does not actually depend on $n$ so it is unlikely that this family will be useful.
Finally, and that's actually a hint, you haven't used the fact that $X$ is compact.
answered Jul 20 at 23:05
Arnaud Mortier
19k22159
It is very wrong to say that $$xneq yimplies|x|neq |y|$$
Note that the way $f_n$ is defined, whether or not $f_n(x)=f_n(y)$ does not actually depend on $n$ so it is unlikely that this family will be useful.
Finally, and that's actually a hint, you haven't used the fact that $X$ is compact.
answered Jul 20 at 23:05
Arnaud Mortier
19k22159
answered Jul 20 at 23:05
Arnaud Mortier
19k22159
answered Jul 20 at 23:05
Arnaud Mortier
19k22159
answered Jul 20 at 23:05
Arnaud Mortier
19k22159
19k22159
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858089%2flet-x-be-a-compact-metric-space-prove-that-cx-contains-a-countable-subset%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
If $x ne y$ it can still be $|x| = |y|$.
– mechanodroid
Jul 20 at 23:03
If $X$ is not a vector space (or contained within one), then we can't equip it with a norm
– Omnomnomnom
Jul 20 at 23:05
Thank you, I knew I overlooked something silly.
– Bojack Horseman
Jul 20 at 23:06
It's not a deep result so don't worry about the complexity of the proof
– DanielWainfleet
Jul 21 at 4:37