Mixing
Does the set of Continuous functions always separate points and vanish nowhere?
$begingroup$
Let $X$ be an arbitrary metric space. Does $C(X;mathbb{R})$ always separate points and vanish nowhere?
I think this is true because the functions $f(x) = x$ and $g(x) = 1$ are both in $C(X;mathbb{R})$ and separate points and does not vanish.
Thank you for the help.
metric-spaces
edited Feb 1 at 13:30
David C. Ullrich
61.7k44095
asked Feb 1 at 5:19
MatthieuMatthieu
1007
$endgroup$
add a comment |
$begingroup$
Let $X$ be an arbitrary metric space. Does $C(X;mathbb{R})$ always separate points and vanish nowhere?
I think this is true because the functions $f(x) = x$ and $g(x) = 1$ are both in $C(X;mathbb{R})$ and separate points and does not vanish.
Thank you for the help.
metric-spaces
edited Feb 1 at 13:30
David C. Ullrich
61.7k44095
asked Feb 1 at 5:19
MatthieuMatthieu
1007
$endgroup$
1
$begingroup$
$f(x)=x$ is not an element of $C(X;mathbb R)$ for a general metric space $X$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 6:04
$begingroup$
@KaviRamaMurthy Yes, I now understand this. Thank you.
$endgroup$
– Matthieu
Feb 1 at 6:16
add a comment |
$begingroup$
Let $X$ be an arbitrary metric space. Does $C(X;mathbb{R})$ always separate points and vanish nowhere?
I think this is true because the functions $f(x) = x$ and $g(x) = 1$ are both in $C(X;mathbb{R})$ and separate points and does not vanish.
Thank you for the help.
metric-spaces
edited Feb 1 at 13:30
David C. Ullrich
61.7k44095
asked Feb 1 at 5:19
MatthieuMatthieu
1007
$endgroup$
Let $X$ be an arbitrary metric space. Does $C(X;mathbb{R})$ always separate points and vanish nowhere?
I think this is true because the functions $f(x) = x$ and $g(x) = 1$ are both in $C(X;mathbb{R})$ and separate points and does not vanish.
Thank you for the help.
metric-spaces
metric-spaces
edited Feb 1 at 13:30
David C. Ullrich
61.7k44095
asked Feb 1 at 5:19
MatthieuMatthieu
1007
edited Feb 1 at 13:30
David C. Ullrich
61.7k44095
asked Feb 1 at 5:19
MatthieuMatthieu
1007
edited Feb 1 at 13:30
David C. Ullrich
61.7k44095
edited Feb 1 at 13:30
David C. Ullrich
61.7k44095
edited Feb 1 at 13:30
David C. Ullrich
61.7k44095
61.7k44095
asked Feb 1 at 5:19
MatthieuMatthieu
1007
asked Feb 1 at 5:19
MatthieuMatthieu
1007
asked Feb 1 at 5:19
MatthieuMatthieu
1007
1007
1
$begingroup$
$f(x)=x$ is not an element of $C(X;mathbb R)$ for a general metric space $X$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 6:04
$begingroup$
@KaviRamaMurthy Yes, I now understand this. Thank you.
$endgroup$
– Matthieu
Feb 1 at 6:16
add a comment |
1
$begingroup$
$f(x)=x$ is not an element of $C(X;mathbb R)$ for a general metric space $X$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 6:04
$begingroup$
@KaviRamaMurthy Yes, I now understand this. Thank you.
$endgroup$
– Matthieu
Feb 1 at 6:16
1
1
$begingroup$
$f(x)=x$ is not an element of $C(X;mathbb R)$ for a general metric space $X$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 6:04
$begingroup$
$f(x)=x$ is not an element of $C(X;mathbb R)$ for a general metric space $X$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 6:04
$begingroup$
@KaviRamaMurthy Yes, I now understand this. Thank you.
$endgroup$
– Matthieu
Feb 1 at 6:16
$begingroup$
@KaviRamaMurthy Yes, I now understand this. Thank you.
$endgroup$
– Matthieu
Feb 1 at 6:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, these two functions do suffice to show that for $Xsubseteq mathbb{R}$. Or use $f$ and $g(x)=x+1$, as $f$ and $g$ never simultaneously are $0$. But the definition of $f$ assumes all points of $X$ are reals.
For general metric $X$ consider using $f_p(x) = d(x,p)$ for $p in X$ as continuous functions, and also $g$.
answered Feb 1 at 5:30
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
would you mind elaborating on your last statement for a general metric?
$endgroup$
– Matthieu
Feb 1 at 5:33
$begingroup$
@MatteoLepur If $x neq y$, what are $f_x(x)$ and $f_x(y)$?
$endgroup$
– Henno Brandsma
Feb 1 at 5:34
$begingroup$
$f_x(x) = 0$ and $f_x(y)$ $not=$ $0$. I don't see how this helps us. Thanks for the help.
$endgroup$
– Matthieu
Feb 1 at 5:38
$begingroup$
@hennobrandsma you're saying since there are at least two different points in the metric space then $ f_p(x)= d(x,y) $ is not equal to zero for some x in X, do I understand that correctly?
$endgroup$
– Kaan Yolsever
Feb 1 at 5:39
2
$begingroup$
@KaanYolsever search for “metric continuous” on this site. Plenty of answers. But basically the triangle inequality implies $|f_p(x) - f_p(y) le d(x,y)$ which implies continuity right away.
$endgroup$
– Henno Brandsma
Feb 1 at 6:04
|
show 5 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Yes, these two functions do suffice to show that for $Xsubseteq mathbb{R}$. Or use $f$ and $g(x)=x+1$, as $f$ and $g$ never simultaneously are $0$. But the definition of $f$ assumes all points of $X$ are reals.
For general metric $X$ consider using $f_p(x) = d(x,p)$ for $p in X$ as continuous functions, and also $g$.
answered Feb 1 at 5:30
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
would you mind elaborating on your last statement for a general metric?
$endgroup$
– Matthieu
Feb 1 at 5:33
$begingroup$
@MatteoLepur If $x neq y$, what are $f_x(x)$ and $f_x(y)$?
$endgroup$
– Henno Brandsma
Feb 1 at 5:34
$begingroup$
$f_x(x) = 0$ and $f_x(y)$ $not=$ $0$. I don't see how this helps us. Thanks for the help.
$endgroup$
– Matthieu
Feb 1 at 5:38
$begingroup$
@hennobrandsma you're saying since there are at least two different points in the metric space then $ f_p(x)= d(x,y) $ is not equal to zero for some x in X, do I understand that correctly?
$endgroup$
– Kaan Yolsever
Feb 1 at 5:39
2
$begingroup$
@KaanYolsever search for “metric continuous” on this site. Plenty of answers. But basically the triangle inequality implies $|f_p(x) - f_p(y) le d(x,y)$ which implies continuity right away.
$endgroup$
– Henno Brandsma
Feb 1 at 6:04
|
show 5 more comments
$begingroup$
Yes, these two functions do suffice to show that for $Xsubseteq mathbb{R}$. Or use $f$ and $g(x)=x+1$, as $f$ and $g$ never simultaneously are $0$. But the definition of $f$ assumes all points of $X$ are reals.
For general metric $X$ consider using $f_p(x) = d(x,p)$ for $p in X$ as continuous functions, and also $g$.
answered Feb 1 at 5:30
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
would you mind elaborating on your last statement for a general metric?
$endgroup$
– Matthieu
Feb 1 at 5:33
$begingroup$
@MatteoLepur If $x neq y$, what are $f_x(x)$ and $f_x(y)$?
$endgroup$
– Henno Brandsma
Feb 1 at 5:34
$begingroup$
$f_x(x) = 0$ and $f_x(y)$ $not=$ $0$. I don't see how this helps us. Thanks for the help.
$endgroup$
– Matthieu
Feb 1 at 5:38
$begingroup$
@hennobrandsma you're saying since there are at least two different points in the metric space then $ f_p(x)= d(x,y) $ is not equal to zero for some x in X, do I understand that correctly?
$endgroup$
– Kaan Yolsever
Feb 1 at 5:39
2
$begingroup$
@KaanYolsever search for “metric continuous” on this site. Plenty of answers. But basically the triangle inequality implies $|f_p(x) - f_p(y) le d(x,y)$ which implies continuity right away.
$endgroup$
– Henno Brandsma
Feb 1 at 6:04
|
show 5 more comments
$begingroup$
Yes, these two functions do suffice to show that for $Xsubseteq mathbb{R}$. Or use $f$ and $g(x)=x+1$, as $f$ and $g$ never simultaneously are $0$. But the definition of $f$ assumes all points of $X$ are reals.
For general metric $X$ consider using $f_p(x) = d(x,p)$ for $p in X$ as continuous functions, and also $g$.
answered Feb 1 at 5:30
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
Yes, these two functions do suffice to show that for $Xsubseteq mathbb{R}$. Or use $f$ and $g(x)=x+1$, as $f$ and $g$ never simultaneously are $0$. But the definition of $f$ assumes all points of $X$ are reals.
For general metric $X$ consider using $f_p(x) = d(x,p)$ for $p in X$ as continuous functions, and also $g$.
answered Feb 1 at 5:30
Henno BrandsmaHenno Brandsma
116k349127
edited Feb 1 at 9:01
answered Feb 1 at 5:30
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 5:30
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 5:30
Henno BrandsmaHenno Brandsma
116k349127
116k349127
$begingroup$
would you mind elaborating on your last statement for a general metric?
$endgroup$
– Matthieu
Feb 1 at 5:33
$begingroup$
@MatteoLepur If $x neq y$, what are $f_x(x)$ and $f_x(y)$?
$endgroup$
– Henno Brandsma
Feb 1 at 5:34
$begingroup$
$f_x(x) = 0$ and $f_x(y)$ $not=$ $0$. I don't see how this helps us. Thanks for the help.
$endgroup$
– Matthieu
Feb 1 at 5:38
$begingroup$
@hennobrandsma you're saying since there are at least two different points in the metric space then $ f_p(x)= d(x,y) $ is not equal to zero for some x in X, do I understand that correctly?
$endgroup$
– Kaan Yolsever
Feb 1 at 5:39
2
$begingroup$
@KaanYolsever search for “metric continuous” on this site. Plenty of answers. But basically the triangle inequality implies $|f_p(x) - f_p(y) le d(x,y)$ which implies continuity right away.
$endgroup$
– Henno Brandsma
Feb 1 at 6:04
|
show 5 more comments
$begingroup$
would you mind elaborating on your last statement for a general metric?
$endgroup$
– Matthieu
Feb 1 at 5:33
$begingroup$
@MatteoLepur If $x neq y$, what are $f_x(x)$ and $f_x(y)$?
$endgroup$
– Henno Brandsma
Feb 1 at 5:34
$begingroup$
$f_x(x) = 0$ and $f_x(y)$ $not=$ $0$. I don't see how this helps us. Thanks for the help.
$endgroup$
– Matthieu
Feb 1 at 5:38
$begingroup$
@hennobrandsma you're saying since there are at least two different points in the metric space then $ f_p(x)= d(x,y) $ is not equal to zero for some x in X, do I understand that correctly?
$endgroup$
– Kaan Yolsever
Feb 1 at 5:39
2
$begingroup$
@KaanYolsever search for “metric continuous” on this site. Plenty of answers. But basically the triangle inequality implies $|f_p(x) - f_p(y) le d(x,y)$ which implies continuity right away.
$endgroup$
– Henno Brandsma
Feb 1 at 6:04
$begingroup$
would you mind elaborating on your last statement for a general metric?
$endgroup$
– Matthieu
Feb 1 at 5:33
$begingroup$
would you mind elaborating on your last statement for a general metric?
$endgroup$
– Matthieu
Feb 1 at 5:33
$begingroup$
@MatteoLepur If $x neq y$, what are $f_x(x)$ and $f_x(y)$?
$endgroup$
– Henno Brandsma
Feb 1 at 5:34
$begingroup$
@MatteoLepur If $x neq y$, what are $f_x(x)$ and $f_x(y)$?
$endgroup$
– Henno Brandsma
Feb 1 at 5:34
$begingroup$
$f_x(x) = 0$ and $f_x(y)$ $not=$ $0$. I don't see how this helps us. Thanks for the help.
$endgroup$
– Matthieu
Feb 1 at 5:38
$begingroup$
$f_x(x) = 0$ and $f_x(y)$ $not=$ $0$. I don't see how this helps us. Thanks for the help.
$endgroup$
– Matthieu
Feb 1 at 5:38
$begingroup$
@hennobrandsma you're saying since there are at least two different points in the metric space then $ f_p(x)= d(x,y) $ is not equal to zero for some x in X, do I understand that correctly?
$endgroup$
– Kaan Yolsever
Feb 1 at 5:39
$begingroup$
@hennobrandsma you're saying since there are at least two different points in the metric space then $ f_p(x)= d(x,y) $ is not equal to zero for some x in X, do I understand that correctly?
$endgroup$
– Kaan Yolsever
Feb 1 at 5:39
2
2
$begingroup$
@KaanYolsever search for “metric continuous” on this site. Plenty of answers. But basically the triangle inequality implies $|f_p(x) - f_p(y) le d(x,y)$ which implies continuity right away.
$endgroup$
– Henno Brandsma
Feb 1 at 6:04
$begingroup$
@KaanYolsever search for “metric continuous” on this site. Plenty of answers. But basically the triangle inequality implies $|f_p(x) - f_p(y) le d(x,y)$ which implies continuity right away.
$endgroup$
– Henno Brandsma
Feb 1 at 6:04
|
show 5 more comments
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1
$begingroup$
$f(x)=x$ is not an element of $C(X;mathbb R)$ for a general metric space $X$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 6:04
$begingroup$
@KaviRamaMurthy Yes, I now understand this. Thank you.
$endgroup$
– Matthieu
Feb 1 at 6:16