How can we show that $C_c^infty(mathbb R)$ strongly separates points?
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Let $C_b(mathbb R)$ denote the set of bounded continuous function from $mathbb R$ to $mathbb R$. We say that $Msubseteq C_b(mathbb R)$
separates points $:Leftrightarrow$ $$forall x,yinmathbb R:xne yRightarrowexists fin M:f(x)ne f(y)tag1$$
strongly separates points $:Leftrightarrow$ $$forall xinmathbb R,delta>0:exists kinmathbb N,left{f_1,ldots,f_kright}subseteq M:inf_{y:::d(x,y):ge:delta}max_{1le ile k}|f_i(x)-f_i(y)|>0tag2$$
How can we show that $C_c^infty(mathbb R)$ strongly separates points?
It's clear that $C_c^infty(mathbb R)$ separates points.
general-topology functional-analysis analysis
asked Nov 28 at 16:48
0xbadf00d
1,70241429
|
show 1 more comment
up vote
1
down vote
favorite
Let $C_b(mathbb R)$ denote the set of bounded continuous function from $mathbb R$ to $mathbb R$. We say that $Msubseteq C_b(mathbb R)$
separates points $:Leftrightarrow$ $$forall x,yinmathbb R:xne yRightarrowexists fin M:f(x)ne f(y)tag1$$
strongly separates points $:Leftrightarrow$ $$forall xinmathbb R,delta>0:exists kinmathbb N,left{f_1,ldots,f_kright}subseteq M:inf_{y:::d(x,y):ge:delta}max_{1le ile k}|f_i(x)-f_i(y)|>0tag2$$
How can we show that $C_c^infty(mathbb R)$ strongly separates points?
It's clear that $C_c^infty(mathbb R)$ separates points.
general-topology functional-analysis analysis
asked Nov 28 at 16:48
0xbadf00d
1,70241429
Do you mean $C_b^infty$ in the places you've written $C_c^infty$?
– Trevor Gunn
Nov 28 at 16:53
Do you mean $f_i$ instead of $h_i$?
– Paul Frost
Nov 28 at 16:55
@PaulFrost Sorry, fixed that.
– 0xbadf00d
Nov 28 at 16:56
@TrevorGunn No, I mean $C_c^infty$.
– 0xbadf00d
Nov 28 at 16:56
2
Cant you just take a single $fin C^infty_c(mathbb R)$ with $f(x)=1$, $mathrm{supp}(f)subset B(x,delta)$? Then you would have $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
– Federico
Nov 28 at 16:58
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $C_b(mathbb R)$ denote the set of bounded continuous function from $mathbb R$ to $mathbb R$. We say that $Msubseteq C_b(mathbb R)$
separates points $:Leftrightarrow$ $$forall x,yinmathbb R:xne yRightarrowexists fin M:f(x)ne f(y)tag1$$
strongly separates points $:Leftrightarrow$ $$forall xinmathbb R,delta>0:exists kinmathbb N,left{f_1,ldots,f_kright}subseteq M:inf_{y:::d(x,y):ge:delta}max_{1le ile k}|f_i(x)-f_i(y)|>0tag2$$
How can we show that $C_c^infty(mathbb R)$ strongly separates points?
It's clear that $C_c^infty(mathbb R)$ separates points.
general-topology functional-analysis analysis
asked Nov 28 at 16:48
0xbadf00d
1,70241429
Let $C_b(mathbb R)$ denote the set of bounded continuous function from $mathbb R$ to $mathbb R$. We say that $Msubseteq C_b(mathbb R)$
separates points $:Leftrightarrow$ $$forall x,yinmathbb R:xne yRightarrowexists fin M:f(x)ne f(y)tag1$$
strongly separates points $:Leftrightarrow$ $$forall xinmathbb R,delta>0:exists kinmathbb N,left{f_1,ldots,f_kright}subseteq M:inf_{y:::d(x,y):ge:delta}max_{1le ile k}|f_i(x)-f_i(y)|>0tag2$$
How can we show that $C_c^infty(mathbb R)$ strongly separates points?
It's clear that $C_c^infty(mathbb R)$ separates points.
general-topology functional-analysis analysis
general-topology functional-analysis analysis
asked Nov 28 at 16:48
0xbadf00d
1,70241429
asked Nov 28 at 16:48
0xbadf00d
1,70241429
edited Nov 28 at 16:56
asked Nov 28 at 16:48
0xbadf00d
1,70241429
asked Nov 28 at 16:48
0xbadf00d
1,70241429
asked Nov 28 at 16:48
0xbadf00d
1,70241429
1,70241429
Do you mean $C_b^infty$ in the places you've written $C_c^infty$?
– Trevor Gunn
Nov 28 at 16:53
Do you mean $f_i$ instead of $h_i$?
– Paul Frost
Nov 28 at 16:55
@PaulFrost Sorry, fixed that.
– 0xbadf00d
Nov 28 at 16:56
@TrevorGunn No, I mean $C_c^infty$.
– 0xbadf00d
Nov 28 at 16:56
2
Cant you just take a single $fin C^infty_c(mathbb R)$ with $f(x)=1$, $mathrm{supp}(f)subset B(x,delta)$? Then you would have $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
– Federico
Nov 28 at 16:58
|
show 1 more comment
Do you mean $C_b^infty$ in the places you've written $C_c^infty$?
– Trevor Gunn
Nov 28 at 16:53
Do you mean $f_i$ instead of $h_i$?
– Paul Frost
Nov 28 at 16:55
@PaulFrost Sorry, fixed that.
– 0xbadf00d
Nov 28 at 16:56
@TrevorGunn No, I mean $C_c^infty$.
– 0xbadf00d
Nov 28 at 16:56
2
Cant you just take a single $fin C^infty_c(mathbb R)$ with $f(x)=1$, $mathrm{supp}(f)subset B(x,delta)$? Then you would have $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
– Federico
Nov 28 at 16:58
Do you mean $C_b^infty$ in the places you've written $C_c^infty$?
– Trevor Gunn
Nov 28 at 16:53
Do you mean $C_b^infty$ in the places you've written $C_c^infty$?
– Trevor Gunn
Nov 28 at 16:53
Do you mean $f_i$ instead of $h_i$?
– Paul Frost
Nov 28 at 16:55
Do you mean $f_i$ instead of $h_i$?
– Paul Frost
Nov 28 at 16:55
@PaulFrost Sorry, fixed that.
– 0xbadf00d
Nov 28 at 16:56
@PaulFrost Sorry, fixed that.
– 0xbadf00d
Nov 28 at 16:56
@TrevorGunn No, I mean $C_c^infty$.
– 0xbadf00d
Nov 28 at 16:56
@TrevorGunn No, I mean $C_c^infty$.
– 0xbadf00d
Nov 28 at 16:56
2
2
Cant you just take a single $fin C^infty_c(mathbb R)$ with $f(x)=1$, $mathrm{supp}(f)subset B(x,delta)$? Then you would have $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
– Federico
Nov 28 at 16:58
Cant you just take a single $fin C^infty_c(mathbb R)$ with $f(x)=1$, $mathrm{supp}(f)subset B(x,delta)$? Then you would have $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
– Federico
Nov 28 at 16:58
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Given $xinmathbb R$ and $delta>0$, take a single function $fin C^infty_c(mathbb R)$ with $f(x)=1$ and $mathrm{supp}(f)subset B(x,delta)$.
Then $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
answered Nov 28 at 17:01
Federico
4,272512
Oops, way to simple.
– 0xbadf00d
Nov 28 at 17:10
Ah those lucky days when we can actually solve something! So satisfying... :-)
– Federico
Nov 28 at 17:11
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Given $xinmathbb R$ and $delta>0$, take a single function $fin C^infty_c(mathbb R)$ with $f(x)=1$ and $mathrm{supp}(f)subset B(x,delta)$.
Then $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
answered Nov 28 at 17:01
Federico
4,272512
Oops, way to simple.
– 0xbadf00d
Nov 28 at 17:10
Ah those lucky days when we can actually solve something! So satisfying... :-)
– Federico
Nov 28 at 17:11
add a comment |
up vote
2
down vote
accepted
Given $xinmathbb R$ and $delta>0$, take a single function $fin C^infty_c(mathbb R)$ with $f(x)=1$ and $mathrm{supp}(f)subset B(x,delta)$.
Then $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
answered Nov 28 at 17:01
Federico
4,272512
Oops, way to simple.
– 0xbadf00d
Nov 28 at 17:10
Ah those lucky days when we can actually solve something! So satisfying... :-)
– Federico
Nov 28 at 17:11
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Given $xinmathbb R$ and $delta>0$, take a single function $fin C^infty_c(mathbb R)$ with $f(x)=1$ and $mathrm{supp}(f)subset B(x,delta)$.
Then $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
answered Nov 28 at 17:01
Federico
4,272512
Given $xinmathbb R$ and $delta>0$, take a single function $fin C^infty_c(mathbb R)$ with $f(x)=1$ and $mathrm{supp}(f)subset B(x,delta)$.
Then $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
answered Nov 28 at 17:01
Federico
4,272512
answered Nov 28 at 17:01
Federico
4,272512
answered Nov 28 at 17:01
Federico
4,272512
answered Nov 28 at 17:01
Federico
4,272512
4,272512
Oops, way to simple.
– 0xbadf00d
Nov 28 at 17:10
Ah those lucky days when we can actually solve something! So satisfying... :-)
– Federico
Nov 28 at 17:11
add a comment |
Oops, way to simple.
– 0xbadf00d
Nov 28 at 17:10
Ah those lucky days when we can actually solve something! So satisfying... :-)
– Federico
Nov 28 at 17:11
Oops, way to simple.
– 0xbadf00d
Nov 28 at 17:10
Oops, way to simple.
– 0xbadf00d
Nov 28 at 17:10
Ah those lucky days when we can actually solve something! So satisfying... :-)
– Federico
Nov 28 at 17:11
Ah those lucky days when we can actually solve something! So satisfying... :-)
– Federico
Nov 28 at 17:11
add a comment |
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Do you mean $C_b^infty$ in the places you've written $C_c^infty$?
– Trevor Gunn
Nov 28 at 16:53
Do you mean $f_i$ instead of $h_i$?
– Paul Frost
Nov 28 at 16:55
@PaulFrost Sorry, fixed that.
– 0xbadf00d
Nov 28 at 16:56
@TrevorGunn No, I mean $C_c^infty$.
– 0xbadf00d
Nov 28 at 16:56
2
Cant you just take a single $fin C^infty_c(mathbb R)$ with $f(x)=1$, $mathrm{supp}(f)subset B(x,delta)$? Then you would have $f(x)-f(y)=1$ for $d(x,y)geqdelta$.
– Federico
Nov 28 at 16:58