Why does the support must be closed?
$begingroup$
Apparently it is important that the support is defined as the closure of ${f neq 0}$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of ${f neq 0}$?
The exercise:
Let (X, $mathcal{T}$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbb{R} $,
$(eta cdot g): X rightarrow mathbb{R}$,
$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and
$(eta cdot g)(x) = 0$ if $x notin U$
is continous. Show that this statement fails if we only assume that ${f neq 0} subset U$.
I have been able to show that the map $g : U rightarrow mathbb{R} $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.
Can anyone help me?
general-topology
asked yesterday
HK4HK4
354
$endgroup$
add a comment |
$begingroup$
Apparently it is important that the support is defined as the closure of ${f neq 0}$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of ${f neq 0}$?
The exercise:
Let (X, $mathcal{T}$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbb{R} $,
$(eta cdot g): X rightarrow mathbb{R}$,
$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and
$(eta cdot g)(x) = 0$ if $x notin U$
is continous. Show that this statement fails if we only assume that ${f neq 0} subset U$.
I have been able to show that the map $g : U rightarrow mathbb{R} $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.
Can anyone help me?
general-topology
asked yesterday
HK4HK4
354
$endgroup$
1
$begingroup$
You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
$endgroup$
– Chessanator
yesterday
$begingroup$
I meant that $(eta cdot g)$ is continuous yes!
$endgroup$
– HK4
yesterday
add a comment |
$begingroup$
Apparently it is important that the support is defined as the closure of ${f neq 0}$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of ${f neq 0}$?
The exercise:
Let (X, $mathcal{T}$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbb{R} $,
$(eta cdot g): X rightarrow mathbb{R}$,
$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and
$(eta cdot g)(x) = 0$ if $x notin U$
is continous. Show that this statement fails if we only assume that ${f neq 0} subset U$.
I have been able to show that the map $g : U rightarrow mathbb{R} $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.
Can anyone help me?
general-topology
asked yesterday
HK4HK4
354
$endgroup$
Apparently it is important that the support is defined as the closure of ${f neq 0}$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of ${f neq 0}$?
The exercise:
Let (X, $mathcal{T}$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbb{R} $,
$(eta cdot g): X rightarrow mathbb{R}$,
$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and
$(eta cdot g)(x) = 0$ if $x notin U$
is continous. Show that this statement fails if we only assume that ${f neq 0} subset U$.
I have been able to show that the map $g : U rightarrow mathbb{R} $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.
Can anyone help me?
general-topology
general-topology
asked yesterday
HK4HK4
354
asked yesterday
HK4HK4
354
asked yesterday
HK4HK4
354
asked yesterday
HK4HK4
354
asked yesterday
HK4HK4
354
354
1
$begingroup$
You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
$endgroup$
– Chessanator
yesterday
$begingroup$
I meant that $(eta cdot g)$ is continuous yes!
$endgroup$
– HK4
yesterday
add a comment |
1
$begingroup$
You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
$endgroup$
– Chessanator
yesterday
$begingroup$
I meant that $(eta cdot g)$ is continuous yes!
$endgroup$
– HK4
yesterday
1
1
$begingroup$
You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
$endgroup$
– Chessanator
yesterday
$begingroup$
You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
$endgroup$
– Chessanator
yesterday
$begingroup$
I meant that $(eta cdot g)$ is continuous yes!
$endgroup$
– HK4
yesterday
$begingroup$
I meant that $(eta cdot g)$ is continuous yes!
$endgroup$
– HK4
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
$$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$
Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.
To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
$$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
0 & |x| geq 1. end{cases}$$
Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.
answered yesterday
levaplevap
47k23273
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add a comment |
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$begingroup$
Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
$$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$
Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.
To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
$$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
0 & |x| geq 1. end{cases}$$
Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.
answered yesterday
levaplevap
47k23273
$endgroup$
add a comment |
$begingroup$
Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
$$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$
Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.
To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
$$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
0 & |x| geq 1. end{cases}$$
Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.
answered yesterday
levaplevap
47k23273
$endgroup$
add a comment |
$begingroup$
Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
$$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$
Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.
To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
$$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
0 & |x| geq 1. end{cases}$$
Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.
answered yesterday
levaplevap
47k23273
$endgroup$
Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy ${ x in X , | , eta(x) neq 0 } = U subseteq U$ but it won't satisfy
$$ overline{ { x in X , | , eta(x) neq 0 } }= overline{U} subseteq U. $$
Take $g colon U rightarrow mathbb{R}$ to be the function $g(x) = frac{1}{eta(x)}$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.
To see a concrete example, take $X = mathbb{R}, U = (-1,1)$ and
$$ eta(x) = begin{cases} 1 - |x| & |x| < 1,\
0 & |x| geq 1. end{cases}$$
Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.
answered yesterday
levaplevap
47k23273
answered yesterday
levaplevap
47k23273
answered yesterday
levaplevap
47k23273
answered yesterday
levaplevap
47k23273
47k23273
add a comment |
add a comment |
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$begingroup$
You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $eta cdot g$ is continuous?
$endgroup$
– Chessanator
yesterday
$begingroup$
I meant that $(eta cdot g)$ is continuous yes!
$endgroup$
– HK4
yesterday