Showing there exists no sequence of indicator functions on $mathbb{R}^mathbb{R}$ that converges to the...
$begingroup$
Definition: Let $(Omega, tau)$ be a topological space. The indicator function $1_A: Omega to mathbb{R}$ on a subset $A subset Omega$ is defined by $1_A(x) = 1$ if $x in A$ and $1_A(x) = 0$ if $x notin A$.
Let $Omega = mathbb{R}^{mathbb{R}}$ and $M = {1_A: A subset mathbb{R} vert text{$A$ is finite}}$. Show that $1_mathbb{R} in overline{M}$ and prove that there exists no sequence ${1_{A_n}}_{n in mathbb{N}} subset M$ such that $1_{A_n} to mathbb{1_mathbb{R}}$. Here we are using the product topology.
We know that if $U = displaystyle{prod_{alpha in mathbb{R}} U_alpha} $ is a basic open set of $mathbb{R}^mathbb{R}$ in the product topology containing $1_mathbb{R}$, then $U_alpha = mathbb{R}$ for all but finitely many $alpha$. Let $F subset mathbb{R}$ be the set such that $U_alpha neq mathbb{R} forall alpha in F$. Since $1_mathbb{R} in U$, then $1 in U_{alpha_i} forall alpha_i in F$. Then $1_F$ witnesses $U cap M neq emptyset$, therefore $1_{mathbb{R}} in overline{M}$.
The second part is where I got stuck. My only ideas were:
-- construct an open set that contains $1_{mathbb{R}}$ but doesn't contain any tail of any sequence ${1_{A_n}}_{n in mathbb{N}}$
-- somehow use the fact that $1_mathbb{R}$ is a limit point of $M$?
But I haven't had any success with either of those. Can anyone help?
general-topology proof-verification
asked Jan 22 at 23:06
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
add a comment |
$begingroup$
Definition: Let $(Omega, tau)$ be a topological space. The indicator function $1_A: Omega to mathbb{R}$ on a subset $A subset Omega$ is defined by $1_A(x) = 1$ if $x in A$ and $1_A(x) = 0$ if $x notin A$.
Let $Omega = mathbb{R}^{mathbb{R}}$ and $M = {1_A: A subset mathbb{R} vert text{$A$ is finite}}$. Show that $1_mathbb{R} in overline{M}$ and prove that there exists no sequence ${1_{A_n}}_{n in mathbb{N}} subset M$ such that $1_{A_n} to mathbb{1_mathbb{R}}$. Here we are using the product topology.
We know that if $U = displaystyle{prod_{alpha in mathbb{R}} U_alpha} $ is a basic open set of $mathbb{R}^mathbb{R}$ in the product topology containing $1_mathbb{R}$, then $U_alpha = mathbb{R}$ for all but finitely many $alpha$. Let $F subset mathbb{R}$ be the set such that $U_alpha neq mathbb{R} forall alpha in F$. Since $1_mathbb{R} in U$, then $1 in U_{alpha_i} forall alpha_i in F$. Then $1_F$ witnesses $U cap M neq emptyset$, therefore $1_{mathbb{R}} in overline{M}$.
The second part is where I got stuck. My only ideas were:
-- construct an open set that contains $1_{mathbb{R}}$ but doesn't contain any tail of any sequence ${1_{A_n}}_{n in mathbb{N}}$
-- somehow use the fact that $1_mathbb{R}$ is a limit point of $M$?
But I haven't had any success with either of those. Can anyone help?
general-topology proof-verification
asked Jan 22 at 23:06
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
add a comment |
$begingroup$
Definition: Let $(Omega, tau)$ be a topological space. The indicator function $1_A: Omega to mathbb{R}$ on a subset $A subset Omega$ is defined by $1_A(x) = 1$ if $x in A$ and $1_A(x) = 0$ if $x notin A$.
Let $Omega = mathbb{R}^{mathbb{R}}$ and $M = {1_A: A subset mathbb{R} vert text{$A$ is finite}}$. Show that $1_mathbb{R} in overline{M}$ and prove that there exists no sequence ${1_{A_n}}_{n in mathbb{N}} subset M$ such that $1_{A_n} to mathbb{1_mathbb{R}}$. Here we are using the product topology.
We know that if $U = displaystyle{prod_{alpha in mathbb{R}} U_alpha} $ is a basic open set of $mathbb{R}^mathbb{R}$ in the product topology containing $1_mathbb{R}$, then $U_alpha = mathbb{R}$ for all but finitely many $alpha$. Let $F subset mathbb{R}$ be the set such that $U_alpha neq mathbb{R} forall alpha in F$. Since $1_mathbb{R} in U$, then $1 in U_{alpha_i} forall alpha_i in F$. Then $1_F$ witnesses $U cap M neq emptyset$, therefore $1_{mathbb{R}} in overline{M}$.
The second part is where I got stuck. My only ideas were:
-- construct an open set that contains $1_{mathbb{R}}$ but doesn't contain any tail of any sequence ${1_{A_n}}_{n in mathbb{N}}$
-- somehow use the fact that $1_mathbb{R}$ is a limit point of $M$?
But I haven't had any success with either of those. Can anyone help?
general-topology proof-verification
asked Jan 22 at 23:06
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
Definition: Let $(Omega, tau)$ be a topological space. The indicator function $1_A: Omega to mathbb{R}$ on a subset $A subset Omega$ is defined by $1_A(x) = 1$ if $x in A$ and $1_A(x) = 0$ if $x notin A$.
Let $Omega = mathbb{R}^{mathbb{R}}$ and $M = {1_A: A subset mathbb{R} vert text{$A$ is finite}}$. Show that $1_mathbb{R} in overline{M}$ and prove that there exists no sequence ${1_{A_n}}_{n in mathbb{N}} subset M$ such that $1_{A_n} to mathbb{1_mathbb{R}}$. Here we are using the product topology.
We know that if $U = displaystyle{prod_{alpha in mathbb{R}} U_alpha} $ is a basic open set of $mathbb{R}^mathbb{R}$ in the product topology containing $1_mathbb{R}$, then $U_alpha = mathbb{R}$ for all but finitely many $alpha$. Let $F subset mathbb{R}$ be the set such that $U_alpha neq mathbb{R} forall alpha in F$. Since $1_mathbb{R} in U$, then $1 in U_{alpha_i} forall alpha_i in F$. Then $1_F$ witnesses $U cap M neq emptyset$, therefore $1_{mathbb{R}} in overline{M}$.
The second part is where I got stuck. My only ideas were:
-- construct an open set that contains $1_{mathbb{R}}$ but doesn't contain any tail of any sequence ${1_{A_n}}_{n in mathbb{N}}$
-- somehow use the fact that $1_mathbb{R}$ is a limit point of $M$?
But I haven't had any success with either of those. Can anyone help?
general-topology proof-verification
general-topology proof-verification
asked Jan 22 at 23:06
Matheus AndradeMatheus Andrade
1,374418
asked Jan 22 at 23:06
Matheus AndradeMatheus Andrade
1,374418
edited Jan 22 at 23:46
Matheus Andrade
asked Jan 22 at 23:06
Matheus AndradeMatheus Andrade
1,374418
asked Jan 22 at 23:06
Matheus AndradeMatheus Andrade
1,374418
asked Jan 22 at 23:06
Matheus AndradeMatheus Andrade
1,374418
1,374418
add a comment |
add a comment |
1 Answer
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All you need for this is the fact that convergence in $mathbb R^{mathbb R}$ implies convergence of each coordinate. $cup_n A_n$ is a countable subset of $mathbb R$ so there is point $x$ in $mathbb Rsetminus cup_n A_n$. If $I_{A_n} to I_{mathbb R}$ then Then $I_{A_n}(x) to 0 neq I_{mathbb R} (x)$.
answered Jan 22 at 23:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$begingroup$
Beautiful and simple, can't believe I didn't see that before. Thanks!
$endgroup$
– Matheus Andrade
Jan 22 at 23:53
add a comment |
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$begingroup$
All you need for this is the fact that convergence in $mathbb R^{mathbb R}$ implies convergence of each coordinate. $cup_n A_n$ is a countable subset of $mathbb R$ so there is point $x$ in $mathbb Rsetminus cup_n A_n$. If $I_{A_n} to I_{mathbb R}$ then Then $I_{A_n}(x) to 0 neq I_{mathbb R} (x)$.
answered Jan 22 at 23:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$begingroup$
Beautiful and simple, can't believe I didn't see that before. Thanks!
$endgroup$
– Matheus Andrade
Jan 22 at 23:53
add a comment |
$begingroup$
All you need for this is the fact that convergence in $mathbb R^{mathbb R}$ implies convergence of each coordinate. $cup_n A_n$ is a countable subset of $mathbb R$ so there is point $x$ in $mathbb Rsetminus cup_n A_n$. If $I_{A_n} to I_{mathbb R}$ then Then $I_{A_n}(x) to 0 neq I_{mathbb R} (x)$.
answered Jan 22 at 23:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$begingroup$
Beautiful and simple, can't believe I didn't see that before. Thanks!
$endgroup$
– Matheus Andrade
Jan 22 at 23:53
add a comment |
$begingroup$
All you need for this is the fact that convergence in $mathbb R^{mathbb R}$ implies convergence of each coordinate. $cup_n A_n$ is a countable subset of $mathbb R$ so there is point $x$ in $mathbb Rsetminus cup_n A_n$. If $I_{A_n} to I_{mathbb R}$ then Then $I_{A_n}(x) to 0 neq I_{mathbb R} (x)$.
answered Jan 22 at 23:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
All you need for this is the fact that convergence in $mathbb R^{mathbb R}$ implies convergence of each coordinate. $cup_n A_n$ is a countable subset of $mathbb R$ so there is point $x$ in $mathbb Rsetminus cup_n A_n$. If $I_{A_n} to I_{mathbb R}$ then Then $I_{A_n}(x) to 0 neq I_{mathbb R} (x)$.
answered Jan 22 at 23:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 22 at 23:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 22 at 23:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 22 at 23:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
63.7k42463
$begingroup$
Beautiful and simple, can't believe I didn't see that before. Thanks!
$endgroup$
– Matheus Andrade
Jan 22 at 23:53
add a comment |
$begingroup$
Beautiful and simple, can't believe I didn't see that before. Thanks!
$endgroup$
– Matheus Andrade
Jan 22 at 23:53
$begingroup$
Beautiful and simple, can't believe I didn't see that before. Thanks!
$endgroup$
– Matheus Andrade
Jan 22 at 23:53
$begingroup$
Beautiful and simple, can't believe I didn't see that before. Thanks!
$endgroup$
– Matheus Andrade
Jan 22 at 23:53
add a comment |
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