Sequential characterization of topology
$begingroup$
Let $X$ be any set, and consider the following two structures on $X$:
- A topology $T$ on $X$;
- For every sequence $(x_n) in {^mathbb{N}X}$, a subset $L(x_n) subseteq X$ of limits of $(x_n)$, satisfying the inclusion property: if $(y_n)$ can be identified with a subsequence of $(x_n)$ after finitely many terms, then $L(x_n) subseteq L(y_n)$.
It is clear that a topology implies a structure of the second kind. My question is, given such structure as in (2), does a topology exists satisfying it, and if so, is it unique?
Progress so far: Let's denote the structure (2) simply by $L$, and say that $L leq L'$ if $L(x_n) subseteq L'(x_n)$ for every sequence $(x_n)$ (it is a partial ordering). I'm trying to mime the case for metric spaces for proving existence.
Given $L$, we say some subset $U$ of $X$ is open if for every $x in U$ and every sequence $(x_n)$ s.t. $x in L(x_n)$, there exists some $n_0$ such as that $x_n in U$ for all $n > n_0$. We sure have $varnothing, X in T$. Closure under arbitrary unions is trivial, and so is closure under finite intersections by taking maxima over the indexes. So it definitly defines a topology $T$ on $X$. Now, let $L_T$ be the structure (2) induced by this topology $T$. By construction we have $L leq L_T$. I'm not sure yet how to prove the reverse inequality (though I'm sure that here the inclusion property comes into play, as it is necessary for the structure (2) to be induced by a topology).
I'm still not convinced that the inclusion property alone is powerful enough to prove uniqueness, though I still couldn't think of a counterexample and a sequent verifiable reformulation.
EDIT: we can have $L(x_n) = varnothing$ for all $(x_n)$, which is certainly a contradiction for $X neq varnothing$. An extra necessary hypothesis must be included $x in L(x_n)$ when $x_n = x$ is constant. Maybe it is enough now?
sequences-and-series general-topology limits
asked Dec 23 '18 at 18:01
Henrique Augusto SouzaHenrique Augusto Souza
1,350514
$endgroup$
add a comment |
$begingroup$
Let $X$ be any set, and consider the following two structures on $X$:
- A topology $T$ on $X$;
- For every sequence $(x_n) in {^mathbb{N}X}$, a subset $L(x_n) subseteq X$ of limits of $(x_n)$, satisfying the inclusion property: if $(y_n)$ can be identified with a subsequence of $(x_n)$ after finitely many terms, then $L(x_n) subseteq L(y_n)$.
It is clear that a topology implies a structure of the second kind. My question is, given such structure as in (2), does a topology exists satisfying it, and if so, is it unique?
Progress so far: Let's denote the structure (2) simply by $L$, and say that $L leq L'$ if $L(x_n) subseteq L'(x_n)$ for every sequence $(x_n)$ (it is a partial ordering). I'm trying to mime the case for metric spaces for proving existence.
Given $L$, we say some subset $U$ of $X$ is open if for every $x in U$ and every sequence $(x_n)$ s.t. $x in L(x_n)$, there exists some $n_0$ such as that $x_n in U$ for all $n > n_0$. We sure have $varnothing, X in T$. Closure under arbitrary unions is trivial, and so is closure under finite intersections by taking maxima over the indexes. So it definitly defines a topology $T$ on $X$. Now, let $L_T$ be the structure (2) induced by this topology $T$. By construction we have $L leq L_T$. I'm not sure yet how to prove the reverse inequality (though I'm sure that here the inclusion property comes into play, as it is necessary for the structure (2) to be induced by a topology).
I'm still not convinced that the inclusion property alone is powerful enough to prove uniqueness, though I still couldn't think of a counterexample and a sequent verifiable reformulation.
EDIT: we can have $L(x_n) = varnothing$ for all $(x_n)$, which is certainly a contradiction for $X neq varnothing$. An extra necessary hypothesis must be included $x in L(x_n)$ when $x_n = x$ is constant. Maybe it is enough now?
sequences-and-series general-topology limits
asked Dec 23 '18 at 18:01
Henrique Augusto SouzaHenrique Augusto Souza
1,350514
$endgroup$
$begingroup$
Hope seems to be lost: mat.unimi.it/users/zanco/AnFunzionale/…
$endgroup$
– Henrique Augusto Souza
Dec 23 '18 at 18:33
2
$begingroup$
Henrique: your very useful link shows that you aren't guaranteed to recover a topology from its notion of sequential convergence. That means there is no hope for uniqueness, but it leaves hope for an existence proof.
$endgroup$
– Rob Arthan
Dec 23 '18 at 19:10
1
$begingroup$
this answer of mine might give you some pointers.
$endgroup$
– Henno Brandsma
Dec 23 '18 at 23:05
add a comment |
$begingroup$
Let $X$ be any set, and consider the following two structures on $X$:
- A topology $T$ on $X$;
- For every sequence $(x_n) in {^mathbb{N}X}$, a subset $L(x_n) subseteq X$ of limits of $(x_n)$, satisfying the inclusion property: if $(y_n)$ can be identified with a subsequence of $(x_n)$ after finitely many terms, then $L(x_n) subseteq L(y_n)$.
It is clear that a topology implies a structure of the second kind. My question is, given such structure as in (2), does a topology exists satisfying it, and if so, is it unique?
Progress so far: Let's denote the structure (2) simply by $L$, and say that $L leq L'$ if $L(x_n) subseteq L'(x_n)$ for every sequence $(x_n)$ (it is a partial ordering). I'm trying to mime the case for metric spaces for proving existence.
Given $L$, we say some subset $U$ of $X$ is open if for every $x in U$ and every sequence $(x_n)$ s.t. $x in L(x_n)$, there exists some $n_0$ such as that $x_n in U$ for all $n > n_0$. We sure have $varnothing, X in T$. Closure under arbitrary unions is trivial, and so is closure under finite intersections by taking maxima over the indexes. So it definitly defines a topology $T$ on $X$. Now, let $L_T$ be the structure (2) induced by this topology $T$. By construction we have $L leq L_T$. I'm not sure yet how to prove the reverse inequality (though I'm sure that here the inclusion property comes into play, as it is necessary for the structure (2) to be induced by a topology).
I'm still not convinced that the inclusion property alone is powerful enough to prove uniqueness, though I still couldn't think of a counterexample and a sequent verifiable reformulation.
EDIT: we can have $L(x_n) = varnothing$ for all $(x_n)$, which is certainly a contradiction for $X neq varnothing$. An extra necessary hypothesis must be included $x in L(x_n)$ when $x_n = x$ is constant. Maybe it is enough now?
sequences-and-series general-topology limits
asked Dec 23 '18 at 18:01
Henrique Augusto SouzaHenrique Augusto Souza
1,350514
$endgroup$
Let $X$ be any set, and consider the following two structures on $X$:
- A topology $T$ on $X$;
- For every sequence $(x_n) in {^mathbb{N}X}$, a subset $L(x_n) subseteq X$ of limits of $(x_n)$, satisfying the inclusion property: if $(y_n)$ can be identified with a subsequence of $(x_n)$ after finitely many terms, then $L(x_n) subseteq L(y_n)$.
It is clear that a topology implies a structure of the second kind. My question is, given such structure as in (2), does a topology exists satisfying it, and if so, is it unique?
Progress so far: Let's denote the structure (2) simply by $L$, and say that $L leq L'$ if $L(x_n) subseteq L'(x_n)$ for every sequence $(x_n)$ (it is a partial ordering). I'm trying to mime the case for metric spaces for proving existence.
Given $L$, we say some subset $U$ of $X$ is open if for every $x in U$ and every sequence $(x_n)$ s.t. $x in L(x_n)$, there exists some $n_0$ such as that $x_n in U$ for all $n > n_0$. We sure have $varnothing, X in T$. Closure under arbitrary unions is trivial, and so is closure under finite intersections by taking maxima over the indexes. So it definitly defines a topology $T$ on $X$. Now, let $L_T$ be the structure (2) induced by this topology $T$. By construction we have $L leq L_T$. I'm not sure yet how to prove the reverse inequality (though I'm sure that here the inclusion property comes into play, as it is necessary for the structure (2) to be induced by a topology).
I'm still not convinced that the inclusion property alone is powerful enough to prove uniqueness, though I still couldn't think of a counterexample and a sequent verifiable reformulation.
EDIT: we can have $L(x_n) = varnothing$ for all $(x_n)$, which is certainly a contradiction for $X neq varnothing$. An extra necessary hypothesis must be included $x in L(x_n)$ when $x_n = x$ is constant. Maybe it is enough now?
sequences-and-series general-topology limits
sequences-and-series general-topology limits
asked Dec 23 '18 at 18:01
Henrique Augusto SouzaHenrique Augusto Souza
1,350514
asked Dec 23 '18 at 18:01
Henrique Augusto SouzaHenrique Augusto Souza
1,350514
edited Dec 23 '18 at 18:14
Henrique Augusto Souza
asked Dec 23 '18 at 18:01
Henrique Augusto SouzaHenrique Augusto Souza
1,350514
asked Dec 23 '18 at 18:01
Henrique Augusto SouzaHenrique Augusto Souza
1,350514
asked Dec 23 '18 at 18:01
Henrique Augusto SouzaHenrique Augusto Souza
1,350514
1,350514
$begingroup$
Hope seems to be lost: mat.unimi.it/users/zanco/AnFunzionale/…
$endgroup$
– Henrique Augusto Souza
Dec 23 '18 at 18:33
2
$begingroup$
Henrique: your very useful link shows that you aren't guaranteed to recover a topology from its notion of sequential convergence. That means there is no hope for uniqueness, but it leaves hope for an existence proof.
$endgroup$
– Rob Arthan
Dec 23 '18 at 19:10
1
$begingroup$
this answer of mine might give you some pointers.
$endgroup$
– Henno Brandsma
Dec 23 '18 at 23:05
add a comment |
$begingroup$
Hope seems to be lost: mat.unimi.it/users/zanco/AnFunzionale/…
$endgroup$
– Henrique Augusto Souza
Dec 23 '18 at 18:33
2
$begingroup$
Henrique: your very useful link shows that you aren't guaranteed to recover a topology from its notion of sequential convergence. That means there is no hope for uniqueness, but it leaves hope for an existence proof.
$endgroup$
– Rob Arthan
Dec 23 '18 at 19:10
1
$begingroup$
this answer of mine might give you some pointers.
$endgroup$
– Henno Brandsma
Dec 23 '18 at 23:05
$begingroup$
Hope seems to be lost: mat.unimi.it/users/zanco/AnFunzionale/…
$endgroup$
– Henrique Augusto Souza
Dec 23 '18 at 18:33
$begingroup$
Hope seems to be lost: mat.unimi.it/users/zanco/AnFunzionale/…
$endgroup$
– Henrique Augusto Souza
Dec 23 '18 at 18:33
2
2
$begingroup$
Henrique: your very useful link shows that you aren't guaranteed to recover a topology from its notion of sequential convergence. That means there is no hope for uniqueness, but it leaves hope for an existence proof.
$endgroup$
– Rob Arthan
Dec 23 '18 at 19:10
$begingroup$
Henrique: your very useful link shows that you aren't guaranteed to recover a topology from its notion of sequential convergence. That means there is no hope for uniqueness, but it leaves hope for an existence proof.
$endgroup$
– Rob Arthan
Dec 23 '18 at 19:10
1
1
$begingroup$
this answer of mine might give you some pointers.
$endgroup$
– Henno Brandsma
Dec 23 '18 at 23:05
$begingroup$
this answer of mine might give you some pointers.
$endgroup$
– Henno Brandsma
Dec 23 '18 at 23:05
add a comment |
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$begingroup$
Hope seems to be lost: mat.unimi.it/users/zanco/AnFunzionale/…
$endgroup$
– Henrique Augusto Souza
Dec 23 '18 at 18:33
2
$begingroup$
Henrique: your very useful link shows that you aren't guaranteed to recover a topology from its notion of sequential convergence. That means there is no hope for uniqueness, but it leaves hope for an existence proof.
$endgroup$
– Rob Arthan
Dec 23 '18 at 19:10
1
$begingroup$
this answer of mine might give you some pointers.
$endgroup$
– Henno Brandsma
Dec 23 '18 at 23:05