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My motivation for this question is as follows.

Consider the set $D$ of cadlag functions on $(0,1)$ and its subset $D^\uparrow$ of non-decreasing cadlag functions. Each $f \in D$ has at most countably many points of discontinuity. One can equip $D$ with the Skorokhod $M_1$ topology which turns both $D$ and its subspace $D^\uparrow$ into Polish spaces. Convergence $f_n \to f$ in $D^\uparrow$ under the $M_1$-topology is equivalent to $f_n(x) \to f(x)$ for each continuity point $x$ of $f$.

On $D$ such a pointwise characterization of the $M_1$-convergence does not hold. I was asking myself, if there is some relation between such a pointwise convergence at points of continuity to the $M_1$-topology.

Let us start more abstractly and consider a topological space $X$ ($X = \mathbb{R}$ is enough for our considerations). Let $F \subseteq X^{[0,1]}$ be some function space and define convergence as follows: $f_n \to' f$ if and only if there exists a dense subset $D \subseteq X$ such that $f_n(x) \to f(x)$ for all $x \in D$. Before speaking about the induced (sequential) topology, it is necessary to check whether this convergence defines at least a limit space (synonymously a subsequential space or an $L$-space), i.e. one has to check (i) and (ii) from the following axioms:

(i) for constant sequence $f_n = f$ we have $f_n \to' f$

(ii) if $f_n \to' f$ then $f_{n_k} \to' f$ for every subsequence $f_{n_k}$

(iii) (subsequence principle) if every subsequence $f_{n_k}$ has a subsubsequence $f_{n_{k_l}}$ with $f_{n_{k_l}} \to' f$ then $f_n \to' f$.

Since (i) and (ii) trivially hold for $\to'$ it follows that $\to'$ induces a sequential topology by defining a set $F$ as closed whenever it contains all limits of $\to'$-convergent sequences $f_n$ with $f_n \in F$. This sequential topology in turn defines another notion of convergence $\to$ which satisfies (i), (ii) and (iii). If $\to'$ already satisfies (iii) then $\to'$-convergence is equivalent to $\to$-convergence, i.e. $\to'$ is precisely the notion of convergence of its established sequential topology.

However, I can't show (iii) for $\to'$ in such a generality and I think that it does not hold.

If it is wrong in general, what about the special case of such a definition of convergence (or even hopefully of a (sequential) topology) for the space $F = D$ of cadlag functions as above?

What happens if we restrict to define $\to'$ to be pointwise convergence outside a countable set (for $X = \mathbb{R}$ the complement of a countable set is dense).

(Maybe there is some similarity to the fact that on measure spaces convergence a.e. does not define a topological space but it does at least define a limit space.)

EDIT: In order not to open a new thread, I continue here with the following newly arising question. In the comments below, Nate Eldredge has shown that for $X = \mathbb{R}$ the $\to'$-convergence is not topological. Since $\to'$ satisfies properties (i) and (ii) it defines an $L$-space. We can define a set $F$ as closed if it contains all limits of all convergent sequences in $C$. The family of all closed sets defines a topology which is in addition sequential. How does this topology look like? Is it some familiar topology?

There is a similar construction in probability theory: The convergence almost surely defines an $L$-space and the induced topology is precisely the topology of convergence in probability.

My motivation for this question is as follows.

Consider the set $D$ of cadlag functions on $(0,1)$ and its subset $D^\uparrow$ of non-decreasing cadlag functions. Each $f \in D$ has at most countably many points of discontinuity. One can equip $D$ with the Skorokhod $M_1$ topology which turns both $D$ and its subspace $D^\uparrow$ into Polish spaces. Convergence $f_n \to f$ in $D^\uparrow$ under the $M_1$-topology is equivalent to $f_n(x) \to f(x)$ for each continuity point $x$ of $f$.

On $D$ such a pointwise characterization of the $M_1$-convergence does not hold. I was asking myself, if there is some relation between such a pointwise convergence at points of continuity to the $M_1$-topology.

Let us start more abstractly and consider a topological space $X$ ($X = \mathbb{R}$ is enough for our considerations). Let $F \subseteq X^{[0,1]}$ be some function space and define convergence as follows: $f_n \to' f$ if and only if there exists a dense subset $D \subseteq X$ such that $f_n(x) \to f(x)$ for all $x \in D$. Before speaking about the induced (sequential) topology, it is necessary to check whether this convergence defines at least a limit space (synonymously a subsequential space or an $L$-space), i.e. one has to check (i) and (ii) from the following axioms:

(i) for constant sequence $f_n = f$ we have $f_n \to' f$

(ii) if $f_n \to' f$ then $f_{n_k} \to' f$ for every subsequence $f_{n_k}$

(iii) (subsequence principle) if every subsequence $f_{n_k}$ has a subsubsequence $f_{n_{k_l}}$ with $f_{n_{k_l}} \to' f$ then $f_n \to' f$.

Since (i) and (ii) trivially hold for $\to'$ it follows that $\to'$ induces a sequential topology by defining a set $F$ as closed whenever it contains all limits of $\to'$-convergent sequences $f_n$ with $f_n \in F$. This sequential topology in turn defines another notion of convergence $\to$ which satisfies (i), (ii) and (iii). If $\to'$ already satisfies (iii) then $\to'$-convergence is equivalent to $\to$-convergence, i.e. $\to'$ is precisely the notion of convergence of its established sequential topology.

However, I can't show (iii) for $\to'$ in such a generality and I think that it does not hold.

If it is wrong in general, what about the special case of such a definition of convergence (or even hopefully of a (sequential) topology) for the space $F = D$ of cadlag functions as above?

What happens if we restrict to define $\to'$ to be pointwise convergence outside a countable set (for $X = \mathbb{R}$ the complement of a countable set is dense).

(Maybe there is some similarity to the fact that on measure spaces convergence a.e. does not define a topological space but it does at least define a limit space.)

EDIT: In order not to open a new thread, I continue here with the following newly arising question. In the comments below, Nate Eldredge has shown that for $X = \mathbb{R}$ the $\to'$-convergence is not topological. Since $\to'$ satisfies properties (i) and (ii) it defines an $L$-space. We can define a set $C$ as closed if it contains all limits of all convergent sequences in $C$. The family of all closed sets defines a topology which is in addition sequential. How does this topology look like? Is it some familiar topology?

There is a similar construction in probability theory: The convergence almost surely defines an $L$-space and the induced topology is precisely the topology of convergence in probability.

My motivation for this question is as follows.

Consider the set $D$ of cadlag functions on $(0,1)$ and its subset $D^\uparrow$ of non-decreasing cadlag functions. Each $f \in D$ has at most countably many points of discontinuity. One can equip $D$ with the Skorokhod $M_1$ topology which turns both $D$ and its subspace $D^\uparrow$ into Polish spaces. Convergence $f_n \to f$ in $D^\uparrow$ under the $M_1$-topology is equivalent to $f_n(x) \to f(x)$ for each continuity point $x$ of $f$.

On $D$ such a pointwise characterization of the $M_1$-convergence does not hold. I was asking myself, if there is some relation between such a pointwise convergence at points of continuity to the $M_1$-topology.

Let us start more abstractly and consider a topological space $X$ ($X = \mathbb{R}$ is enough for our considerations). Let $F \subseteq X^{[0,1]}$ be some function space and define convergence as follows: $f_n \to' f$ if and only if there exists a dense subset $D \subseteq X$ such that $f_n(x) \to f(x)$ for all $x \in D$. Before speaking about the induced (sequential) topology, it is necessary to check whether this convergence defines at least a limit space (synonymously a subsequential space or an $L$-space), i.e. one has to check (i) and (ii) from the following axioms:

(i) for constant sequence $f_n = f$ we have $f_n \to' f$

(ii) if $f_n \to' f$ then $f_{n_k} \to' f$ for every subsequence $f_{n_k}$

(iii) (subsequence principle) if every subsequence $f_{n_k}$ has a subsubsequence $f_{n_{k_l}}$ with $f_{n_{k_l}} \to' f$ then $f_n \to' f$.

Since (i) and (ii) trivially hold for $\to'$ it follows that $\to'$ induces a sequential topology by defining a set $F$ as closed whenever it contains all limits of $\to'$-convergent sequences $f_n$ with $f_n \in F$. This sequential topology in turn defines another notion of convergence $\to$ which satisfies (i), (ii) and (iii). If $\to'$ already satisfies (iii) then $\to'$-convergence is equivalent to $\to$-convergence, i.e. $\to'$ is precisely the notion of convergence of its established sequential topology.

However, I can't show (iii) for $\to'$ in such a generality and I think that it does not hold.

If it is wrong in general, what about the special case of such a definition of convergence (or even hopefully of a (sequential) topology) for the space $F = D$ of cadlag functions as above?

What happens if we restrict to define $\to'$ to be pointwise convergence outside a countable set (for $X = \mathbb{R}$ the complement of a countable set is dense).

(Maybe there is some similarity to the fact that on measure spaces convergence a.e. does not define a topological space but it does at least define a limit space.)

My motivation for this question is as follows.

Consider the set $D$ of cadlag functions on $(0,1)$ and its subset $D^\uparrow$ of non-decreasing cadlag functions. Each $f \in D$ has at most countably many points of discontinuity. One can equip $D$ with the Skorokhod $M_1$ topology which turns both $D$ and its subspace $D^\uparrow$ into Polish spaces. Convergence $f_n \to f$ in $D^\uparrow$ under the $M_1$-topology is equivalent to $f_n(x) \to f(x)$ for each continuity point $x$ of $f$.

On $D$ such a pointwise characterization of the $M_1$-convergence does not hold. I was asking myself, if there is some relation between such a pointwise convergence at points of continuity to the $M_1$-topology.

Let us start more abstractly and consider a topological space $X$ ($X = \mathbb{R}$ is enough for our considerations). Let $F \subseteq X^{[0,1]}$ be some function space and define convergence as follows: $f_n \to' f$ if and only if there exists a dense subset $D \subseteq X$ such that $f_n(x) \to f(x)$ for all $x \in D$. Before speaking about the induced (sequential) topology, it is necessary to check whether this convergence defines at least a limit space (synonymously a subsequential space or an $L$-space), i.e. one has to check (i) and (ii) from the following axioms:

(i) for constant sequence $f_n = f$ we have $f_n \to' f$

(ii) if $f_n \to' f$ then $f_{n_k} \to' f$ for every subsequence $f_{n_k}$

(iii) (subsequence principle) if every subsequence $f_{n_k}$ has a subsubsequence $f_{n_{k_l}}$ with $f_{n_{k_l}} \to' f$ then $f_n \to' f$.

Since (i) and (ii) trivially hold for $\to'$ it follows that $\to'$ induces a sequential topology by defining a set $F$ as closed whenever it contains all limits of $\to'$-convergent sequences $f_n$ with $f_n \in F$. This sequential topology in turn defines another notion of convergence $\to$ which satisfies (i), (ii) and (iii). If $\to'$ already satisfies (iii) then $\to'$-convergence is equivalent to $\to$-convergence, i.e. $\to'$ is precisely the notion of convergence of its established sequential topology.

However, I can't show (iii) for $\to'$ in such a generality and I think that it does not hold.

If it is wrong in general, what about the special case of such a definition of convergence (or even hopefully of a (sequential) topology) for the space $F = D$ of cadlag functions as above?

What happens if we restrict to define $\to'$ to be pointwise convergence outside a countable set (for $X = \mathbb{R}$ the complement of a countable set is dense).

(Maybe there is some similarity to the fact that on measure spaces convergence a.e. does not define a topological space but it does at least define a limit space.)

EDIT: In order not to open a new thread, I continue here with the following newly arising question. In the comments below, Nate Eldredge has shown that for $X = \mathbb{R}$ the $\to'$-convergence is not topological. Since $\to'$ satisfies properties (i) and (ii) it defines an $L$-space. We can define a set $F$ as closed if it contains all limits of all convergent sequences in $C$. The family of all closed sets defines a topology which is in addition sequential. How does this topology look like? Is it some familiar topology?

There is a similar construction in probability theory: The convergence almost surely defines an $L$-space and the induced topology is precisely the topology of convergence in probability.

My motivation for this question is as follows.

Consider the set $D$ of cadlag functions on $(0,1)$ and its subset $D^\uparrow$ of non-decreasing cadlag functions. Each $f \in D$ has at most countably many points of discontinuity. One can equip $D$ with the Skorokhod $M_1$ topology which turns both $D$ and its subspace $D^\uparrow$ into Polish spaces. Convergence $f_n \to f$ in $D^\uparrow$ under the $M_1$-topology is equivalent to $f_n(x) \to f(x)$ for each continuity point $x$ of $f$.

On $D$ such a pointwise characterization of the $M_1$-convergence does not hold. I was asking myself, if there is some relation between such a pointwise convergence at points of continuity to the $M_1$-topology.

Let us start more abstractly and consider a topological space $X$ ($X = \mathbb{R}$ is enough for our considerations). Let $F \subseteq X^{[0,1]}$ be some function space and define convergence as follows: $f_n \to' f$ if and only if there exists a dense subset $D \subseteq X$ such that $f_n(x) \to f(x)$ for all $x \in D$. Before speaking about the induced (sequential) topology, it is necessary to check whether this convergence defines a limit space (synonymously a subsequential space or an $L$-space), i.e. one has to check the Kuratowski axioms (without uniqueness of limit) which are:

(i) for constant sequence $f_n = f$ we have $f_n \to' f$

(ii) if $f_n \to' f$ then $f_{n_k} \to' f$ for every subsequence $f_{n_k}$

(iii) (subsequence principle) if every subsequence $f_{n_k}$ has a subsubsequence $f_{n_{k_l}}$ with $f_{n_{k_l}} \to' f$ then $f_n \to' f$.

Properties (i) and (ii) do trivially hold but I can't show (iii) in such a generality and I think that it does not hold for $\to'$.

If it is wrong in general, what about the special case of such a definition of convergence (or even hopefully of a (sequential) topology) for the space $F = D$ of cadlag functions as above?

What happens if we restrict to define $\to'$ to be pointwise convergence outside a countable set (for $X = \mathbb{R}$ the complement of a countable set is dense).

(Maybe there is some similarity to the fact that on measure spaces convergence a.e. does not define a topological space but it does at least define a limit space.)

My motivation for this question is as follows.

Consider the set $D$ of cadlag functions on $(0,1)$ and its subset $D^\uparrow$ of non-decreasing cadlag functions. Each $f \in D$ has at most countably many points of discontinuity. One can equip $D$ with the Skorokhod $M_1$ topology which turns both $D$ and its subspace $D^\uparrow$ into Polish spaces. Convergence $f_n \to f$ in $D^\uparrow$ under the $M_1$-topology is equivalent to $f_n(x) \to f(x)$ for each continuity point $x$ of $f$.

On $D$ such a pointwise characterization of the $M_1$-convergence does not hold. I was asking myself, if there is some relation between such a pointwise convergence at points of continuity to the $M_1$-topology.

Let us start more abstractly and consider a topological space $X$ ($X = \mathbb{R}$ is enough for our considerations). Let $F \subseteq X^{[0,1]}$ be some function space and define convergence as follows: $f_n \to' f$ if and only if there exists a dense subset $D \subseteq X$ such that $f_n(x) \to f(x)$ for all $x \in D$. Before speaking about the induced (sequential) topology, it is necessary to check whether this convergence defines at least a limit space (synonymously a subsequential space or an $L$-space), i.e. one has to check (i) and (ii) from the following axioms:

(i) for constant sequence $f_n = f$ we have $f_n \to' f$

(ii) if $f_n \to' f$ then $f_{n_k} \to' f$ for every subsequence $f_{n_k}$

(iii) (subsequence principle) if every subsequence $f_{n_k}$ has a subsubsequence $f_{n_{k_l}}$ with $f_{n_{k_l}} \to' f$ then $f_n \to' f$.

Since (i) and (ii) trivially hold for $\to'$ it follows that $\to'$ induces a sequential topology by defining a set $F$ as closed whenever it contains all limits of $\to'$-convergent sequences $f_n$ with $f_n \in F$. This sequential topology in turn defines another notion of convergence $\to$ which satisfies (i), (ii) and (iii). If $\to'$ already satisfies (iii) then $\to'$-convergence is equivalent to $\to$-convergence, i.e. $\to'$ is precisely the notion of convergence of its established sequential topology.

However, I can't show (iii) for $\to'$ in such a generality and I think that it does not hold.

If it is wrong in general, what about the special case of such a definition of convergence (or even hopefully of a (sequential) topology) for the space $F = D$ of cadlag functions as above?

What happens if we restrict to define $\to'$ to be pointwise convergence outside a countable set (for $X = \mathbb{R}$ the complement of a countable set is dense).

(Maybe there is some similarity to the fact that on measure spaces convergence a.e. does not define a topological space but it does at least define a limit space.)