How is a sequence not converging usually but $I_{\tau}$ converging in this given paper.

Question

I am reading the paper Pratulananda Das and Ekrem Savas: On I-convergence of nets in locally solid Riesz spaces, Filomat 27:1 (2013), 89–94, DOI: 10.2298/FIL1301089D.

I am stuck at example $3.2$

Here is how far I managed :

Conferring the definitions in p.g.$90$, we have, for a net $\{s_{\alpha}:\alpha \in D\}$ where $D$ is directed set and $s:D\rightarrow X$ is the net. $$D_{\alpha}=\{\beta \in D : \beta \ge \alpha\}\\F_{0}=\{ A\subset D : A\supset D_{\alpha} \text{ for some }\alpha \in D\}\\I_0=\{ A\subset D : A^c \in F_0\}$$ $N_{x_{0}}$ is the nbd system of a fixed point $x_0$ forming a directed set under inclusion. So I take it that means $A\lt B$ iff $A\subset B.$ Then $D_{A}=\{ \text{nbd's of }x_0 \text{ that contains A}\}.$ And $$I_0=\{ A\subset D : A^c \in F_0\}\\=\{ A\subset D : A^c \supset D_{B}\text{ for some }B\in D\}\\=\{ A\subset D : A\subset D_B \text{ for some } B\in D\}\\=\{ A\subset D : A\supset B \text{ for some nbd B of } x_0 \}$$ Now for any ideal $I$ of $D$ that contains $I_0$ , if $C\in I\backslash I_0.$ Then $$C=\{ A\subset D : \text{there is some nbd B of }x_0 \text{ such that }A \text{ does not contain B}\}$$

And the net $\{ s_U\}$ is given by $$s_U\in I ,\forall U\in N_{x_0}\backslash C\\ s_U= y_0 \forall U \in C,\text{ where } y_0 \neq x_0 \text{ is a fixed point. }$$

Then we can easily see that $\{s_U\}$ cannot converge to $x_0$ usually. Because of Hausdorff property we can find disjoint nbd's of $x$ and $y$. And for $U\in C$ $s_U=y_0$ won't be in the nbd of $x_0.$ And we will find infinitely manu such $s_U$'s.

The next thing is to prove it is $I_{\tau}$ convergent. I could not do this at all for I could not comprehend the given equation $$\{ U\in D : s_U- x_0 \lt U\}=C\in I.$$

Help me with this. Also , if there is any mistake in what I have done above , please point out. Thanks.

Answer 1

Let us choose any net $t_U$ on $D=\mathcal N_{x_0}$ (ordered by reverse inclusion) such that $t_U\in U$ for each $U$. Then this net converges to $x_0$: $$t_U\to x_0$$ (In the sense of the usual definition of convergence of nets.)

The modification introduced in the paper can be described as: $$ s_U= \begin{cases} t_U & U\notin C, \\ y_0 & U\in C. \end{cases} $$

Then we can easily see that $\{s_U\}$ cannot converge to $x_0$ usually. Because of Hausdorff property we can find disjoint nbd's of $x$ and $y$. And for $U\in C$ $s_U=y_0$ won't be in the nbd of $x_0.$ And we will find infinitely manu such $s_U$'s.

In the context of this paper the authors clearly mean convergence of the given net by the usual convergence. So we want to show that the net $s_U$ is does not converge to $x_0$.

For this, it suffices to check that for each $U_0\in\mathcal N_{x_0}$ there exists $U\supseteq U_0$, $U\in\mathcal N_{x_0}$, such that $s_U=y_0$. This is equivalent to $C\cap M\ne\emptyset$ for each $M\in F_0$. And this property follows from $C\notin I_0$.

(If it is not clear how $C\notin I_0$ implies that $C$ intersects some element of $F_0$, you can prove this by contradiction. Assume that $C\cap M=\emptyset$ for some $M\in F_0$. This means that $C\subseteq D\setminus M \in I_0$.)

The next thing is to prove it is $I_{\tau}$ convergent.

Let us denote by $F$ the filter dual to $I$.

For any neighborhood $B$ of $x_0$ we have $$\{U\in D; s_U\in B\} \supseteq \{U\in D; t_U\in B\}\setminus C.$$ Since $t_U$ is convergent to $x_0$, we get $\{U\in D; t_U\in B\}\in F_0\subseteq F$. This implies that $\{U\in D; t_U\in B\}\setminus C\in F$ and, consequently, $\{U\in D; s_U\in B\}\in F$.

Since the above is true for every neighborhood of $x_0$, we get that $s_U$ is $I$-convergent to $x_0$.

I could not comprehend the given equation $$\{ U\in D : s_U- x_0 \notin U\}=C\in I.$$

The above equality is certainly not correct. Moreover, the symbol $U$ is used there in two different meanings, which is clearly oversight on the side of authors.

They want to show that $$\{ U\in D : s_U- x_0 \notin V\} \in I$$ for every $V$ which is a neighborhood of zero.

To show this, you could do the following. Denote $B=x_0+V$, which is neighborhood of $x_0$. Moreover, $B$ does not contain $y_0$. Then you have

$$\begin{align} \{ U\in D : s_U- x_0 \notin V\} &= \{ U\in D : s_U \notin B\} \subseteq\\ &\subseteq C \cup \{U\in D; U\not\subseteq B\} =\\ &= C \cup (D\setminus\{U\in D; U\subseteq B\}) \in I. \end{align}$$ (Just notice that $\{U\in D;U \subseteq B\}$ belongs to $F_0$, hence it also belongs to $F$.)

(1) this is just convergence along a filter; (2) this is really just convergence of nets

Since in your comment you seem to object to above comment, I will stress that:

• What the authors call $I_\tau$-convergence is exactly the same as the usual $I$-convergence.
• The $I_0$-convergence is precisely the convergence of the corresponding net.

The only difference is that the authors require $I\supseteq I_0$ (admissible ideals). This is a natural condition in order to study only ideal such that the convergence along the ideal $I$ behaves reasonably to the usual convergence of nets.

Answer 2

There are numerous errors here. First, the general calculation should be

$$\begin{align*} I_0&=\{A\subseteq D:A^c\in F_0\}\\ &=\{A\subseteq D:A^c\supseteq D_\alpha\text{ for some }\alpha\in D\}\\ &=\{A\subseteq D:A\cap D_\alpha=\varnothing\text{ for some }\alpha\in D\}\;; \end{align*}$$

$A^c\supseteq D_\alpha$ is equivalent to $A\cap D_\alpha=\varnothing$, not to $A\subseteq D_\alpha$.

Now you want to specialize it to the nhbd system at $x_0$ considered as a directed set with respect to inclusion. Unfortunately, you have the order backwards: $A\le B$ iff $A\supseteq B$, so if $D=N_{x_0}$, then $D_A$ is the set of nbhds of $x_0$ that are subsets of $A$, and

$$\begin{align*} I_0&=\{\mathscr{A}\subseteq N_{x_0}:\mathscr{A}\cap D_B=\varnothing\text{ for some }B\in N_{x_0}\}\\ &=\{\mathscr{A}\subseteq N_{x_0}:\exists B\in N_{x_0}\,\forall A\in\mathscr{A}\,(A\setminus B\ne\varnothing)\}\;. \end{align*}$$

Note that the elements of $I_0$ are not nbhds of $x_0$: they are families of nbhds of $x_0$. Each such family $\mathscr{A}$ has the property that there is some nbhd $B$ of $x_0$ that does not contain any member of $\mathscr{A}$ as a subset. (Informally, each $A\in\mathscr{A}$ ‘sticks out of’ $B$.) This means that if $I$ is an ideal of $N_{x_0}$ that contains $I_0$, an object in $I\setminus I_0$ is not a nbhd of $x_0$: it’s a whole family of nbhds of $x_0$.

This is why in Example $\mathbf{3.2}$ the authors can talk about $N_{x_0}\setminus C$: $C$ is a family of nbhds of $x_0$. Your description of the net $s_U:N_{x_0}\to\Bbb R^2$, however, is incorrect: when $U\in N_{x_0}\setminus C$, we let $s_U$ be an arbitrary point of the nbhd $U$ of $x_0$, not (as you have it) an element of the ideal $I$. (The later doesn’t even make sense, since elements of $I$ are not points in $\Bbb R^2$.) As you say, for $U\in C$ we set $s_U=y_0$, where $y_0\in\Bbb R^2\setminus\{x_0\}$.

Now go back to Definition $\mathbf{3.1}$: this net is $I_\tau$-convergent to $x_0$, where $\tau$ is the Euclidean topology on $\Bbb R^2$, if and only if for each nbhd $V$ of $\langle 0,0\rangle$, $\{U\in N_{x_0}:s_U-x_0\notin V\}\in I$. This is actually equivalent to saying that for each $V\in N_{x_0}$, $\{U\in N_{x_0}:s_U\notin V\}\in I$, since we can translate nbhds of $\langle 0,0\rangle$ to nbhds of $x_0$ by adding $x_0$ to each point.

Unfortunately, at this point the authors’ explanation devolves into nonsense: In the expression

$$\{U\in D:s_U-x_0\notin U\}$$

they are using $U$ for two completely different things. The first and second instances of $U$ are a dummy variable ranging over nbhds of $x_0$, and the third is some fixed nbhd of $\langle 0,0\rangle$. They apparently wanted to say that if $V\in N_{x_0}$ does not contain $y_0$, then

$$\{U\in N_{x_0}:s_U\notin V\}=C\in I\;.$$

This need not actually be true. It’s true that if $U\in C$, then $s_U=y_0\notin V$, so

$$\{U\in N_{x_0}:s_U\notin V\}\supseteq C\;.$$

However, it’s certainly possible that there are also $U\in N_{x_0}\setminus C$ such that $s_U\notin V$: $s_U$ can be any point of the nbhd $U$. On the other hand, if $U\in D_V$, then certainly $s_U\in U\subseteq V$, so

$$\{U\in N_{x_0}:s_U\notin V\}\subseteq C\cup(N_{x_0}\setminus D_V)\;.$$

And $N_{x_0}\setminus D_V\in I_0\subseteq I$, so

$$\{U\in N_{x_0}:s_U\notin V\}\in I\;,$$

as desired.