Is collection of all functions I-convergent to a point form a ring?

Question

$S$ be a set. $I$ is an ideal of $S.$ $X$ is a topological space. A function $$f: S\rightarrow X$$ is said to be $I$-convergent to a point $x\in X$ if $$f^{-1}(U)=\{ s\in S; f(s)\in U\}\in \mathscr F(I)$$ for any nbd $U$ of $\mathscr a$ where $\mathscr F(I)$ is the dual filter of the ideal $I.$ Now I take the topological space to be $\mathbb R.$ I fix a point $\mathscr a\in \mathbb R.$

Let, $C_{\mathscr a}(\mathbb R)$ be the set of all functions from $S$ to $\mathbb R$ that are $I$-convergent to $\mathscr a.$ Is this set a ring, like the set of all continuous functions $C(\mathbb R)?$

Since the function is real valued, associativity and commutativity are trivial. Now for the $0$ function, i.e. $0 : S\rightarrow \mathbb R$ s.t. $0(s)=0$. For any nbd $U$ of the point $\mathscr a$ $$0^{-1}(U)=S={\Phi}^c\in \mathscr F(I).$$

Now the additive inverse $-f$ defined by $$(-f)(s)=-(f(s)).$$ Now to prove $-f$ is also $I$-convergent to $\mathscr a$ I have to prove that $$(-f)^{-1}(U)=\{ s\in S : (-f)(s)\in U\}\in \mathscr F(I)$$ using the info that $$\{t\in S :f(t)\in U \}\in \mathscr F(I).$$ Now, $$\{t\in S : f(t)\in -U\}=\{t\in S : (-f)(t)\in U\}.$$ So, $$(-f)^{-1}(U)=\{ s\in S : f(s)\in -U\}.$$ Am I really close to the conclusion or a contradiction? How can I show from here that this set $$(-f)^{-1}(U)=\{ s\in S : f(s)\in -U\}\in \mathscr F(I)?$$

Some lead please. Thanks.

P.S. - If $U=(a,b)\text{ then } -U=(-b,-a).$

Answer 1

Since for $a\ne0$ zero function does not belong to $C_a$, the claim is only true for $a=0$.$\newcommand{\Ilim}{\operatorname{I-lim}}$

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To prove that $C_0$ is a ring, to me the most natural way seems to be use the following facts (which are probably already known to you, if you study I-convergence):

• If $\Ilim f=x$ and $\Ilim g=y$, then $f+g$ is $I$-convergent and $\Ilim (f+g)=x+y$.
• If $\Ilim f=x$ and $\Ilim g=y$, then $\Ilim (fg) = xy$.
• If $\Ilim f=x$ and $c\in\mathbb R$, then $\Ilim (cf) = cx$.

Proofs for the special case $S=\mathbb N$ can be found here. The general case can be proved in a similar fashion.

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You also asked for comments on your attempt.

For any nbd $U$ of the point $\mathscr a$ $$0^{-1}(U)=S={\Phi}^c\in \mathscr F(I).$$

This is only true if $a=0$. If $a\ne0$, then there is a neighborhood $U$ of $a$ which does not contain zero. For this neighborhood you have $0^{-1}(U)=\emptyset$.

If $f\in C_0$ then $-f\in C_0$.

The proof you suggested for this part seems ok. But again, it only works for $a=0$.

Basically you are using the fact that if $U$ is a neighborhood of $0$, so is $-U$.

If $f,g\in C_0$, then $f+g\in C_0$.

For any neighborhood $U$ of zero, there exists another neighborhood $V$ of zero such that $V+V\subseteq U$. You have $$(f+g)^{-1}(U) \supseteq f^{-1}(V) \cap g^{-1}(V),$$ which shows that $(f+g)^{-1}(U)\in F(I)$.

You may notice that very similar proof would work for any topological group instead of $\mathbb R$.

If $f,g\in C_0$, then $fg\in C_0$.

Let $U=(-\varepsilon,\varepsilon)$ be any neighborhood of zero. You have $$(fg)^{-1}(U) \supseteq f^{-1}(-1,1) \cap g^{-1}(-\varepsilon,\varepsilon)$$ which shows that $(fg)^{-1}(U)\in F(I)$.

However, all these proofs are simply special cases of proofs of the facts mentioned above for $x=y=0$. Which is why I think that it is better to try to prove the more general facts. (Or simply use them, if you have already seen proofs of those.)