i'm trying to verify that something does not satisfies the ascending chain condition (ACC) and some questions appeared,
Let $C(\mathbb{R})$ denote the ring of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ with additive and multiplication defined by $$(f+g)(x)=f(x)+g(x); (fg)(x)=f(x)g(x), \mbox{ for }f,g\in C(\mathbb{R}),x\in \mathbb{R}.$$ For $n=1,2,3,\ldots,$ define $I_n$ to be the subset of $C(\mathbb{R})$ consisting of those functions that maps every element of the interval $[-\frac{1}{n},\frac{1}{n}]$ to $0$. Then $I_n$ is an ideal of $C(\mathbb{R})$ for each $n$ and $$I_1\subset I_2\subset I_3\subset\cdots$$ is an infinite strictly ascending chain of ideals in $C(\mathbb{R})$. So the ACC is not satisfied in $C(\mathbb{R}).$
Ok, so what i tried is the following. First i want to see that $I_n$ is an ideal of $C(\mathbb{R})$. So i wrote the set $I_n$ as the following set
$$I_n=\left\{f\in C(\mathbb{R})\mid f(x)=0\,\,\,\,\,\forall x\in\left[-\frac{1}{n},\frac{1}{n}\right]\right\}$$
Then choose an arbitrary continuous function $g\in C(\mathbb{R})$, and we get that for $f\in I_n$, $$(f\cdot g)(x)=f(x)g(x)$$ and we see that for $x\in[-\frac{1}{n},\frac{1}{n}]$ we have that $$(f\cdot g)(x)=f(x)g(x)=0$$ therefore since $f\cdot g$ is continuous since $f$ and $g$ are continuous, we get that $f\cdot g$ is in $I_n$.
- This is where the first question appears, that is enough to conclude that $I_n$ is an ideal for $C(\mathbb{R})$ or do i need to prove it too for the addition? (In that case, i don't really know what to do so any help would be appreciated too.)
Assuming that $I_n$ is an ideal of $C(\mathbb{R})$, now we want to show that $$I_1\subset I_2\subset I_3\subset\cdots$$ but i don't get the desired result (or don't understand where i'm missing it) since we have the following sets for example $$I_1=\left\{f\in C(\mathbb{R})\mid f(x)=0\,\,\,\,\,\forall x\in\left[-1,1\right]\right\}$$ $$I_2=\left\{f\in C(\mathbb{R})\mid f(x)=0\,\,\,\,\,\forall x\in\left[-\frac{1}{2},\frac{1}{2}\right]\right\}$$ So, i think the subset would be $I_1\supset I_2$ since all the continuous functions that maps $[-\frac{1}{2},\frac{1}{2}]$ to $0$ would be a subset of $I_1$... so,
- The second question would be, what i am doing wrong? do i didn't write correctly the set $I_n$? or i'm missing something else? (If it's the case if im correct here, would it be enough to change the subsets to $[-n,n]$?)
And that's all, if you could help me i would be really aprreciated, thanks in advance.
To answer your first question, yes you also have to show that $I_n$ is an additive subgroup of $C(\mathbb{R})$, i.e. $0 \in I_n$ and for $f,g \in I_n$, we have $f + g \in I_n$ and $-f \in I_n$.
The inclusion $I_n \subset I_{n+1}$ holds because $$[-\tfrac{1}{n},\tfrac{1}{n}] \supset [\tfrac{-1}{n+1},\tfrac{1}{n+1}].$$ If $f(x) = 0$ on $[-\tfrac{1}{n},\tfrac{1}{n}]$, i.e. $f \in I_n$ then $f(x) = 0$ on $[\tfrac{-1}{n+1},\tfrac{1}{n+1}]$, thus $f \in I_{n+1}$.