# Is there an example of a collection of functions with the following properties?

I am interested to know an example of a collection $S$ of real valued functions $f : \mathbb{R} \to \mathbb{R}$ with the following properties.

1. The set $S$ is not closed under addition. ($h(x) = f(x) + g(x)$).

2. The set $S$ is not closed under multiplication. ($h(x) = f(x)g(x)$).

3. There exists a binary operator under which the set $S$ is closed.

EDIT : This edit is being made after receiving comments and suggestions from Zev Chonoles and joriki.

The condition 3 is modified as below conditions.

3A. There exists a binary operator $\star$ under which the set $S$ is closed.

3B. The operator $\star$ on the set $S$ has associative property.

3C. $\forall f_1,f_2,f_3 \in S$ if $f_1 \star f_2 = f_3$ then $f_1 \ne f_3$ and $f_2 \ne f_3$.

• This is true for any set with properties 1 and 2, since you can define an arbitrary binary operator on it. I think you'll need to say something about what sort of binary operator you want this to be. – joriki Apr 26 '11 at 8:28

Let $S=\{f_n\}_{n\in\mathbb{N}}$ where $$f_n(x)=\begin{cases}n\text{ if }x\neq0\\\textstyle \frac{1}{2}\text{ if }x=0\end{cases}$$ Clearly $S$ is not closed under addition or multiplication, and $S$ is infinite. Define $f_n\star f_m= f_{n+m}$ for all $n,m\in\mathbb{N}$. This operation is associative (because $+$ is associative on $\mathbb{N}$), and is guaranteed to have $f_n\star f_m\neq f_m$ and $f_n\star f_m\neq f_n$ for all $n,m\in\mathbb{N}$ (because $0\notin\mathbb{N}$).
• @joriki : does these conditions help. 1. Third operator ($\star$) should be associative and if $f_1 \star f_2 = f_3$, then $f_3 \ne f_1$ and $f_3 \ne f_2$ for all $f_1,f_2 \in S$. – Rajesh Dachiraju Apr 26 '11 at 8:41
• @Rajesh: Using bump functions, we can construct $f_n$'s that are not only continuous, but $C^\infty$, and have the property that: $f_n(0)=\frac{1}{2}$ for all $n\in\mathbb{N}$, and $f_n(x)=n$ for all $|x|>1$. These functions, along with the same operation in my answer, will do the trick. – Zev Chonoles Apr 26 '11 at 9:18