Thread: Determining if this subset is a subring

1. Determining if this subset is a subring

Hi there. Firstly, sorry if the title wasn't overly descriptive.

I'm working on an assignment question dealing with group/ring theory that is:

Let $R=\{f : [0, 1] \to \mathbb{R}\}$ be the ring of all real-valued functions defined on the closed interval [0, 1]. Decide if the folowing subset of R is a subring of R:

$S = \{f \in R | f(x) = 0$ has an infinite number of solutions in [0, 1] $\}$

I know since it is already a subset of R that I need to check if:

for $g, h \in S$ if

$g - h \in S$ and $gh \in S$

So basically whether

$g - h = 0$ and $gh = 0$

will also have an infinite number of solutions.

Another question was similar but the subset consisted of the functions of R that had only a finite number of solutions in [0,1]. I was able to determine by contradiction using an example that it indeed was not, so my intuition here is that this subset is a subring. I'm just not sure how to start it.

Thanks for any help,
--
Dave

2. Originally Posted by driegert
Hi there. Firstly, sorry if the title wasn't overly descriptive.

I'm working on an assignment question dealing with group/ring theory that is:

Let $R=\{f : [0, 1] \to \mathbb{R}\}$ be the ring of all real-valued functions defined on the closed interval [0, 1]. Decide if the folowing subset of R is a subring of R:

$S = \{f \in R | f(x) = 0$ has an infinite number of solutions in [0, 1] $\}$

I know since it is already a subset of R that I need to check if:

for $g, h \in S$ if

$g - h \in S$ and $gh \in S$

Another question was similar but the subset consisted of the functions of R that had only a finite number of solutions in [0,1]. I was able to determine by contradiction using an example that it indeed was not, so my intuition here is that this subset is a subring. I'm just not sure how to start it.

Thanks for any help,
--
Dave

Take $g(x):=\left\{\begin{array}{ll}0&if\,0\leq x<\frac{1}{2}\\1&if\,\frac{1}{2}\leq x\leq 1\end{array}\right.\,,\,\,h(x):=\left\{\begin{arra y}{ll}1&if\,1\leq x<\frac{1}{2}\\0&if\,\frac{1}{2}\leq x\leq 1\end{array}\right.$ to show S is not a subring.

Obviously you people don't require a subset to contain the ring's unit in other to be considered a candidate

to be a subring, otherwise S is automatically not a subring.

Tonio

3. In order to be a ring doesn't it just have to be associative under addition and multiplication, commutative under addition, 0 and negatives exist, and obey the distributive law? A ring with unity (identity element) is in addition to a basic ring I thought.

I could be completely wrong however and any clarification would help my understanding!

Thank you as well for your help Tonio. Really appreciate it.