Let S=C[0,1] be the set of real-valued continuous functions defined on the closed interval [0,1], where we define f+g and fg, as usual, by (f+g)(x)=f(x)+g(x) and (fg)(x)=f(x)g(x). Let 0 and 1 be the constant functions 0 and 1, respectively. Let a be in the interval [0,1]. Show that the set T={f in S|f(a)=0} is a subring such that fg, gf in T for all f in T and g in S.
I know that I have to show that T is nonempty, closed under subtraction and multiplication. I am just a little confused because there is so much going on in this question.
This is what I did for nonempty: Let f(x)=0 in S, then for all a in [0,1] f(a)=0. Therefore f(x) is in T and T is nonempty.
I am not sure if this part is correct. I am also not sure where to start with showing that T is closed under subtraction and multiplication.
A good start has been made.
$T \ne \emptyset$ since the zero function, $0(\cdot)$ which takes $x \in [0, 1]$ to $0$, $0(x) = 0$, is in $T$, $0(\cdot) \in T$, as our OP NL37 has pointed out. So that is correct.
It is easy to find more functions in $T$; for instance, $x - a \in T$, clearly. And so is any function of the form $(x - a)^kg(x)$, where $k \in \Bbb N$, the naturals, and $g(x) \in S$. There are many more.
As for the rest, the whole thing depends on evaluation at $a$: if $f(x), g(x) \in T$, then $f(a) = g(a) = 0$, so $(f - g)(a) = f(a) - g(a) = 0 - 0 = 0$, whence $(f - g)(x) \in T$; if $f(x) \in T$ and $g(x) \in S$, then $(fg)(a) = f(a)g(a) = 0 \cdot g(a) = 0$, no matter what the value of $g(a)$; so $(fg)(x) = f(x) g(x) \in T$.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!