In a previous homework problem, I proved that $[a,b]\subset \mathbb R$ is closed. Therefore, in the case that $f(x)=g(x)$ for all $x\in [a,b]$, then $T=[a,b]$ which is closed by a previous proof. I'm allowed to cite that problem in my work. Since, the greatest case is $f(x)=g(x)\space \forall x\in[a,b]$ and T would be closed– $T$ can only become a smaller set from this point on. However, I think this problem is trying to get us to prove that any subset of a closed set, is also closed. I don't know if this is true, but this is what my intuition is telling me the problem actually is.
I don't know how to write a proof saying that maybe card$T$ is finite and so its a subset of the greater $T$ mentioned earlier which is closed so it must be closed. However, if we look at $f(x)=x^2$ and $g(x)=x$ which have two intersection points. In this case, $T=\{0,1\}$ and so looking at the definition of a closed set, it requires that every accumulation point of the set, belongs to the set. Digging further, looking at the definition of an accumulation point, $0,1$ can only be accumulation points iff every neighborhood of $0$ or $1$ contain infinitely many points in $T$, but $T$ is not infinite. Leading me to think that maybe every subset of the "greater" closed $T$ mentioned earlier is not closed.
I'm not sure where to go from here.
Best Answer
$T= \{x \in \mathbb R \mid (f-g)(x)=0\} = (f-g)^{-1}(\{0\})$, i.e. $T$ is the inverse image under $f-g$ of the singleton $\{0\}$. As $\{0\}$ is closed (a singleton set is closed) and $f-g$ continuous, $T$ is closed.