I am self studying topology and I was unable to solve this question.
Wayne Patty's Question 2.4.5.
Let $U$ be usual topology on $\mathbb{R}$ and let $T$ be the weak Topology on $\mathbb{R}$ induced by the family of all bounded real valued continuous functions that map $(\mathbb{R},U) $ to $(\mathbb{R}, U)$. Describe $T$.
The problem is, how do I know which topology will be smallest? Also $f_{\alpha}$ 's can be constant functions and more functions also like $\sin x$.
Also, what approach should one follow to find smallest topologies which make the all $f_{\alpha}$'s continuous?
Can you please help me with this?
If I understand your question right, it seems to me that any interval $(a,b)$ is in $T$.
This is because you can have a (bounded) function such as:
$$f(x)=\begin{cases}-1&x<a\\\text{linear: }\frac{2x-(a+b)}{b-a}&a\le x\le b\\1&x>b\end{cases}$$
and $f^{-1}((-1,1))=(a,b)$. So, as the interval $(-1,1)$ is open in $U$, the interval $(a,b)$ must be open in $T$ if $f$ is to stay continuous as a map $(\mathbb R, T)\to(\mathbb R, U)$.
However, then it follows that every open set in $U$ is also in $T$, i.e. $T=U$.