I was unable to solve this question of Exercise 2.4 (Weak Topologies)
Question : Let U be the usual topology on $\mathbb{R}$ . Describe the weak topology T on $\mathbb{R}$ induced by each of the following families of functions.
(a)The family of constant functions that map $\mathbb{R}$ into $(\mathbb{R} ,U)$
(b) The family that consists only of the function $i: \mathbb{R} \to (\mathbb{R}, U)$ defined by i(x)=x.
Now the probelm is that i am not able to use the definition of weak topology in here.
Definition given in the book :Let {$X_{\alpha} , T_{\alpha}$}: $\alpha \in \Lambda$ be a indexed family of topological spaces, and for each $\alpha \in \Lambda$ let $f_{\alpha} : X\to X_{\alpha}$ , and for each $\alpha \in \Lambda$ let $ f_{\alpha}: X \to X_{\alpha} $be a function. The weak topology on X induced by {${ f_{\alpha} : \alpha \in \Lambda}$} is smallest topology for which each $f_{\alpha}$ is continuous.
Consider answering (a) part I am unable to understand what should I take $f_{\alpha}$, should I take $\alpha \in \mathbb{R}$ ? and fuamily of functions by $f_{\alpha} $ by $f_{\alpha}(x)=\alpha$ for all x in real numbers.
then i will get the weak topology =$\mathbb{R}$. Am I right?
For (b), I think it should also be $\mathbb{R}$ by the sam elogic as used for answering (a).
Am I right?
For (a): As I think you were trying to say, the weak topology is indeed $\{\mathbb{R}, \emptyset\}$ (you just said $\mathbb{R}$, but every topology must contain the empty set as well, and moreover the empty set is needed here for continuity: for instance, if $f_0$ is the constant function mapping to $0$, then $f^{-1}(1, 2) = \emptyset$ needs to be open)
For (b): The weak topology is again $U$ itself. For any $U$-open set (meaning an open set in the standard topology $U$) $A \subseteq R$, $i^{-1}(A)$ must be open in the weak topology. But $i^{-1}(A) = A$ itself. So the weak topology must contain all the open sets of $U$. Since clearly $i: (\mathbb{R}, U) \rightarrow (\mathbb{R}, U)$ is continuous, and the weak topology is the smallest topology making $i$ continuous, the weak topology is $U$ itself.