What is the interior of this set? (weak topology) Here the $\mathbb R$ is a set of real numbers. Let the topological space, $T$ of wich basis are $B =\{(a,b] \vert a,b\in \mathbb R\} $
Define $f_1, f_2 : (\mathbb R,T_1) \rightarrow (\mathbb R,T)$  by $f_1(x) =x^2$ and $f_2(x) = -x^2$ respectively. Here the $(\mathbb R ,T_1)$  be the smallest topological spaces defined by both of the $f_1$ and $f_2$ are continuous on $\mathbb R$. Find the interior of the $[0,1)$ in $(\mathbb R, T_1)$
In my solution all the $a$ to $e$  are positive real number. In the $f_1$ case, the open set $G_1$ can be $\{0\}$ and $[-b,-a) \cup (a,b] $ considering the $f_1^{-1}((-a,0])$ and $f_1^{-1}((a^2,b^2])$  set in order.  Vice versa when the $f_2$ case with the same method, open set $G_2$ would be $(-c,c)$ and $(-c,-d] \cup [d,c)$ considering the $f_2^{-1}((-c^2,0])$ and $f_2^{-1}((-c^2,-d^2])$ in sequence.
Consequently, the $T$ should have a subbasis as $G_1$ and $G_2$. The open set form of the $T $ are like the below
(1) $\{0\}$ considering the intersection of $G_1 =\{0\}$ and $G_2 = (-c,c)$
(2) $(-e,-d) \cup (d,e)$ considering the intersection of $G_1 = [-b,-a) \cup (a,b] $ and $G_2= (-c,-d] \cup [d,c) $
So my answer is "interior of the $[0,1) $ is $\{0\}$". But the answer was $\phi$ in my workbook. What did I wrong?

A base for $\mathcal{T}_1$ is given by all sets of the form
$$f_1^{-1}[O]\cap f_2^{-1}[O'] \text{, where } O,O' \in \mathcal{T}$$
So $\{0\} = f_1^{-1}[(-1,0]] \cap f_2^{-1}[(-1,0]] = \{0\}\cap (-1,1)$ is indeed an open neighbourhood of $0$, so the interior of $[0,1)$ contains at least that. So your book's answer of $\emptyset$ is not correct.
