MUnkre topology. Page no : 126 section : 20 question 3. - b).
Let $X'$ denote a space having the same underlying set as X. Show that if $d : X' × X' → R$ is continuous, then the topology of $X'$ is finer than the topology of $X.$
my attempts :Let the $d \colon \mathbb{R}\times\mathbb{R} \to \mathbb{R}$ be defined by $$d (\alpha, \beta) \colon= \begin{cases} 1 \ \mbox{ if } \ \alpha \neq \beta, \\ 0 \ \mbox{ otherwise;} \end{cases} $$ for all $\alpha, \beta \in \mathbb{R}$. Then $d$ is of course a metric on $\mathbb{R}$.
Now let $d_1 \colon \left( \mathbb{R} \times \mathbb{R} \right) \times \left( \mathbb{R} \times \mathbb{R} \right) \to \mathbb{R}$ be defined by $$d_1 \left( x_1 \times y_1, \ x_2 \times y_2 \right) \colon= \max \left\{ d \left( x_1, x_2 \right), \ \left\vert y_1 - y_2 \right\vert \right\} \ \mbox{ for all } \ x_1 \times y_1, \ x_2 \times y_2 \in \mathbb{R} \times \mathbb{R}.$$ I know that this $d$ is a metric on $\mathbb{R} \times \mathbb{R}$.
now $X' = R \times R$ is finer then $ R$ hence proved
IS its correct or not ??
PLiz help me..
Let $U \subseteq \mathbb{R}$ be open. Assume $\operatorname{d}^{-1}(U) \neq \varnothing$.
Let $(x, y) \in \operatorname{d}^{-1}(U)$ and set $\epsilon_1 = \frac{1}{2}(b-\operatorname{d}(x, y))$
Clearly, if $(\alpha, \beta) \in B_d(x, \epsilon_1) \times B_d(y, \epsilon_1)$, then \begin{eqnarray} \operatorname{d}(\alpha, \beta) &\leq& \operatorname{d}(x, \alpha) + \operatorname{d}(x, y) + \operatorname{d}(y, \beta)\\ &<& \operatorname{d}(x, y) + 2\left(\frac{b-\operatorname{d}(x, y)}{2}\right) \\ \end{eqnarray}
That is,
$$\operatorname{d}(\alpha, \beta) < b$$
Similarly, set $\epsilon_2 = \frac{1}{2}(\operatorname{d}(x, y) - a)$.
Let $(\alpha, \beta) \in B_d(x, \epsilon_2) \times B_d(y, \epsilon_2)$. Assume, for the sake of contradiction, that $\operatorname{d}(\alpha, \beta) \leq a$.
Then,
\begin{eqnarray} \operatorname{d}(x, y) &\leq& \operatorname{d}(x, \alpha) + \operatorname{d}(\alpha, \beta) + \operatorname{d}(\beta, y) \\ \operatorname{d}(x, y) &<& \operatorname{d}(x, y)-a + \operatorname{d}(\alpha, \beta) \\ &<& \operatorname{d}(x, y) \end{eqnarray}
Therefore, it must be the case that $$\operatorname{d}(\alpha, \beta) > a$$
Set $\epsilon = \operatorname{min}\{\epsilon_1, \epsilon_2\}$.
Then, if $(\alpha, \beta) \in B_d(x, \epsilon) \times B_d(y, \epsilon)$,
$$a < \operatorname{d}(\alpha, \beta) < b$$ Since any open set $U$ can be written as union of some family of intervals $\{(a_i,b_i):i\in I\}$, then $$ d^{-1}(U)=d^{-1}(\cup_{i\in I}(a_i,b_i))=\cup_{i\in I}d^{-1}((a_i,b_i)) $$ Thus $d^{-1}(U)$ is open as union of open sets