I have to show the following: Given is the standard topology in the field of real numbers, denoted as $$ \{U \subseteq \Bbb R: \forall u \in U \,\exists r\,(u,U) > 0, \text{ with } \{w \in \Bbb R: |u-w| < r\}\}$$ It needs to be proven that the standard topology denoted above, $\mathcal O$, holds that $\mathcal O \subseteq \mathcal O_T$, where $\mathcal O_T$ denotes any topology for which the set $(-\infty, a) \cup (a,\infty) a \in \Bbb Q$, is open. What I did for now: I showed that the intervals $(-\infty, a)$ and $(a,\infty)$ are open in $\mathcal O $. I then defined a topology $\mathcal T$ which holds the same open intervals and now try the following.
$(a,b)$ is a basis for $\mathcal O$. I need to deduce that since $(-\infty, a)$ and $(a,\infty)$ are open in $\mathcal T$, that every interval with rational boundaries is open in $\mathcal T$. Any help? Thanks very much
$O_T$ could be discrete, in which case O subset $O_T,$
or it could be { (-oo,a) $\cup$ (a,oo) : a in R },
which is a counterexample to what you want to prove.