From my point of view I think it isn't but I'm not sure. Here is what I thought:

$(0,1) \bigcap \mathbb{Q}$ is basically $$ A = \{ \frac a b \in \mathbb{Q} \mid \frac a b \gt 0 \text{ and } \frac a b \lt 1 \}$$ and $\forall x \in A$ and $\forall R \gt 0$ there is a $q \in \mathbb{R} \setminus \mathbb{Q}$ such that $q \in \mathcal{B}_R(x)$ (i.e., open ball of center $x$ and radius $R$), so $\forall x \in A, \mathcal{B}_R(x) \nsubseteq \mathbb{Q}$ so $A \notin $ standard topology of $\mathbb{R}$. Conclusion, $A$ is not an open set in the standard topology of $\mathbb{R}$.

Is that correct ?

  • $\begingroup$ Yes that's correct. Great! $\endgroup$ – Error 404 Jan 29 '17 at 15:06
  • 1
    $\begingroup$ Oh for once I did it . Thank you ! $\endgroup$ – Eduard Valentin Jan 29 '17 at 15:07
  • $\begingroup$ Also $q$ depends on $x$ and $R$. So for more clarity you can write $q(x,R).$ $\endgroup$ – Error 404 Jan 29 '17 at 15:08
  • $\begingroup$ Yes, you are right,there are so many different notation for things. $\endgroup$ – Eduard Valentin Jan 29 '17 at 15:11
  • $\begingroup$ Your question (in the header) asks if it is an open set in a topology of $\Bbb R$; the answer to this is "yes": in the discrete topology on $\Bbb R$, all sets are open. I assume that what you intended was "is it open in the standard topology on $\Bbb R$?", in which case your answer is correct. $\endgroup$ – John Hughes Jan 29 '17 at 15:41

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