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I am trying to compare different topologies on the set of rationals. I feel like the following two are distinct (i.e. some sets are open in one and not in the other), but I can't write a formal enough argument unfortunately. Any feedback is very appreciated!

Here are the topologies on $\mathbb{Q}$:

  • $\tau_1$ is the subspace topology inherited from the real numbers (i.e. taking elements as $U \cap \mathbb{Q}$ for $U$ open in euclidean topology on $\mathbb{R}$)
  • $\tau_2$is the euclidean topology on the rationals (i.e. given by the basis of open balls centered on rationals)

Let $c \in \mathbb{R} \setminus \mathbb{Q}, \epsilon \in \mathbb{Q}$. My clain is that "visually" $(c - \epsilon, c + \epsilon) \cap \mathbb{Q} = U$ is open in $\tau_1$ but not in $\tau_2$. I feel like this is not a valid counter-example, however this is the closest I found.

Alf
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  • Why do you think that $(c - \epsilon, c + \epsilon) \cap \mathbb{Q} = U $ is not open in $\tau_2$? Actually, $U$ is the ball of center $c$ and radius $\epsilon$ so it belongs to the basis of $\tau_2$. – Taladris Jan 17 '23 at 13:17
  • I think that might be a union of two balls of real radius in the metric topology? – Alex K Jan 17 '23 at 13:19
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    @Taladris $c$ is not in $\mathbb{Q}$ so it isn't a ball. But I do think it's open – Alex K Jan 17 '23 at 13:20
  • @AlexK. right, thank you for the correction – Taladris Jan 17 '23 at 13:23
  • I might be wrong for sure, maybe those two topologies are actually equal ? I couldn't prove it and after trial and error I thought it was most likely not equivalent, hence my question, but yes I cannot either be sure it is indeed a counter example. Im also not sure it's the union of two balls, as those potential balls could be centered on irrational numbers as well, hence not per see in the euclidean topology on $\mathbb{Q}$ – Alf Jan 17 '23 at 13:40

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Let $(\Bbb R,\tau_e)$, the real numbers with the standard euclidean topology. Since $\Bbb Q$ is dense in $\Bbb R$, it is easy to verify that a base for the euclidean topology is $\mathcal B=$ {$B(y,R)|y\in\Bbb Q,R\in\Bbb Q_{>0}$}.A very trivial fact to verify is that if $\mathcal B$is a base for $X$ and $Y\subset X$, $\mathcal B_y$={$\mathcal B\cap Y|B\in\mathcal B$}is a base for the subspace topology. Another fact is that if $d:X\times X\to[0,+\infty)$ induces a topology on $X$ and given $Y\subseteq X$, if $d_y:Y\times Y\to[0,+\infty)$ is the restriction of $d$ to $Y$, a base for the subspace topology of $Y$ is $\mathcal B=$ {$B(y,R)\cap Y |y\in\Bbb Q,R\in\Bbb Q_{>0}$}. So the subspace of topology inherited from the real numbers is the same as the topology induced by the restriction. So with your notation $\tau_1=\tau_2$. $\square$

giorgiokyn23
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