Standard topology doesn't match with the topology induced by the p-adic metric

Question

I want to show that the standard topology on $\Bbb Q$ doesn't match with the topology that is induced by the p-adic metric. For this I wanna show that for the open set $M:= \{x \in \Bbb Q: |x|_p <\frac{1}{p}\}$ there exist no ball set in the standard topology with only these elements. I know that in $M$ there are only elements that are divisible by $p^2$ in the counter. But now I am stucked....

Answer 1

In $\Bbb Q$ with the $p$-adic metric, $\lim_{n\to\infty}p^n=0$, but not in $\Bbb Q$ with the usual metric.

Answer 2

To work with your set $M=\{x \in \mathbb Q: \lvert x\rvert_p < 1/p \}$, let's just see that it's not open with respect to the standard topology of $\mathbb Q$ (and that is what you have to show: as explained in the comments, it does not suffice to show that it's not a ball in the standard metric).

Well $0 \in M$ and so for $M$ to be open in the standard topology, there must be some $\varepsilon > 0$ such that $M$ contains $(-\varepsilon, \varepsilon) \cap \mathbb Q$. But now choose $n$ big enough so that (for example) $x:= \frac{1}{(p+1)^n} < \varepsilon$. Then $x \in (-\varepsilon, \varepsilon) \cap \mathbb Q$ but $x \notin M$, contradiction, so $M$ is not open in the standard topology.