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Let $X$ be a set and let $\tau_1\subseteq\tau_2$ be topologies on $X$. Suppose that $A\subseteq X$ is nowhere dense in $\left(X, \tau_2\right)$. I was wondering if it follows that $A$ is nowhere dense in $\left(X, \tau_1\right)$ as well.

After trying to prove it, I believe that it is not generally true. I haven't been able to come up with any (preferably non-trivial) counterexample, though. On the other hand, I easily came up with a counterxample showing that being nowhere dense in $\left(X, \tau_1\right)$ does not imply being nowhere dense in $\left(X, \tau_2\right)$.

I would be really grateful for any help. Thank you:)

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Let $\tau_1 = \{\emptyset, X\}$ be the trivial topology. Then any nonempty subset of $X$ is dense in $\tau_1$.

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  • $\begingroup$ Thank you - before trying to come up with some fancy counterexample, I should have tried to find at least some trivial one... However, I'm still looking for some non-trivial counterexample (preferably $X=\mathbb{R}$ or $X=\mathbb{R}^{\omega}$ and $\tau_1$ is at least Hausdorff):) $\endgroup$ – user1321324 Aug 17 '14 at 16:36

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