# Collection of dense subsets as a “fingerprint” for Hausdorff topologies?

Let $(X,\tau)$ be a Hausdorff space and let ${\cal D}$ denote the collection of dense subsets of $(X,\tau)$. Is it possible that there is another Hausdorff topology $\tau_1 \neq \tau$ on $X$ such that the collection of dense subsets of the space $(X,\tau_1)$ also equals ${\cal D}$?

The standard topology and the lower limit topology on $\mathbb{R}$ have the same dense subsets. They are two different topologies(even up to homeomorphism) on the real line.

So the next question could be "Is the collection of open dense subsets a “fingerprint” for Hausdorff topologies?

• FYI, I think yours is a good answer, but I'm on my tablet right now and everything looks different, so I somehow didn't see that you posted an answer while I was typing mine. Now that it is written, I'll leave mine there too as a different example. – Will Brian Feb 8 '16 at 11:13
• The answer to your question is still no. Consider $X = \omega \cup \{p\}$, where every point of $\omega$ is isolated, and $p$ is not. There are many topologies like this, but all of them have the same dense open sets, namely $\omega$ and $X$. – Will Brian Feb 8 '16 at 17:15

Let $\tau_1$ be the usual topology on the real line, and let $\tau_2$ be the finer topology obtained by breaking off the positive reals to form a clopen set: that is, $\tau_2$ has a subbasis consisting of everything in $\tau_1$, plus the positive reals and their complement. It's not too hard to see that $D$ is dense in $\tau_1$ if and only if it is dense in $\tau_2$.

By the following proposition, the only Hausdorff spaces which are completely determined by their dense subsets are the discrete spaces.

Proposition. Let $(X,\mathcal{T})$ be a $T_{0}$ topological space such that there does not exist an $\mathcal{S}$ where

1. $(X,\mathcal{S})$ is a topology

2. $\mathcal{T}\subseteq\mathcal{S}$ and

3. $\mathcal{S}\neq\mathcal{T}$

4. $(X,\mathcal{S})$ and $(X,\mathcal{T})$ have the same dense sets.

Then $(X,\mathcal{T})$ is discrete.

Proof. Suppose to the contrary that $(X,\mathcal{T})$ is not discrete. Then there exists an open set $U$ which is not clopen since the only $T_{0}$ space where every open set is clopen is the discrete space.

Therefore, let $U$ be an open set which is not clopen. Now let $C=\overline{U}$. Let $\mathcal{S}$ be the topology generated by $\mathcal{T}\cup\{C\}$. Then $\mathcal{S}$ is generated by the basis consisting of sets in $\mathcal{T}$ along with the sets of the from $O\cap C$ where $O\in\mathcal{T}$. I claim that $\mathcal{T}$ and $\mathcal{S}$ have the same dense sets. If $D$ is dense in $(X,\mathcal{S})$, then $D$ is clearly dense in $(X,\mathcal{T})$. Conversely, suppose that $D$ is dense in $(X,\mathcal{S})$. Suppose that $O\in\mathcal{T}$ and $O\cap C=O\cap\overline{U}$ is non-empty. Then $O\cap U$ is also non-empty. Therefore, $D\cap O\cap U$ is non-empty, hence $D\cap O\cap C=D\cap O\cap\overline{U}$ is non-empty as well. We therefore conclude that $(X,\mathcal{S})$ and $(X,\mathcal{T})$ both have the same dense sets.