TITLE: $T_2$ topologies that are "as disjoint as possible"
QUESTION [14 upvotes]: Let $X$ be an infinite set. Are there Hausdorff topologies $\tau_1, \tau_2$ on $X$ such that $\tau_1\cap\tau_2 = \{\emptyset\} \cup \{U\subseteq X: X\setminus U\text{ is finite}\}$? 
(That is, the intersection is as small as it can get.) 
And what about the special case $X=\mathbb{R}$ and $\tau_1$ being the topology coming from the Euclidean metric?

REPLY [9 votes]: Let $\varepsilon$ denote the Euclidean topology on $\mathbb R$.
Proposition: There is a 0-dimensional $T_2$ topology $\tau$ on $\mathbb R$
such that $\tau$ and $\varepsilon$ intersect in only the co-finite sets.
Proof: 
We need the following lemma:
Lemma:
Let $Y=(Y,\nu)$ be an infinite  topological space such that
$$|\overline{E}^\nu|\ge 2^{\omega}$$ for all $E\in [Y]^\omega$.
Then there is an injective map $f:\mathbb R\to Y$
such that 
$$
\forall D\in [\mathbb R]^\omega \text{ if }\overline D\ne \mathbb R
\text{ then }
\exists x_D\in (\mathbb R\setminus \overline D) \ f(x_D)\in \overline{f[D]}^\nu.
$$
Proof of the lemma:
Enumerate $[\mathbb R]^\omega$
as $\{D_{\alpha}:{\alpha}<2^{\omega}\}$.
By transfinite induction define 
an increasing continuous sequence $(f_{\alpha}:{\alpha}\le 2^{\omega})$
of injective functions from subsets of $\mathbb R$ into $Y$ such that 
$|f_{\alpha}|\le {\alpha}+{\omega}$,  $D_{{\beta}}\subset dom(f_{\alpha})$
for ${\beta}<{\alpha}$, and 
$$\text{ if ${\beta}<{\alpha}$ and    }\overline D_{\beta}\ne \mathbb R
\text{ then }
\exists x_{\beta}\in (dom(f_{\alpha})\setminus \overline D_{\beta}) \ 
f_{\alpha}(x_{\beta})\in \overline{f_{\alpha}[D_{\beta}]}^\nu.
$$
Assume that ${\alpha}={\beta}+1$, and we have constructed $f_{\beta}$.
Let $g\supset f_{\beta}$ be an injective function from 
$dom(f_{\beta})\cup D_{\beta}$ into $Y$.
If $\overline{D_{\beta}}=\mathbb R$, then$f_{\alpha}=g$ works.
Assume that $U=\mathbb R\setminus \overline{D_{\beta}}\ne \emptyset$.
Since $|U|=2^{\omega}$ and $|\overline{g[D_{\beta}]}^{\nu}|\ge 2^{\omega}$   we can pick $x_{\beta}\in U\setminus  dom(g)$ 
and $y_{\beta}\in \overline{g[D_{\beta}]}^{\nu}\setminus ran(g)$.
Let 
$$f_{\alpha}=g\cup\{(x_{\beta},y_{\beta})\}$$
This completes the inductive construction.
QED.
Pick a 0-dimensional $T_2$  space $(Y,\nu)$ which meets the requirements of the lemma. 
(For example, $Y=\omega^*$ works)
Apply the lemma  to obtain an injective 
$f:\mathbb R\to Y$.
Define the topology $\tau$ by declaring that $f$
 is a homeomorphism between $(\mathbb R,\tau)$ and $(f[\mathbb R],{\nu})$.
To show that $\tau $ is as required assume that $\emptyset\ne U\in \varepsilon $ such that $F=\mathbb R\setminus U$
is infinite. Let $D$ be a countable $\varepsilon$-dense subset of $F$.
Then there is  $x_D\in \mathbb R\setminus \overline D$ such that
$f(x_D)\in\overline{f[D]}^{\nu}$.
Since $f$ is a homeomorphism, $x_D\in\overline{D}^{\tau}$.
Thus $D\subset F$ implies that $x_D\in \overline{F}^{\tau}\setminus F$,
and so $F$ is not $\tau$-closed. Thus $U\notin\tau$.
Thus we proved the proposition.