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?



$T_2$ topologies that are "as disjoint as possible"
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.


