Let $\ |X|\ge 4,\ $ e.g. $\ X := \mathbb Z/4.\ $ Let $\ j : X\rightarrow X\ $ be an involution without any fixed point, i.e. $\ (j\circ j)(x) = x \ne j(x)\ $ for every $\ x\in X;\ $ e.g. $\ j(x) = x+2 \mod 4\ $. Let

$$ R\ :=\ \{ (x\ y)\in X^2\ :\ y\ne j(x) \} $$

Then there does not exist a topology $\ T\ $ in $\ X\ $ such that $\ R\ $ is the closure of the diagonal $\ D := \{ (x\ x):\ x\in X\}\ \subseteq X^2$.

**PROOF** By a contradiction, let $\ T\ $ be a topology in $\ X\ $ as described above. The functions $\ I_X : X\rightarrow X\ $ and $\ \imath_x : \{x\} \rightarrow X\ $ such $\ I(t):=t\ $ for every $\ t\in X\ $ and $\ \imath_x(x):=x\ $ are continuous. Thus the diagonal product $\ I_X\Delta\imath_x : X\rightarrow X^2\ $ is continuous; and so is the projection $\ \pi_1:X^2\rightarrow X\ $ given by $\ \pi_1(t\ u) := t.\ $ It follows that the canonical bijection

$$\ b_x := \pi_1|X\times\{x\} :\ X\times \{x\} \rightarrow X$$

is a homeomorphism. Thus $\ X\setminus\{\imath(y)\} = \pi_1(R\ \cap\ X\times \{y\}\ $ is closed in $\ X\ $ for each $\ y\in X,\ $ hence $\ \{x\}\ $ is open in $\ X\ $ for $\ y:=j(x);\ $ this means that $\ \{x\} $ is open for every $\ x\in X,\ $ i.e. $\ X\ $ would be Hausdorff. However the closure $\ R\ $ of the diagonal is different from the diagonal--a contradiction. (*so much for an intro to Gen. Top.; sorry to be boring*).

discretebut it can be anything because the would be space would not be a topological space. $\endgroup$ – Włodzimierz Holsztyński Sep 2 '14 at 13:46