If $(X,\tau)$ is a topological space, we can consider the product topology on $X\times X$ and take the closure of the diagonal $\Delta_X = \{(x,x): x\in X\}$, which we denote by $\mathrm{cl}(\Delta_X)$. Obviously, $\mathrm{cl}(\Delta_X)$ is a symmetric binary relation.

Now we can take things upside down: Let $X$ be a set and let $R\subseteq X\times X$ be a reflexive and symmetric relation. Is there a topology $\tau$ on $X$ such that $\mathrm{cl}(\Delta_X)=R$?

  • $\begingroup$ That's right - I just edited my post accordingly. Thanks! $\endgroup$ – Dominic van der Zypen Sep 2 '14 at 12:46
  • $\begingroup$ Consider at least 4-point set $X$, and an involution $\ i:X\rightarrow X\ $ without any fixed point (i.e. $\ (i\circ i)(x)=x\ne i(x)\ $ for every $\ x\in X).\ $ Let $\ R:= \{(x\ y)\in X\times x: y\ne x \}.\ $ Then on one hand $X$ would be discrete, while on the other hand $R$ is different from the diagonal $\ \{(x\ x):x\in X\}.$ $\endgroup$ – Włodzimierz Holsztyński Sep 2 '14 at 13:38
  • $\begingroup$ I said discrete but it can be anything because the would be space would not be a topological space. $\endgroup$ – Włodzimierz Holsztyński Sep 2 '14 at 13:46
  • $\begingroup$ Just for non-general-topologists, Bourbaki has proved that a topological space $\ X\ $ is Hausdorff $\ \Leftrightarrow\ $ the diagonal of $\ X\times X\ $ is closed. Thus indeed for each fixed point free involution in any at least 4-point space we gat a counterexample. $\endgroup$ – Włodzimierz Holsztyński Sep 2 '14 at 14:02
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    $\begingroup$ @WlodzimierzHolsztynski I suggest that you post your comment as an answer. $\endgroup$ – Joel David Hamkins Sep 2 '14 at 14:20

The answer is no, not necessarily. For a counterexample, let $X=\mathbb{R}$ and let $aRb\iff a=b \text{ or } |a-b|\geq 1$, the "equal or differ by at least one" relation. This is symmetric and reflexive. Suppose $\tau$ is a topology on $\mathbb{R}$ with $\text{cl}(\Delta)=R$. For any real number $k$, the $k^{\rm th}$ slice in the plane leads to $(k-1,k)\cup(k,k+1)$ being open with respect to $\tau$. By taking intersections of such sets, we get tiny open intervals $(a,b)$ being open, and this violates $\text{cl}(\Delta)=R$.

  • $\begingroup$ Thanks! - Is it also possible to find an equivalence relation $R$ on a set $X$ such that there is no topology $\tau$ on $X$ such that $\mathrm{cl}(\Delta_X) = R$? $\endgroup$ – Dominic van der Zypen Sep 2 '14 at 13:29
  • $\begingroup$ Sorry, I’m missing something. How can you get a tiny interval by taking unions of unbounded sets? $\endgroup$ – Emil Jeřábek Sep 2 '14 at 13:30
  • $\begingroup$ No, I believe every equivalence relation will be the closure of the diagonal, if you take the discrete topology on the classes, with the indiscrete topology inside each class. The relation $R$ looks like a diagonal bunch of squares. $\endgroup$ – Joel David Hamkins Sep 2 '14 at 13:30
  • $\begingroup$ Emil, oops, I have confused two versions of examples I was looking at. I edited with a different argument--does it work now? $\endgroup$ – Joel David Hamkins Sep 2 '14 at 13:32
  • $\begingroup$ Hi @dominiczypen!, I didn´t know you were on MO! I just mentioned you in my answer, what are the odds? $\endgroup$ – Ramiro de la Vega Sep 2 '14 at 13:37

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).


According to this article of Colasante and Van Der Zypen, it is an open problem to characterize those reflexive and symmetric relations for which there is such a topology. If we restrict ourselves to $T_2$ topologies then we all know that only the identity relation admitts one. In the paper I just mentioned, Colasnate and Van Der Zypen show that an equivalence relation on an infinite $X$ admits a $T_1$ topology if and only if it has infinitely many equivalence classes or every finite equivalence class is a singleton.


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