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Let $X$ be a topological space, not necessarily nice, let $R$ be a subspace of $X \times X$, and let $\overline{R}$ be the closure of $R$ in $X \times X$. Then:

  • If $R$ is a reflexive binary relation on $X$, then $\overline{R}$ is also a reflexive binary relation on $X$.
  • If $R$ is a symmetric binary relation on $X$, then $\overline{R}$ is also a symmetric binary relation on $X$.

On the other hand, the following claim is false:

  • If $R$ is an equivalence relation on $X$, then $\overline{R}$ is also an equivalence relation on $X$.

Indeed, we could take $X = \mathbb{R}$ and $$R = \left\{ (x_0, x_1) \in \mathbb{R} \,\middle\vert\, \exists z \in \mathbb{Z} . -\frac{1}{2} \le x_0 - z < \frac{1}{2}, -\frac{1}{2} \le x_1 - z < \frac{1}{2} \right\}$$ in which case $\overline{R}$ fails to be a transitive binary relation. (Note that $X / R$ is indiscrete, so the smallest closed equivalence relation containing $R$ is $X \times X$ itself!)

Of course, this cannot happen if $X$ is either discrete or indiscrete. What I would like to know is this:

Question. What are some necessary or sufficient conditions on $X$ that ensure that the closure of an equivalence relation on $X$ is always an equivalence relation?

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This partial answer solves the problem for connected spaces.

Using the idea from the example in the question you can show that if a connected space $X$ has the property that the closure of every equivalence relation on $X$ is an equivalence relation, then $X$ is hyperconnected.

If $X$ is connected but not hyperconnected, let $U \subseteq X$ be a nonempty open set whose complement, $U^{\mathtt{C}}$, has nonempty interior. Let $R$ be the equivalence relation on $X$ whose equivalence classes are $U,U^{\mathtt{C}}$ (i.e., $R = ( U \times U) \cup ( U^{\mathtt{C}} \times U^{\mathtt{C}} )$). As $U$ is not closed, some $v \in U^{\mathtt{C}}$ is a limit point of $U$. Taking $u \in U$ and $w \in \operatorname{Int} (U^{\mathtt{C}})$ it follows that $u \mathrel{\overline{R}} v$ and $v \mathrel{\overline{R}} w$, however $u \not\mathrel{\overline{R}} w$.

Conversely, every hyperconnected space $X$ satisfies the property since the diagonal is dense in $X \times X$ (so that the closure of every reflexive relation on $X$ is $X \times X$).

Expanding this, for a general space to satisfy the property every open connected subspace must be hyperconnected. This condition is not sufficient, as restricting the example from the question to the set $\mathbb{Q}$ of rationals gives an equivalence relation on a totally disconnected space without isolated points (which has no open connected subspaces) whose closure is not transitive.

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