Continuing the topic of Topological "closure" of a binary relation:

Let $\mu$, $\nu$ be binary relations on a set $U$.

Topology $T \mu = \{ E \in \mathscr{P} U \mid \mu [E] \subseteq E \}$ (here $\mu[E]$ is the image of a set $E$ by binary relation $\mu$).

By definition, a function $f$ (on $U$) is a continuous function from $\mu$ to $\nu$ iff $f\circ\mu\subseteq\nu\circ f$. (It is the definition of discrete continuity if we consider $\mu$ and $\nu$ as directed graphs.)

Conjecture If $f$ is a continuous function from $\mu$ to $\nu$ then $f$ is a continuous function from the topology $T\mu$ to the the topology $T\nu$.


I will identify functions with their "transpose graphs", and not their usual graphs: $$f=\left\{(f(x),x):x\in U\right\}$$ and composition as $$(z,y)(y,x)=(z,x)$$ so that composition of relations and functions becomes $$\nu\circ f=\left\{(z,x):(z,f(x))\in \nu\right\}$$ $$f\circ\mu=\left\{(f(y),x):(y,x)\in\mu\right\}$$ (this is actually more suitable for usual functional notation, where "functions act on the left of elements", but I digress).

First assume $f\circ\mu\subseteq\nu\circ f$, and suppose $E\in T_\nu$, that is, $\nu[E]\subseteq E$. We will prove that $f^{-1}(E)\in T_\mu$, that is, $\mu[f^{-1}(E)]\subseteq f^{-1}(E)$.

Let $y\in\mu[f^{-1}(E)]$. Then there is $x\in U$ with $x\in f^{-1}(E)$ and $(y,x)\in\mu$, thus $$(f(y),y)\in f\quad\text{and}\quad (y,x)\in\mu\quad \text{which implies}\quad (f(y),x)\in f\circ\mu\subseteq\nu\circ f$$ and this means that $(f(y),f(x))\in\nu$. But since $x\in f^{-1}(E)$ (i.e. $f(x)\in E$) then $f(y)\in\nu[E]\subseteq E$, so $y\in f^{-1}(E)$.

This proves that $f$ is $(T_\mu,T_\nu)$-continuous.

The reverse is not true if $\nu$ is not transitive. Let $U=\left\{1,2,3\right\}$, $\mu=\left\{(3,1)\right\}$, $\nu=\left\{(3,2),(2,1)\right\}$. The $\nu$-invariant subsets are $U$, $\left\{2,3\right\}$, $\left\{3\right\}$ and $\varnothing$, and these are also $\mu$-invariant, so the identity map $f=\operatorname{id}_U$ is $(T_\mu,T_\nu)$-continuous, but $$\operatorname{id}_U\circ\mu=\mu\not\subseteq\nu=\nu\circ\operatorname{id}_U.$$

If $\nu$ is transitive then the reverse is true but I'll leave the details to you: An element of $f\circ\mu$ is of the form $(f(y),x)$ where $(y,x)\in\mu$. Let $E_0=\left\{f(x)\right\}$, $E_{n+1}=\nu[E_n]$ and $E=\bigcup_{n\geq 0}E_n$. Then $E$ is $\nu$-invariant so $f^{-1}(E)$ is $\mu$-invariant (by continuity) and contains $x$. Use transitivity of $\nu$ and the definition of $E$ to conclude $(f(y),f(x))\in \nu$ and therefore $(f(y),x)\in\nu\circ f$.

  • $\begingroup$ Why $x\in E$? do you mean $y\in E$? if it is $x\in E$, how follows $f(y)\in\nu[E]$? $\endgroup$ – porton May 24 '18 at 12:47
  • $\begingroup$ I (yet) don't see a reason for either $x\in E$ or $y\in E$ $\endgroup$ – porton May 24 '18 at 12:51
  • $\begingroup$ @porton I got confused at some point and wrote $x\in E$ instead of $x\in f^{-1}(E)$, that's all.... $\endgroup$ – Luiz Cordeiro May 24 '18 at 13:47
  • 1
    $\begingroup$ Do you want me to refer to this answer or mention your name in the proof of a generalization of the questioned conjecture? $\endgroup$ – porton May 24 '18 at 14:35
  • $\begingroup$ @porton You could refer to this answer. It is always good practice to make references when appropriate. Thanks $\endgroup$ – Luiz Cordeiro May 24 '18 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.