Let $(X,\tau_1)$ and $(X,\tau_2)$ be two given infinite topological spaces (i.e., here $X$ is an infinite set) such that $\tau_2\subset \tau_1$ and $\tau_1$ is not discreet and $\tau_2$ is not indiscreet. My questions are,

  • Does there always exist an open map $f:(X,\tau_1)\to (X,\tau_2)$ such that it is not continuous?

  • Does there always exist a surjective open map $f:(X,\tau_1)\to (X,\tau_2)$ such that it is not continuous?

  • Does there always exist an injective open map $f:(X,\tau_1)\to (X,\tau_2)$ such that it is not continuous?

Although I think that each of the question can be answered in the negative, I can't find any example to confirm my guess.

Can anyone help me?

  • 1
    $\begingroup$ I think you will find this question of interest $\endgroup$ – MPW Mar 2 '17 at 13:43
  • $\begingroup$ So you think there are no such maps that are injective or surjective? at least for some pair of $\tau_1$ and $\tau_2$.? $\endgroup$ – Henno Brandsma Mar 4 '17 at 6:07

If $X$ is any infinite set, consider $\tau_1$ the discrete topology. $\tau_2$ the indiscrete one (just $\{\emptyset, X\}$. Clearly $\tau_2 \subset \tau_1$.

If $f:(X, \tau_1) \rightarrow (X, \tau_2)$ is any function (injective/surjective or not), $f$ is always continuous. So for this specific pair of topologies, no such map can exist. If $f$ is constant it cannot be open (as $f[X] \subsetneq X$), and if $f$ is not constant, it assumes two values $f(x) \neq f(y)$, but then $U = \{x\}$ is open in $\tau_1$ but $f(y) \notin f[U] = \{f(x)\}$, so $X \neq f[U]$ is not open. So no function $f$ between these topologies can be open either. So the answer for this pair on any set is no.

Unless I misunderstand the question and you want a positive construction for some pair of topologies? As you pose it, no there isn't always such $f$. There might be for some pair of topologies on some infinite set $X$, I'd have to think about it. Switching the topologies in domain an codomain the answer would always be yes, by using the identity function (and I misread it that way at first); this might have triggered the question, perhaps.

  • $\begingroup$ The inclusion is strict. However, I don't understand your last line. To show that $f$ is not continuous we need to show that there exists an open $U\in \tau_2$ such that $f^{-1}(U)\not\in \tau_1$ (where $f^{-1}(U)$ according to your definition is just $U$). Where have you done that? $\endgroup$ – user 170039 Mar 4 '17 at 3:19
  • $\begingroup$ I still don't understand how $O\in\tau_1\setminus \tau_2$ is exactly the open set I need to disprove continuity. For example if I have $O$ is only closed in $\tau_2$ then we are done. But if that's not the case I don't understand how your comment should be interpreted. $\endgroup$ – user 170039 Mar 4 '17 at 6:31
  • $\begingroup$ @user170039 You saw I totally rewrote my answer? $\endgroup$ – Henno Brandsma Mar 4 '17 at 7:03
  • $\begingroup$ Oh, sorry then. Since you didn't remove your comment, I thought that it still applies to the current answer. $\endgroup$ – user 170039 Mar 4 '17 at 7:11

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