We say that a $T_2$-space has the open extension property (OEP) if for any open set $U$ and continuous map $f:U\to U$ there is a continous map $g:X\to X$ such that $g|_U = f$.

The space $\mathbb{R}$ with the Euclidean topology does not have this property: consider $(0,1)\cup(1,2)$ and the map $f$ sending $(0,1)$ to $\frac{1}{2}$ and $(1,2)$ to $\frac{3}{2}$.

We say that a space $(X,\tau)$ is totally disconnected if for $x\neq y\in X$ there is a clopen (closed and open) set $U$ such that $x\in U$ and $y\notin U$.

Question. Does (OEP) imply total disconnectedness?

  • 1
    $\begingroup$ In fact OEP implies extremally disconnected (which is a lot stronger than totally disconnected), essentially only using your argument that $\mathbb{R}$ is not OEP. $\endgroup$ – Ramiro de la Vega Feb 25 at 16:48
  • 3
    $\begingroup$ Is there a non-discrete OEP? $\endgroup$ – Ramiro de la Vega Feb 25 at 16:53
  • 3
    $\begingroup$ @RamirodelaVega: $\beta \omega$ $\endgroup$ – Will Brian Feb 26 at 13:55
  • $\begingroup$ @WillBrian, Of course, thanks! $\endgroup$ – Ramiro de la Vega Feb 26 at 14:56

Take any pair of distinct points $x, y$. Define a partial order on the set of pairs of disjoint open sets $U \ni x, V \ni y$ by double inclusion (i.e. $(U_1, V_1) \preceq (U_2, V_2)$ if $U_1 \subseteq U_2, V_1 \subseteq V_2$). The set of pairs is nonempty, as $X$ is $T_2$. Then by Zorn's lemma, there is a maximal such pair $(U, V)$. I claim that if $X$ has the OEP, then $X = U \cup V$, and therefore $U$ is clopen.

Assume otherwise, and that there is some $z \notin U \cup V$. Then $z \in \bar{U}$, as otherwise, there would be some open set $W \ni z$ not meeting $U$, and $(U, V \cup W)$ would be a larger pair than $(U, V)$. Similarly, $z \in \bar{V}$.

Edit: Zorn's lemma is unnecessary. Let $U_1 \ni x, V_1 \ni y$ be disjoint open sets (as guaranteed by $T_2$). Then let $U = \overline{\text{int}(\overline{U}^c)}^c$, and let $V = \text{int}(\overline{U}^c)$. Note that $U = \text{int}(V^c)$ and that $V = \text{int}(U^c)$.

Define $f: U \cup V \rightarrow U \cup V$ by $f(U) = x, f(V) = y$. This is clearly continuous (as the preimages are $\emptyset, U, V,$ and $U \cup V$), so by the OEP, there must be some $g: X \rightarrow X$ extending $f$. But then because $X$ is $T_2$, we have that $f(z) = x, f(z) = y$ - a contradiction. Therefore, there can be no such $z$, so $X = U \cup V$. Therefore, $U = V^c$ is clopen, and $x \in U \not\ni y$, so we are done.

There's not much that requires the "target space" of the maps to be $X$; this proof works for any $T_2$ space. It also doesn't require $f$ to have outputs in $U$ specifically; having more than one point (and so can distinguish between $U$ and $V$) is enough.

  • $\begingroup$ Very nice use of ZL and great answer - thanks! $\endgroup$ – Dominic van der Zypen Feb 25 at 16:12

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.