# Extending a continuous self-map of an open subset to the whole space

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?

• 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. – Ramiro de la Vega Feb 25 at 16:48
• Is there a non-discrete OEP? – Ramiro de la Vega Feb 25 at 16:53
• @RamirodelaVega: $\beta \omega$ – Will Brian Feb 26 at 13:55
• @WillBrian, Of course, thanks! – 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.