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












3












$begingroup$


We say that a $T_2$-space has the open extension property (OEP) if for any open set $U$ and continuous map $f:Uto U$ there is a continous map $g:Xto X$ 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 $xneq yin X$ there is a clopen (closed and open) set $U$ such that $xin U$ and $ynotin U$.



Question. Does (OEP) imply total disconnectedness?






gn.general-topology







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$endgroup$








  • 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
    yesterday








  • 3




    $begingroup$
    Is there a non-discrete OEP?
    $endgroup$
    – Ramiro de la Vega
    yesterday






  • 2




    $begingroup$
    @RamirodelaVega: $beta omega$
    $endgroup$
    – Will Brian
    17 hours ago










  • $begingroup$
    @WillBrian, Of course, thanks!
    $endgroup$
    – Ramiro de la Vega
    16 hours ago
















3












$begingroup$


We say that a $T_2$-space has the open extension property (OEP) if for any open set $U$ and continuous map $f:Uto U$ there is a continous map $g:Xto X$ 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 $xneq yin X$ there is a clopen (closed and open) set $U$ such that $xin U$ and $ynotin U$.



Question. Does (OEP) imply total disconnectedness?






gn.general-topology







share|cite|improve this question











$endgroup$








  • 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
    yesterday








  • 3




    $begingroup$
    Is there a non-discrete OEP?
    $endgroup$
    – Ramiro de la Vega
    yesterday






  • 2




    $begingroup$
    @RamirodelaVega: $beta omega$
    $endgroup$
    – Will Brian
    17 hours ago










  • $begingroup$
    @WillBrian, Of course, thanks!
    $endgroup$
    – Ramiro de la Vega
    16 hours ago














3












3








3





$begingroup$


We say that a $T_2$-space has the open extension property (OEP) if for any open set $U$ and continuous map $f:Uto U$ there is a continous map $g:Xto X$ 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 $xneq yin X$ there is a clopen (closed and open) set $U$ such that $xin U$ and $ynotin U$.



Question. Does (OEP) imply total disconnectedness?






gn.general-topology







share|cite|improve this question











$endgroup$




We say that a $T_2$-space has the open extension property (OEP) if for any open set $U$ and continuous map $f:Uto U$ there is a continous map $g:Xto X$ 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 $xneq yin X$ there is a clopen (closed and open) set $U$ such that $xin U$ and $ynotin U$.



Question. Does (OEP) imply total disconnectedness?






gn.general-topology




gn.general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Ramiro de la Vega

9,83313248




9,83313248










asked yesterday









Dominic van der ZypenDominic van der Zypen

15k43280




15k43280








  • 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
    yesterday








  • 3




    $begingroup$
    Is there a non-discrete OEP?
    $endgroup$
    – Ramiro de la Vega
    yesterday






  • 2




    $begingroup$
    @RamirodelaVega: $beta omega$
    $endgroup$
    – Will Brian
    17 hours ago










  • $begingroup$
    @WillBrian, Of course, thanks!
    $endgroup$
    – Ramiro de la Vega
    16 hours ago














  • 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
    yesterday








  • 3




    $begingroup$
    Is there a non-discrete OEP?
    $endgroup$
    – Ramiro de la Vega
    yesterday






  • 2




    $begingroup$
    @RamirodelaVega: $beta omega$
    $endgroup$
    – Will Brian
    17 hours ago










  • $begingroup$
    @WillBrian, Of course, thanks!
    $endgroup$
    – Ramiro de la Vega
    16 hours ago








1




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
yesterday






$begingroup$
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
yesterday






3




3




$begingroup$
Is there a non-discrete OEP?
$endgroup$
– Ramiro de la Vega
yesterday




$begingroup$
Is there a non-discrete OEP?
$endgroup$
– Ramiro de la Vega
yesterday




2




2




$begingroup$
@RamirodelaVega: $beta omega$
$endgroup$
– Will Brian
17 hours ago




$begingroup$
@RamirodelaVega: $beta omega$
$endgroup$
– Will Brian
17 hours ago












$begingroup$
@WillBrian, Of course, thanks!
$endgroup$
– Ramiro de la Vega
16 hours ago




$begingroup$
@WillBrian, Of course, thanks!
$endgroup$
– Ramiro de la Vega
16 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

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, $z in bar{V}$. 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 $x in U notni y$, 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Very nice use of ZL and great answer - thanks!
    $endgroup$
    – Dominic van der Zypen
    yesterday











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1 Answer
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active

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6












$begingroup$

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, $z in bar{V}$. 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 $x in U notni y$, 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Very nice use of ZL and great answer - thanks!
    $endgroup$
    – Dominic van der Zypen
    yesterday
















6












$begingroup$

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, $z in bar{V}$. 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 $x in U notni y$, 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Very nice use of ZL and great answer - thanks!
    $endgroup$
    – Dominic van der Zypen
    yesterday














6












6








6





$begingroup$

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, $z in bar{V}$. 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 $x in U notni y$, 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.






share|cite|improve this answer











$endgroup$



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, $z in bar{V}$. 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 $x in U notni y$, 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









user44191user44191

3,0631431




3,0631431












  • $begingroup$
    Very nice use of ZL and great answer - thanks!
    $endgroup$
    – Dominic van der Zypen
    yesterday


















  • $begingroup$
    Very nice use of ZL and great answer - thanks!
    $endgroup$
    – Dominic van der Zypen
    yesterday
















$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
yesterday




$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
yesterday


















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