Extending a continuous self-map of an open subset to the whole spaceParacompact zero-dimensional space...
Extending a continuous self-map of an open subset to the whole space
Paracompact zero-dimensional space without clopen partition refinementCo-HausdorffificationMinimal totally separated spacesTwo notions of zero-dimensionality for topological spacesMinimal zero-dimensional Hausdorff spacesCompactification of order-disconnected spacesA specific weak analog of the Baire category theorem for a 'continuously indexed' family of setsEmbedding into $Ctimes [0,1]$$T_2$-spaces where all non-empty open sets are homeomorphicAre homogeneous $T_2$-spaces flexible?
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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
edited 5 hours ago
Ramiro de la Vega
9,83313248
asked 6 hours ago
Dominic van der ZypenDominic van der Zypen
15k43280
$endgroup$
add a comment |
$begingroup$
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
edited 5 hours ago
Ramiro de la Vega
9,83313248
asked 6 hours ago
Dominic van der ZypenDominic van der Zypen
15k43280
$endgroup$
$begingroup$
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
5 hours ago
$begingroup$
Is there a non-discrete OEP?
$endgroup$
– Ramiro de la Vega
5 hours ago
add a comment |
$begingroup$
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
edited 5 hours ago
Ramiro de la Vega
9,83313248
asked 6 hours ago
Dominic van der ZypenDominic van der Zypen
15k43280
$endgroup$
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
edited 5 hours ago
Ramiro de la Vega
9,83313248
asked 6 hours ago
Dominic van der ZypenDominic van der Zypen
15k43280
edited 5 hours ago
Ramiro de la Vega
9,83313248
asked 6 hours ago
Dominic van der ZypenDominic van der Zypen
15k43280
edited 5 hours ago
Ramiro de la Vega
9,83313248
edited 5 hours ago
Ramiro de la Vega
9,83313248
edited 5 hours ago
Ramiro de la Vega
9,83313248
9,83313248
asked 6 hours ago
Dominic van der ZypenDominic van der Zypen
15k43280
asked 6 hours ago
Dominic van der ZypenDominic van der Zypen
15k43280
asked 6 hours ago
Dominic van der ZypenDominic van der Zypen
15k43280
15k43280
$begingroup$
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
5 hours ago
$begingroup$
Is there a non-discrete OEP?
$endgroup$
– Ramiro de la Vega
5 hours ago
add a comment |
$begingroup$
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
5 hours ago
$begingroup$
Is there a non-discrete OEP?
$endgroup$
– Ramiro de la Vega
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
Is there a non-discrete OEP?
$endgroup$
– Ramiro de la Vega
5 hours ago
$begingroup$
Is there a non-discrete OEP?
$endgroup$
– Ramiro de la Vega
5 hours ago
add a comment |
1 Answer
1
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oldest
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$begingroup$
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 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.
answered 6 hours ago
user44191user44191
3,0331429
$endgroup$
$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
5 hours ago
add a comment |
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$begingroup$
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 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.
answered 6 hours ago
user44191user44191
3,0331429
$endgroup$
$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
5 hours ago
add a comment |
$begingroup$
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 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.
answered 6 hours ago
user44191user44191
3,0331429
$endgroup$
$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
5 hours ago
add a comment |
$begingroup$
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 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.
answered 6 hours ago
user44191user44191
3,0331429
$endgroup$
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 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.
answered 6 hours ago
user44191user44191
3,0331429
edited 6 hours ago
answered 6 hours ago
user44191user44191
3,0331429
answered 6 hours ago
user44191user44191
3,0331429
answered 6 hours ago
user44191user44191
3,0331429
3,0331429
$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
5 hours ago
add a comment |
$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
5 hours ago
$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
5 hours ago
$begingroup$
Very nice use of ZL and great answer - thanks!
$endgroup$
– Dominic van der Zypen
5 hours ago
add a comment |
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$begingroup$
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
5 hours ago
$begingroup$
Is there a non-discrete OEP?
$endgroup$
– Ramiro de la Vega
5 hours ago