Open Mapping Theorem proof
$begingroup$
In my notes on complex analysis we have this version of the Open Mapping Theorem
The image of an open set under a non-constant holomorphic map is open.
The proof begins by stating that it is enough to show that the image of an open disc is open under a non-constant holomorphic map.
I do not understand why this is the case.
complex-analysis
asked Jan 9 at 14:59
pureundergradpureundergrad
535210
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add a comment |
$begingroup$
In my notes on complex analysis we have this version of the Open Mapping Theorem
The image of an open set under a non-constant holomorphic map is open.
The proof begins by stating that it is enough to show that the image of an open disc is open under a non-constant holomorphic map.
I do not understand why this is the case.
complex-analysis
asked Jan 9 at 14:59
pureundergradpureundergrad
535210
$endgroup$
add a comment |
$begingroup$
In my notes on complex analysis we have this version of the Open Mapping Theorem
The image of an open set under a non-constant holomorphic map is open.
The proof begins by stating that it is enough to show that the image of an open disc is open under a non-constant holomorphic map.
I do not understand why this is the case.
complex-analysis
asked Jan 9 at 14:59
pureundergradpureundergrad
535210
$endgroup$
In my notes on complex analysis we have this version of the Open Mapping Theorem
The image of an open set under a non-constant holomorphic map is open.
The proof begins by stating that it is enough to show that the image of an open disc is open under a non-constant holomorphic map.
I do not understand why this is the case.
complex-analysis
complex-analysis
asked Jan 9 at 14:59
pureundergradpureundergrad
535210
asked Jan 9 at 14:59
pureundergradpureundergrad
535210
asked Jan 9 at 14:59
pureundergradpureundergrad
535210
asked Jan 9 at 14:59
pureundergradpureundergrad
535210
asked Jan 9 at 14:59
pureundergradpureundergrad
535210
535210
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Every open set is an union of open discs. If you prove that the image of an open disc is open, then, for every open set $U$ we have
$$U=bigcup_{iin I}D_i$$
$$f(U)=fleft(bigcup_{iin I}D_iright)=bigcup_{iin I}f(D_i)$$
So if each $f(D_i)$ is open, then $f(U)$ is open.
answered Jan 9 at 15:04
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
$begingroup$
since the open discs form a basis of the topology, every open set can be written as the union of open discs.
So lets assume we have an Open $U$ then $U= bigcup_{iin I} D_i$ for $D_i$ open disks (observe that $I$ is an arbitrary index set!). Now $$f(U)=f(bigcup_{iin I} D_i)=bigcup_{iin I}f(D_i)$$
and so since the righthandside is open, the lefthandside is as well.
Furthermore, since every open disk is open we have that a map $f:mathbb{C} to X$ is open if and only if every open disk gets sent to an open disk.
answered Jan 9 at 15:06
EnkiduEnkidu
1,28319
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add a comment |
$begingroup$
You can cover any non-empty open set of the plane by open discs. Then any point in the image of this open set will be in the image of such a disc, and therefore will have an open neighborhood included in the image.
answered Jan 9 at 15:05
A. BailleulA. Bailleul
1794
$endgroup$
$begingroup$
"cover by" is not what you need here; in fact you can cover any subset of the plane by open disks...
$endgroup$
– David C. Ullrich
Jan 9 at 15:41
$begingroup$
You are right, what I meant is that such an open set can be written as the union of such open discs.
$endgroup$
– A. Bailleul
Jan 9 at 15:53
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Every open set is an union of open discs. If you prove that the image of an open disc is open, then, for every open set $U$ we have
$$U=bigcup_{iin I}D_i$$
$$f(U)=fleft(bigcup_{iin I}D_iright)=bigcup_{iin I}f(D_i)$$
So if each $f(D_i)$ is open, then $f(U)$ is open.
answered Jan 9 at 15:04
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
$begingroup$
Every open set is an union of open discs. If you prove that the image of an open disc is open, then, for every open set $U$ we have
$$U=bigcup_{iin I}D_i$$
$$f(U)=fleft(bigcup_{iin I}D_iright)=bigcup_{iin I}f(D_i)$$
So if each $f(D_i)$ is open, then $f(U)$ is open.
answered Jan 9 at 15:04
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
$begingroup$
Every open set is an union of open discs. If you prove that the image of an open disc is open, then, for every open set $U$ we have
$$U=bigcup_{iin I}D_i$$
$$f(U)=fleft(bigcup_{iin I}D_iright)=bigcup_{iin I}f(D_i)$$
So if each $f(D_i)$ is open, then $f(U)$ is open.
answered Jan 9 at 15:04
ajotatxeajotatxe
53.7k23890
$endgroup$
Every open set is an union of open discs. If you prove that the image of an open disc is open, then, for every open set $U$ we have
$$U=bigcup_{iin I}D_i$$
$$f(U)=fleft(bigcup_{iin I}D_iright)=bigcup_{iin I}f(D_i)$$
So if each $f(D_i)$ is open, then $f(U)$ is open.
answered Jan 9 at 15:04
ajotatxeajotatxe
53.7k23890
answered Jan 9 at 15:04
ajotatxeajotatxe
53.7k23890
answered Jan 9 at 15:04
ajotatxeajotatxe
53.7k23890
answered Jan 9 at 15:04
ajotatxeajotatxe
53.7k23890
53.7k23890
add a comment |
add a comment |
$begingroup$
since the open discs form a basis of the topology, every open set can be written as the union of open discs.
So lets assume we have an Open $U$ then $U= bigcup_{iin I} D_i$ for $D_i$ open disks (observe that $I$ is an arbitrary index set!). Now $$f(U)=f(bigcup_{iin I} D_i)=bigcup_{iin I}f(D_i)$$
and so since the righthandside is open, the lefthandside is as well.
Furthermore, since every open disk is open we have that a map $f:mathbb{C} to X$ is open if and only if every open disk gets sent to an open disk.
answered Jan 9 at 15:06
EnkiduEnkidu
1,28319
$endgroup$
add a comment |
$begingroup$
since the open discs form a basis of the topology, every open set can be written as the union of open discs.
So lets assume we have an Open $U$ then $U= bigcup_{iin I} D_i$ for $D_i$ open disks (observe that $I$ is an arbitrary index set!). Now $$f(U)=f(bigcup_{iin I} D_i)=bigcup_{iin I}f(D_i)$$
and so since the righthandside is open, the lefthandside is as well.
Furthermore, since every open disk is open we have that a map $f:mathbb{C} to X$ is open if and only if every open disk gets sent to an open disk.
answered Jan 9 at 15:06
EnkiduEnkidu
1,28319
$endgroup$
add a comment |
$begingroup$
since the open discs form a basis of the topology, every open set can be written as the union of open discs.
So lets assume we have an Open $U$ then $U= bigcup_{iin I} D_i$ for $D_i$ open disks (observe that $I$ is an arbitrary index set!). Now $$f(U)=f(bigcup_{iin I} D_i)=bigcup_{iin I}f(D_i)$$
and so since the righthandside is open, the lefthandside is as well.
Furthermore, since every open disk is open we have that a map $f:mathbb{C} to X$ is open if and only if every open disk gets sent to an open disk.
answered Jan 9 at 15:06
EnkiduEnkidu
1,28319
$endgroup$
since the open discs form a basis of the topology, every open set can be written as the union of open discs.
So lets assume we have an Open $U$ then $U= bigcup_{iin I} D_i$ for $D_i$ open disks (observe that $I$ is an arbitrary index set!). Now $$f(U)=f(bigcup_{iin I} D_i)=bigcup_{iin I}f(D_i)$$
and so since the righthandside is open, the lefthandside is as well.
Furthermore, since every open disk is open we have that a map $f:mathbb{C} to X$ is open if and only if every open disk gets sent to an open disk.
answered Jan 9 at 15:06
EnkiduEnkidu
1,28319
answered Jan 9 at 15:06
EnkiduEnkidu
1,28319
answered Jan 9 at 15:06
EnkiduEnkidu
1,28319
answered Jan 9 at 15:06
EnkiduEnkidu
1,28319
1,28319
add a comment |
add a comment |
$begingroup$
You can cover any non-empty open set of the plane by open discs. Then any point in the image of this open set will be in the image of such a disc, and therefore will have an open neighborhood included in the image.
answered Jan 9 at 15:05
A. BailleulA. Bailleul
1794
$endgroup$
$begingroup$
"cover by" is not what you need here; in fact you can cover any subset of the plane by open disks...
$endgroup$
– David C. Ullrich
Jan 9 at 15:41
$begingroup$
You are right, what I meant is that such an open set can be written as the union of such open discs.
$endgroup$
– A. Bailleul
Jan 9 at 15:53
add a comment |
$begingroup$
You can cover any non-empty open set of the plane by open discs. Then any point in the image of this open set will be in the image of such a disc, and therefore will have an open neighborhood included in the image.
answered Jan 9 at 15:05
A. BailleulA. Bailleul
1794
$endgroup$
$begingroup$
"cover by" is not what you need here; in fact you can cover any subset of the plane by open disks...
$endgroup$
– David C. Ullrich
Jan 9 at 15:41
$begingroup$
You are right, what I meant is that such an open set can be written as the union of such open discs.
$endgroup$
– A. Bailleul
Jan 9 at 15:53
add a comment |
$begingroup$
You can cover any non-empty open set of the plane by open discs. Then any point in the image of this open set will be in the image of such a disc, and therefore will have an open neighborhood included in the image.
answered Jan 9 at 15:05
A. BailleulA. Bailleul
1794
$endgroup$
You can cover any non-empty open set of the plane by open discs. Then any point in the image of this open set will be in the image of such a disc, and therefore will have an open neighborhood included in the image.
answered Jan 9 at 15:05
A. BailleulA. Bailleul
1794
answered Jan 9 at 15:05
A. BailleulA. Bailleul
1794
answered Jan 9 at 15:05
A. BailleulA. Bailleul
1794
answered Jan 9 at 15:05
A. BailleulA. Bailleul
1794
1794
$begingroup$
"cover by" is not what you need here; in fact you can cover any subset of the plane by open disks...
$endgroup$
– David C. Ullrich
Jan 9 at 15:41
$begingroup$
You are right, what I meant is that such an open set can be written as the union of such open discs.
$endgroup$
– A. Bailleul
Jan 9 at 15:53
add a comment |
$begingroup$
"cover by" is not what you need here; in fact you can cover any subset of the plane by open disks...
$endgroup$
– David C. Ullrich
Jan 9 at 15:41
$begingroup$
You are right, what I meant is that such an open set can be written as the union of such open discs.
$endgroup$
– A. Bailleul
Jan 9 at 15:53
$begingroup$
"cover by" is not what you need here; in fact you can cover any subset of the plane by open disks...
$endgroup$
– David C. Ullrich
Jan 9 at 15:41
$begingroup$
"cover by" is not what you need here; in fact you can cover any subset of the plane by open disks...
$endgroup$
– David C. Ullrich
Jan 9 at 15:41
$begingroup$
You are right, what I meant is that such an open set can be written as the union of such open discs.
$endgroup$
– A. Bailleul
Jan 9 at 15:53
$begingroup$
You are right, what I meant is that such an open set can be written as the union of such open discs.
$endgroup$
– A. Bailleul
Jan 9 at 15:53
add a comment |
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