Let $A: X rightarrow Y$ a bijective continous map between two Banach spaces X and Y. Then, $A^-1$ is also continous? The 2019 Stack Overflow Developer Survey Results Are InIs it true that the unit ball is compact in a normed linear space iff the space is finite-dimensional?Existence of a continuous function which does not achieve a maximum.Unbounded continuous function on non-compact metric spaceTopology of bijective functions between Banach spacesContinuity criterions for linear maps between Banach spaces?When linear map is bounded and bijective then $X$ is reflexive.proving that the quotient linear map of a continuous linear map is also continuous (normed spaces)Can natural quotient map between Banach spaces be closed?The image of an bounded and linear operator between two banach spaces is closed if the image is finite codimension?Does any continuous map between two Banach spaces over a nonarchimedian field is a closed map?bijective continuous map from X to Y and Y to XFind an homeomorphism between two topological spacesMap which is bijective and continuous, but not open
Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?
Why didn't the Event Horizon Telescope team mention Sagittarius A*?
What is the most efficient way to store a numeric range?
Output the Arecibo Message
Can there be female White Walkers?
I am an eight letter word. What am I?
The phrase "to the numbers born"?
Geography at the pixel level
How to translate "being like"?
Why are there uneven bright areas in this photo of black hole?
Does adding complexity mean a more secure cipher?
For what reasons would an animal species NOT cross a *horizontal* land bridge?
writing variables above the numbers in tikz picture
If I can cast sorceries at instant speed, can I use sorcery-speed activated abilities at instant speed?
Deal with toxic manager when you can't quit
How come people say “Would of”?
Will it cause any balance problems to have PCs level up and gain the benefits of a long rest mid-fight?
What is the motivation for a law requiring 2 parties to consent for recording a conversation
Loose spokes after only a few rides
Match Roman Numerals
Why can't devices on different VLANs, but on the same subnet, communicate?
Is an up-to-date browser secure on an out-of-date OS?
Likelihood that a superbug or lethal virus could come from a landfill
Can we generate random numbers using irrational numbers like π and e?
Let $A: X rightarrow Y$ a bijective continous map between two Banach spaces X and Y. Then, $A^-1$ is also continous?
The 2019 Stack Overflow Developer Survey Results Are InIs it true that the unit ball is compact in a normed linear space iff the space is finite-dimensional?Existence of a continuous function which does not achieve a maximum.Unbounded continuous function on non-compact metric spaceTopology of bijective functions between Banach spacesContinuity criterions for linear maps between Banach spaces?When linear map is bounded and bijective then $X$ is reflexive.proving that the quotient linear map of a continuous linear map is also continuous (normed spaces)Can natural quotient map between Banach spaces be closed?The image of an bounded and linear operator between two banach spaces is closed if the image is finite codimension?Does any continuous map between two Banach spaces over a nonarchimedian field is a closed map?bijective continuous map from X to Y and Y to XFind an homeomorphism between two topological spacesMap which is bijective and continuous, but not open
$begingroup$
We know if A is a map continuous, bijective and linear then the answer is yes, $A^-1$ is continuous. But, if $A$ is no linear then $A^-1$ is also continuous ?
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
$endgroup$
add a comment |
$begingroup$
We know if A is a map continuous, bijective and linear then the answer is yes, $A^-1$ is continuous. But, if $A$ is no linear then $A^-1$ is also continuous ?
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
$endgroup$
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
add a comment |
$begingroup$
We know if A is a map continuous, bijective and linear then the answer is yes, $A^-1$ is continuous. But, if $A$ is no linear then $A^-1$ is also continuous ?
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
$endgroup$
We know if A is a map continuous, bijective and linear then the answer is yes, $A^-1$ is continuous. But, if $A$ is no linear then $A^-1$ is also continuous ?
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
edited Mar 30 at 22:12
Paul Frost
12.6k31035
edited Mar 30 at 22:12
Paul Frost
12.6k31035
edited Mar 30 at 22:12
Paul Frost
12.6k31035
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
asked Mar 30 at 21:59
jugartecjugartec
111
asked Mar 30 at 21:59
jugartecjugartec
111
111
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
add a comment |
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
2
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168826%2flet-a-x-rightarrow-y-a-bijective-continous-map-between-two-banach-spaces-x-a%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
add a comment |
$begingroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
add a comment |
$begingroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
edited Apr 1 at 17:48
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
48.5k344110
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168826%2flet-a-x-rightarrow-y-a-bijective-continous-map-between-two-banach-spaces-x-a%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25