On the completeness of topologically isomorphic spaces
Let $(E_1,tau_1)$ be a locally convex space and let $(E_2,tau_2)$ be a complete locally convex space. Suppose that $T:(E_1,tau_1) longrightarrow (E_2,tau_2)$ is a topological isomorphism (that is, $T$ is linear, bijective, continuous and its inverse $T^{-1}$ is continuous, too).
Is it true that the space $(E_1,tau_1)$ is necessarily complete?
Thanks for any hints/comments.
functional-analysis locally-convex-spaces
asked Nov 28 at 7:29
mathmax
1896
add a comment |
Let $(E_1,tau_1)$ be a locally convex space and let $(E_2,tau_2)$ be a complete locally convex space. Suppose that $T:(E_1,tau_1) longrightarrow (E_2,tau_2)$ is a topological isomorphism (that is, $T$ is linear, bijective, continuous and its inverse $T^{-1}$ is continuous, too).
Is it true that the space $(E_1,tau_1)$ is necessarily complete?
Thanks for any hints/comments.
functional-analysis locally-convex-spaces
asked Nov 28 at 7:29
mathmax
1896
add a comment |
Let $(E_1,tau_1)$ be a locally convex space and let $(E_2,tau_2)$ be a complete locally convex space. Suppose that $T:(E_1,tau_1) longrightarrow (E_2,tau_2)$ is a topological isomorphism (that is, $T$ is linear, bijective, continuous and its inverse $T^{-1}$ is continuous, too).
Is it true that the space $(E_1,tau_1)$ is necessarily complete?
Thanks for any hints/comments.
functional-analysis locally-convex-spaces
asked Nov 28 at 7:29
mathmax
1896
Let $(E_1,tau_1)$ be a locally convex space and let $(E_2,tau_2)$ be a complete locally convex space. Suppose that $T:(E_1,tau_1) longrightarrow (E_2,tau_2)$ is a topological isomorphism (that is, $T$ is linear, bijective, continuous and its inverse $T^{-1}$ is continuous, too).
Is it true that the space $(E_1,tau_1)$ is necessarily complete?
Thanks for any hints/comments.
functional-analysis locally-convex-spaces
functional-analysis locally-convex-spaces
asked Nov 28 at 7:29
mathmax
1896
asked Nov 28 at 7:29
mathmax
1896
asked Nov 28 at 7:29
mathmax
1896
asked Nov 28 at 7:29
mathmax
1896
asked Nov 28 at 7:29
mathmax
1896
1896
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Yes, of course (a property which is not stable under isomorphisms of a category is typically not very useful). The proof is as it should be: If $(x_i)_{iin I}$ is a Cauchy net in $E_1$ then $T(x_i)$ is a Cauchy net in $E_2$ because linear continuous maps are uniformly continuous. If $y$ is a limit of $T(x_i)$ then $x_i$ converges to $T^{-1}(y)$ because of the continuity of $T^{-1}$.
answered Nov 28 at 14:32
Jochen
6,630922
Thanks for the answer. What I did not consider was the uniform continuity of $T$.
– mathmax
Nov 28 at 16:31
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%2f3016856%2fon-the-completeness-of-topologically-isomorphic-spaces%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
Yes, of course (a property which is not stable under isomorphisms of a category is typically not very useful). The proof is as it should be: If $(x_i)_{iin I}$ is a Cauchy net in $E_1$ then $T(x_i)$ is a Cauchy net in $E_2$ because linear continuous maps are uniformly continuous. If $y$ is a limit of $T(x_i)$ then $x_i$ converges to $T^{-1}(y)$ because of the continuity of $T^{-1}$.
answered Nov 28 at 14:32
Jochen
6,630922
Thanks for the answer. What I did not consider was the uniform continuity of $T$.
– mathmax
Nov 28 at 16:31
add a comment |
Yes, of course (a property which is not stable under isomorphisms of a category is typically not very useful). The proof is as it should be: If $(x_i)_{iin I}$ is a Cauchy net in $E_1$ then $T(x_i)$ is a Cauchy net in $E_2$ because linear continuous maps are uniformly continuous. If $y$ is a limit of $T(x_i)$ then $x_i$ converges to $T^{-1}(y)$ because of the continuity of $T^{-1}$.
answered Nov 28 at 14:32
Jochen
6,630922
Thanks for the answer. What I did not consider was the uniform continuity of $T$.
– mathmax
Nov 28 at 16:31
add a comment |
Yes, of course (a property which is not stable under isomorphisms of a category is typically not very useful). The proof is as it should be: If $(x_i)_{iin I}$ is a Cauchy net in $E_1$ then $T(x_i)$ is a Cauchy net in $E_2$ because linear continuous maps are uniformly continuous. If $y$ is a limit of $T(x_i)$ then $x_i$ converges to $T^{-1}(y)$ because of the continuity of $T^{-1}$.
answered Nov 28 at 14:32
Jochen
6,630922
Yes, of course (a property which is not stable under isomorphisms of a category is typically not very useful). The proof is as it should be: If $(x_i)_{iin I}$ is a Cauchy net in $E_1$ then $T(x_i)$ is a Cauchy net in $E_2$ because linear continuous maps are uniformly continuous. If $y$ is a limit of $T(x_i)$ then $x_i$ converges to $T^{-1}(y)$ because of the continuity of $T^{-1}$.
answered Nov 28 at 14:32
Jochen
6,630922
answered Nov 28 at 14:32
Jochen
6,630922
answered Nov 28 at 14:32
Jochen
6,630922
answered Nov 28 at 14:32
Jochen
6,630922
6,630922
Thanks for the answer. What I did not consider was the uniform continuity of $T$.
– mathmax
Nov 28 at 16:31
add a comment |
Thanks for the answer. What I did not consider was the uniform continuity of $T$.
– mathmax
Nov 28 at 16:31
Thanks for the answer. What I did not consider was the uniform continuity of $T$.
– mathmax
Nov 28 at 16:31
Thanks for the answer. What I did not consider was the uniform continuity of $T$.
– mathmax
Nov 28 at 16:31
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3016856%2fon-the-completeness-of-topologically-isomorphic-spaces%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
