Subspace topology is the unique topology that satisfies the Characteristic Property
$begingroup$
Lee has as an exercise that the subspace topology is the unique topology that satisfies the characteristic property which is for $Ssubset X$ a function $F:Y to S$ is continuous iff the composition $i_Scirc F$ is continuous.
I think that he means that the only topology that this is true is the subspace topology, but I’m not sure how to prove that no other topology satisfies the property. My first question is what needs to be shown in a proof to say that it is the unique topology, and once I know this I should be able to apply it to the exercise.
general-topology
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
232
$endgroup$
add a comment |
$begingroup$
Lee has as an exercise that the subspace topology is the unique topology that satisfies the characteristic property which is for $Ssubset X$ a function $F:Y to S$ is continuous iff the composition $i_Scirc F$ is continuous.
I think that he means that the only topology that this is true is the subspace topology, but I’m not sure how to prove that no other topology satisfies the property. My first question is what needs to be shown in a proof to say that it is the unique topology, and once I know this I should be able to apply it to the exercise.
general-topology
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
232
$endgroup$
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
add a comment |
$begingroup$
Lee has as an exercise that the subspace topology is the unique topology that satisfies the characteristic property which is for $Ssubset X$ a function $F:Y to S$ is continuous iff the composition $i_Scirc F$ is continuous.
I think that he means that the only topology that this is true is the subspace topology, but I’m not sure how to prove that no other topology satisfies the property. My first question is what needs to be shown in a proof to say that it is the unique topology, and once I know this I should be able to apply it to the exercise.
general-topology
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
232
$endgroup$
Lee has as an exercise that the subspace topology is the unique topology that satisfies the characteristic property which is for $Ssubset X$ a function $F:Y to S$ is continuous iff the composition $i_Scirc F$ is continuous.
I think that he means that the only topology that this is true is the subspace topology, but I’m not sure how to prove that no other topology satisfies the property. My first question is what needs to be shown in a proof to say that it is the unique topology, and once I know this I should be able to apply it to the exercise.
general-topology
general-topology
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
232
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
232
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
232
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
232
asked Jan 4 at 3:15
Anthony SalibAnthony Salib
232
232
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
add a comment |
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
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%2f3061280%2fsubspace-topology-is-the-unique-topology-that-satisfies-the-characteristic-prope%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
$begingroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
Suppose that $tau$ is any topology on $S$ that satisfies the "characteristic property".
Then let $1_S$ be the identity on $(S, tau)$, which is always continuous (for any topology on $S$) so we know that $i_S circ 1_S = i_S$ is continuous (using the "characteristic property" in one direction). This implies that if $U$ is open in $X$ then $i_S^{-1}[U] = U cap S$ is open in $tau$. So the subspace topology $tau_S$ obeys $tau_S subseteq tau$.
Define the function $f(x)=x$ from $(S,tau_S)$ to $(S,tau)$, and note that $i_S = i_S circ f$ as a function $ (S, tau_S) to X$ is continuous (we know the subspace topology $tau_S$ makes this map continuous) and so $f$ is continuous (by the other direction of the "characteristic property"), which implies that for all $U in tau$, $f^{-1}[U]=U in tau_S$ or $tau subseteq tau_S$ and so $tau = tau_S$ and we have unicity of $tau_S$.
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
108k347114
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
108k347114
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
108k347114
answered Jan 4 at 5:37
Henno BrandsmaHenno Brandsma
108k347114
108k347114
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%2f3061280%2fsubspace-topology-is-the-unique-topology-that-satisfies-the-characteristic-prope%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
$begingroup$
Show that any topology $tau$ on $S$ that makes this property true must be such that any $Uintau$ is the intersection of a subset of $S$ and an open set in $X$.
$endgroup$
– John Douma
Jan 4 at 3:58
$begingroup$
It is useful to note that for any $U subset X$, $i_S^{-1}(U) = U cap S$.
$endgroup$
– Omnomnomnom
Jan 4 at 4:03