Id $: mathbb{R} rightarrow mathbb{ R}$ is the identity mapping then choose the correct statement
$begingroup$
Let $τ_1$ be the topology on $mathbb R$ generated by the base $mathcal B = {[a, b):a,binmathbb{R}text{ and }a<b}$.
Let $τ_0$ be the standard topology on $mathbb{R}$ and let Id $: mathbb{R} rightarrow mathbb{ R}$ be the identity mapping.
Choose the correct statement:
$(a)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is continuous
$(b)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is an open mapping
My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.
Is it true ?
Any hints/solution will be appreciated.
Thank you.
general-topology
edited Jan 6 at 12:12
drhab
103k545136
asked Jan 6 at 11:48
jasminejasmine
1,890418
$endgroup$
add a comment |
$begingroup$
Let $τ_1$ be the topology on $mathbb R$ generated by the base $mathcal B = {[a, b):a,binmathbb{R}text{ and }a<b}$.
Let $τ_0$ be the standard topology on $mathbb{R}$ and let Id $: mathbb{R} rightarrow mathbb{ R}$ be the identity mapping.
Choose the correct statement:
$(a)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is continuous
$(b)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is an open mapping
My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.
Is it true ?
Any hints/solution will be appreciated.
Thank you.
general-topology
edited Jan 6 at 12:12
drhab
103k545136
asked Jan 6 at 11:48
jasminejasmine
1,890418
$endgroup$
2
$begingroup$
For (b) push forward $[0,1)$.
$endgroup$
– Randall
Jan 6 at 11:56
add a comment |
$begingroup$
Let $τ_1$ be the topology on $mathbb R$ generated by the base $mathcal B = {[a, b):a,binmathbb{R}text{ and }a<b}$.
Let $τ_0$ be the standard topology on $mathbb{R}$ and let Id $: mathbb{R} rightarrow mathbb{ R}$ be the identity mapping.
Choose the correct statement:
$(a)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is continuous
$(b)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is an open mapping
My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.
Is it true ?
Any hints/solution will be appreciated.
Thank you.
general-topology
edited Jan 6 at 12:12
drhab
103k545136
asked Jan 6 at 11:48
jasminejasmine
1,890418
$endgroup$
Let $τ_1$ be the topology on $mathbb R$ generated by the base $mathcal B = {[a, b):a,binmathbb{R}text{ and }a<b}$.
Let $τ_0$ be the standard topology on $mathbb{R}$ and let Id $: mathbb{R} rightarrow mathbb{ R}$ be the identity mapping.
Choose the correct statement:
$(a)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is continuous
$(b)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is an open mapping
My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.
Is it true ?
Any hints/solution will be appreciated.
Thank you.
general-topology
general-topology
edited Jan 6 at 12:12
drhab
103k545136
asked Jan 6 at 11:48
jasminejasmine
1,890418
edited Jan 6 at 12:12
drhab
103k545136
asked Jan 6 at 11:48
jasminejasmine
1,890418
edited Jan 6 at 12:12
drhab
103k545136
edited Jan 6 at 12:12
drhab
103k545136
edited Jan 6 at 12:12
drhab
103k545136
103k545136
asked Jan 6 at 11:48
jasminejasmine
1,890418
asked Jan 6 at 11:48
jasminejasmine
1,890418
asked Jan 6 at 11:48
jasminejasmine
1,890418
1,890418
2
$begingroup$
For (b) push forward $[0,1)$.
$endgroup$
– Randall
Jan 6 at 11:56
add a comment |
2
$begingroup$
For (b) push forward $[0,1)$.
$endgroup$
– Randall
Jan 6 at 11:56
2
2
$begingroup$
For (b) push forward $[0,1)$.
$endgroup$
– Randall
Jan 6 at 11:56
$begingroup$
For (b) push forward $[0,1)$.
$endgroup$
– Randall
Jan 6 at 11:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.
answered Jan 6 at 11:56
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
$begingroup$
(a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?
answered Jan 6 at 11:50
Matt SamuelMatt Samuel
38.9k63769
$endgroup$
add a comment |
$begingroup$
In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
Draw conclusions.
answered Jan 6 at 12:05
drhabdrhab
103k545136
$endgroup$
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%2f3063760%2fid-mathbbr-rightarrow-mathbb-r-is-the-identity-mapping-then-choose-th%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.
answered Jan 6 at 11:56
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
$begingroup$
If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.
answered Jan 6 at 11:56
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
$begingroup$
If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.
answered Jan 6 at 11:56
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.
answered Jan 6 at 11:56
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Jan 6 at 11:56
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Jan 6 at 11:56
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Jan 6 at 11:56
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
69.1k53169
add a comment |
add a comment |
$begingroup$
(a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?
answered Jan 6 at 11:50
Matt SamuelMatt Samuel
38.9k63769
$endgroup$
add a comment |
$begingroup$
(a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?
answered Jan 6 at 11:50
Matt SamuelMatt Samuel
38.9k63769
$endgroup$
add a comment |
$begingroup$
(a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?
answered Jan 6 at 11:50
Matt SamuelMatt Samuel
38.9k63769
$endgroup$
(a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?
answered Jan 6 at 11:50
Matt SamuelMatt Samuel
38.9k63769
answered Jan 6 at 11:50
Matt SamuelMatt Samuel
38.9k63769
answered Jan 6 at 11:50
Matt SamuelMatt Samuel
38.9k63769
answered Jan 6 at 11:50
Matt SamuelMatt Samuel
38.9k63769
38.9k63769
add a comment |
add a comment |
$begingroup$
In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
Draw conclusions.
answered Jan 6 at 12:05
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
Draw conclusions.
answered Jan 6 at 12:05
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
Draw conclusions.
answered Jan 6 at 12:05
drhabdrhab
103k545136
$endgroup$
In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
Draw conclusions.
answered Jan 6 at 12:05
drhabdrhab
103k545136
answered Jan 6 at 12:05
drhabdrhab
103k545136
answered Jan 6 at 12:05
drhabdrhab
103k545136
answered Jan 6 at 12:05
drhabdrhab
103k545136
103k545136
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%2f3063760%2fid-mathbbr-rightarrow-mathbb-r-is-the-identity-mapping-then-choose-th%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
$begingroup$
For (b) push forward $[0,1)$.
$endgroup$
– Randall
Jan 6 at 11:56