Show that the domain of locally constant and continuous functions are connected [on hold]Prove a fact involving continuous functions and connected spaces.Show that $M$ is $not$ $connected$ if and only if there exists a non-constant continuous function $g$ : $M$ → $0,1$Is a function that maps every connected set to a connected set continuous?Can locally “a.e. constant” function on a connected subset $U$ of $mathbbR^n$ be constant a.e. in $U$?Show that if every continuous function $X rightarrow mathbbZ$ is constant then X is connected$nabla U=0 implies U=mathrmconstant$ with $A$ connected and $U$ continuous?Non-constant continuous function that is locally constant almost everywherePathological Continua which are Path Connected and Locally Path Connected.Show that $(X,tau)$ is Locally Compact and Locally Path Connected but not CompactConnected sets. Connected by paths. Locally constant functions. Incomplete solution.
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Show that the domain of locally constant and continuous functions are connected [on hold]
Prove a fact involving continuous functions and connected spaces.Show that $M$ is $not$ $connected$ if and only if there exists a non-constant continuous function $g$ : $M$ → $0,1$Is a function that maps every connected set to a connected set continuous?Can locally “a.e. constant” function on a connected subset $U$ of $mathbbR^n$ be constant a.e. in $U$?Show that if every continuous function $X rightarrow mathbbZ$ is constant then X is connected$nabla U=0 implies U=mathrmconstant$ with $A$ connected and $U$ continuous?Non-constant continuous function that is locally constant almost everywherePathological Continua which are Path Connected and Locally Path Connected.Show that $(X,tau)$ is Locally Compact and Locally Path Connected but not CompactConnected sets. Connected by paths. Locally constant functions. Incomplete solution.
$begingroup$
Given that every continuous, locally constant function f: I -> R is constant function, how to show that I is connected?
real-analysis analysis
asked Mar 11 at 19:32
Yilin LiYilin Li
1
New contributor
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by darij grinberg, user90369, José Carlos Santos, Alex Provost, hardmath Mar 12 at 17:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – darij grinberg, user90369, José Carlos Santos, Alex Provost, hardmath
add a comment |
$begingroup$
Given that every continuous, locally constant function f: I -> R is constant function, how to show that I is connected?
real-analysis analysis
asked Mar 11 at 19:32
Yilin LiYilin Li
1
New contributor
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by darij grinberg, user90369, José Carlos Santos, Alex Provost, hardmath Mar 12 at 17:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – darij grinberg, user90369, José Carlos Santos, Alex Provost, hardmath
2
$begingroup$
On m.se we state what we have tried before submitting the question. For example, did you try assuming that $I$ is disconnecting, and somehow building a locally constant function that is not constant by exploiting this disconnectedness?
$endgroup$
– darij grinberg
Mar 11 at 19:40
$begingroup$
NB: The problem should require $I$ to be nonempty.
$endgroup$
– darij grinberg
Mar 11 at 19:41
1
$begingroup$
Have you thought about what happens when $I$ is not connected? In fact, consider the case of $I$ being the discrete space $1,2$. Can you construct a locally constant $f$ on $I$ which is not constant?
$endgroup$
– Paul Sinclair
Mar 12 at 3:22
add a comment |
$begingroup$
Given that every continuous, locally constant function f: I -> R is constant function, how to show that I is connected?
real-analysis analysis
asked Mar 11 at 19:32
Yilin LiYilin Li
1
New contributor
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Given that every continuous, locally constant function f: I -> R is constant function, how to show that I is connected?
real-analysis analysis
real-analysis analysis
asked Mar 11 at 19:32
Yilin LiYilin Li
1
New contributor
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 19:32
Yilin LiYilin Li
1
New contributor
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 19:32
Yilin LiYilin Li
1
New contributor
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 19:32
Yilin LiYilin Li
1
asked Mar 11 at 19:32
Yilin LiYilin Li
1
1
New contributor
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yilin Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by darij grinberg, user90369, José Carlos Santos, Alex Provost, hardmath Mar 12 at 17:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – darij grinberg, user90369, José Carlos Santos, Alex Provost, hardmath
put on hold as off-topic by darij grinberg, user90369, José Carlos Santos, Alex Provost, hardmath Mar 12 at 17:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – darij grinberg, user90369, José Carlos Santos, Alex Provost, hardmath
2
$begingroup$
On m.se we state what we have tried before submitting the question. For example, did you try assuming that $I$ is disconnecting, and somehow building a locally constant function that is not constant by exploiting this disconnectedness?
$endgroup$
– darij grinberg
Mar 11 at 19:40
$begingroup$
NB: The problem should require $I$ to be nonempty.
$endgroup$
– darij grinberg
Mar 11 at 19:41
1
$begingroup$
Have you thought about what happens when $I$ is not connected? In fact, consider the case of $I$ being the discrete space $1,2$. Can you construct a locally constant $f$ on $I$ which is not constant?
$endgroup$
– Paul Sinclair
Mar 12 at 3:22
add a comment |
2
$begingroup$
On m.se we state what we have tried before submitting the question. For example, did you try assuming that $I$ is disconnecting, and somehow building a locally constant function that is not constant by exploiting this disconnectedness?
$endgroup$
– darij grinberg
Mar 11 at 19:40
$begingroup$
NB: The problem should require $I$ to be nonempty.
$endgroup$
– darij grinberg
Mar 11 at 19:41
1
$begingroup$
Have you thought about what happens when $I$ is not connected? In fact, consider the case of $I$ being the discrete space $1,2$. Can you construct a locally constant $f$ on $I$ which is not constant?
$endgroup$
– Paul Sinclair
Mar 12 at 3:22
2
2
$begingroup$
On m.se we state what we have tried before submitting the question. For example, did you try assuming that $I$ is disconnecting, and somehow building a locally constant function that is not constant by exploiting this disconnectedness?
$endgroup$
– darij grinberg
Mar 11 at 19:40
$begingroup$
On m.se we state what we have tried before submitting the question. For example, did you try assuming that $I$ is disconnecting, and somehow building a locally constant function that is not constant by exploiting this disconnectedness?
$endgroup$
– darij grinberg
Mar 11 at 19:40
$begingroup$
NB: The problem should require $I$ to be nonempty.
$endgroup$
– darij grinberg
Mar 11 at 19:41
$begingroup$
NB: The problem should require $I$ to be nonempty.
$endgroup$
– darij grinberg
Mar 11 at 19:41
1
1
$begingroup$
Have you thought about what happens when $I$ is not connected? In fact, consider the case of $I$ being the discrete space $1,2$. Can you construct a locally constant $f$ on $I$ which is not constant?
$endgroup$
– Paul Sinclair
Mar 12 at 3:22
$begingroup$
Have you thought about what happens when $I$ is not connected? In fact, consider the case of $I$ being the discrete space $1,2$. Can you construct a locally constant $f$ on $I$ which is not constant?
$endgroup$
– Paul Sinclair
Mar 12 at 3:22
add a comment |
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$begingroup$
On m.se we state what we have tried before submitting the question. For example, did you try assuming that $I$ is disconnecting, and somehow building a locally constant function that is not constant by exploiting this disconnectedness?
$endgroup$
– darij grinberg
Mar 11 at 19:40
$begingroup$
NB: The problem should require $I$ to be nonempty.
$endgroup$
– darij grinberg
Mar 11 at 19:41
1
$begingroup$
Have you thought about what happens when $I$ is not connected? In fact, consider the case of $I$ being the discrete space $1,2$. Can you construct a locally constant $f$ on $I$ which is not constant?
$endgroup$
– Paul Sinclair
Mar 12 at 3:22