Is every uniformly continuous function 1-1 and onto?
$begingroup$
Let $f : (X,d)rightarrow (Y,rho)$. Is $f$ 1-1 and onto if $f$ is a uniformly continuous function on X?
If not, would $X$ being compact change things?
If not, do you know a theorem or something similar to this?
Thanks in advance.
metric-spaces uniform-continuity
asked Jan 7 at 20:46
Anıl Burak GökceAnıl Burak Gökce
82
$endgroup$
add a comment |
$begingroup$
Let $f : (X,d)rightarrow (Y,rho)$. Is $f$ 1-1 and onto if $f$ is a uniformly continuous function on X?
If not, would $X$ being compact change things?
If not, do you know a theorem or something similar to this?
Thanks in advance.
metric-spaces uniform-continuity
asked Jan 7 at 20:46
Anıl Burak GökceAnıl Burak Gökce
82
$endgroup$
4
$begingroup$
Well, a constant function is uniformly continuous...
$endgroup$
– Mindlack
Jan 7 at 20:47
1
$begingroup$
Continuity, and, even more, uniforme continuity, says that $Y$ is, in some sense, less chaotic than $X$. But $X$ can be arbitrarily more chaotic than $Y$. In other words: the existence of a continuous function from $X$ to $Y$ is NOT symmetric.
$endgroup$
– ajotatxe
Jan 7 at 20:50
add a comment |
$begingroup$
Let $f : (X,d)rightarrow (Y,rho)$. Is $f$ 1-1 and onto if $f$ is a uniformly continuous function on X?
If not, would $X$ being compact change things?
If not, do you know a theorem or something similar to this?
Thanks in advance.
metric-spaces uniform-continuity
asked Jan 7 at 20:46
Anıl Burak GökceAnıl Burak Gökce
82
$endgroup$
Let $f : (X,d)rightarrow (Y,rho)$. Is $f$ 1-1 and onto if $f$ is a uniformly continuous function on X?
If not, would $X$ being compact change things?
If not, do you know a theorem or something similar to this?
Thanks in advance.
metric-spaces uniform-continuity
metric-spaces uniform-continuity
asked Jan 7 at 20:46
Anıl Burak GökceAnıl Burak Gökce
82
asked Jan 7 at 20:46
Anıl Burak GökceAnıl Burak Gökce
82
edited Jan 7 at 20:47
Anıl Burak Gökce
asked Jan 7 at 20:46
Anıl Burak GökceAnıl Burak Gökce
82
asked Jan 7 at 20:46
Anıl Burak GökceAnıl Burak Gökce
82
asked Jan 7 at 20:46
Anıl Burak GökceAnıl Burak Gökce
82
82
4
$begingroup$
Well, a constant function is uniformly continuous...
$endgroup$
– Mindlack
Jan 7 at 20:47
1
$begingroup$
Continuity, and, even more, uniforme continuity, says that $Y$ is, in some sense, less chaotic than $X$. But $X$ can be arbitrarily more chaotic than $Y$. In other words: the existence of a continuous function from $X$ to $Y$ is NOT symmetric.
$endgroup$
– ajotatxe
Jan 7 at 20:50
add a comment |
4
$begingroup$
Well, a constant function is uniformly continuous...
$endgroup$
– Mindlack
Jan 7 at 20:47
1
$begingroup$
Continuity, and, even more, uniforme continuity, says that $Y$ is, in some sense, less chaotic than $X$. But $X$ can be arbitrarily more chaotic than $Y$. In other words: the existence of a continuous function from $X$ to $Y$ is NOT symmetric.
$endgroup$
– ajotatxe
Jan 7 at 20:50
4
4
$begingroup$
Well, a constant function is uniformly continuous...
$endgroup$
– Mindlack
Jan 7 at 20:47
$begingroup$
Well, a constant function is uniformly continuous...
$endgroup$
– Mindlack
Jan 7 at 20:47
1
1
$begingroup$
Continuity, and, even more, uniforme continuity, says that $Y$ is, in some sense, less chaotic than $X$. But $X$ can be arbitrarily more chaotic than $Y$. In other words: the existence of a continuous function from $X$ to $Y$ is NOT symmetric.
$endgroup$
– ajotatxe
Jan 7 at 20:50
$begingroup$
Continuity, and, even more, uniforme continuity, says that $Y$ is, in some sense, less chaotic than $X$. But $X$ can be arbitrarily more chaotic than $Y$. In other words: the existence of a continuous function from $X$ to $Y$ is NOT symmetric.
$endgroup$
– ajotatxe
Jan 7 at 20:50
add a comment |
1 Answer
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For a counterexample for both, consider a constant function when the domain and codomain both have more than one element.
answered Jan 7 at 20:48
IanIan
68.2k25388
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$begingroup$
For a counterexample for both, consider a constant function when the domain and codomain both have more than one element.
answered Jan 7 at 20:48
IanIan
68.2k25388
$endgroup$
add a comment |
$begingroup$
For a counterexample for both, consider a constant function when the domain and codomain both have more than one element.
answered Jan 7 at 20:48
IanIan
68.2k25388
$endgroup$
add a comment |
$begingroup$
For a counterexample for both, consider a constant function when the domain and codomain both have more than one element.
answered Jan 7 at 20:48
IanIan
68.2k25388
$endgroup$
For a counterexample for both, consider a constant function when the domain and codomain both have more than one element.
answered Jan 7 at 20:48
IanIan
68.2k25388
answered Jan 7 at 20:48
IanIan
68.2k25388
answered Jan 7 at 20:48
IanIan
68.2k25388
answered Jan 7 at 20:48
IanIan
68.2k25388
68.2k25388
add a comment |
add a comment |
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4
$begingroup$
Well, a constant function is uniformly continuous...
$endgroup$
– Mindlack
Jan 7 at 20:47
1
$begingroup$
Continuity, and, even more, uniforme continuity, says that $Y$ is, in some sense, less chaotic than $X$. But $X$ can be arbitrarily more chaotic than $Y$. In other words: the existence of a continuous function from $X$ to $Y$ is NOT symmetric.
$endgroup$
– ajotatxe
Jan 7 at 20:50