Open and closed subgroups of continous functions with max metric












0












$begingroup$


Looking at the metric space C[-1,1] with the max metric.



Show that only one of the following subgroups is open.

How many of them are closed?



$A = {f in C[-1,1] : f(x)<1 ,,,forall x in [-1,0);, f(x)<0 ,,, forall x in [0,1]}$
$B = {f in C[-1,1] : f(x)<1 ,,,forall x in [-1,0];, f(x)<0 ,,, forall x in (0,1]}$
$C = {f in C[-1,1] : f(x) leq 1 ,,,forall x in [-1,0);, f(x) leq 0 ,,, forall x in [0,1]}$
$D = {f in C[-1,1] : f(x) leq 1 ,,,forall x in [-1,0];, f(x) leq 0 ,,, forall x in (0,1]}$



I think $A,B$ are not closed because the sequence $f_n(x)=-frac{1}{n}$ when $nrightarrow infty$ will converge to $f(x)=0$, which is not in either of them.

I still have no idea as for $C,D$ being closed or not, and which of the subgroups is open.



Thanks.






metric-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: Consider functions with $f(0) = 0$.
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 2:18












  • $begingroup$
    Thanks! I can see how this helps with groups being open / not open. But I'm still now sure about C,D being closed or not
    $endgroup$
    – shahaf finder
    Dec 8 '18 at 16:41








  • 1




    $begingroup$
    if $y_n to y$ and for all $n, y_n le 1$. Is it possible that $y > 1$?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 18:43






  • 1




    $begingroup$
    Let $f_n to f$ with $f_n in C$ for all $n$. What can you prove about $f$?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 19:19






  • 1




    $begingroup$
    In this case $C[-1,1]$ is first-countable, so identifying limit points by sequences is sufficient. But it is not necessary for this argument. I just used sequences as an example to get you thinking about it. You could use more general limits, or turn the argument around and show that every function not in $C$ (resp. $D$) has a neighborhood that misses it.
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 19:35
















0












$begingroup$


Looking at the metric space C[-1,1] with the max metric.



Show that only one of the following subgroups is open.

How many of them are closed?



$A = {f in C[-1,1] : f(x)<1 ,,,forall x in [-1,0);, f(x)<0 ,,, forall x in [0,1]}$
$B = {f in C[-1,1] : f(x)<1 ,,,forall x in [-1,0];, f(x)<0 ,,, forall x in (0,1]}$
$C = {f in C[-1,1] : f(x) leq 1 ,,,forall x in [-1,0);, f(x) leq 0 ,,, forall x in [0,1]}$
$D = {f in C[-1,1] : f(x) leq 1 ,,,forall x in [-1,0];, f(x) leq 0 ,,, forall x in (0,1]}$



I think $A,B$ are not closed because the sequence $f_n(x)=-frac{1}{n}$ when $nrightarrow infty$ will converge to $f(x)=0$, which is not in either of them.

I still have no idea as for $C,D$ being closed or not, and which of the subgroups is open.



Thanks.






metric-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: Consider functions with $f(0) = 0$.
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 2:18












  • $begingroup$
    Thanks! I can see how this helps with groups being open / not open. But I'm still now sure about C,D being closed or not
    $endgroup$
    – shahaf finder
    Dec 8 '18 at 16:41








  • 1




    $begingroup$
    if $y_n to y$ and for all $n, y_n le 1$. Is it possible that $y > 1$?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 18:43






  • 1




    $begingroup$
    Let $f_n to f$ with $f_n in C$ for all $n$. What can you prove about $f$?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 19:19






  • 1




    $begingroup$
    In this case $C[-1,1]$ is first-countable, so identifying limit points by sequences is sufficient. But it is not necessary for this argument. I just used sequences as an example to get you thinking about it. You could use more general limits, or turn the argument around and show that every function not in $C$ (resp. $D$) has a neighborhood that misses it.
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 19:35














0












0








0





$begingroup$


Looking at the metric space C[-1,1] with the max metric.



Show that only one of the following subgroups is open.

How many of them are closed?



$A = {f in C[-1,1] : f(x)<1 ,,,forall x in [-1,0);, f(x)<0 ,,, forall x in [0,1]}$
$B = {f in C[-1,1] : f(x)<1 ,,,forall x in [-1,0];, f(x)<0 ,,, forall x in (0,1]}$
$C = {f in C[-1,1] : f(x) leq 1 ,,,forall x in [-1,0);, f(x) leq 0 ,,, forall x in [0,1]}$
$D = {f in C[-1,1] : f(x) leq 1 ,,,forall x in [-1,0];, f(x) leq 0 ,,, forall x in (0,1]}$



I think $A,B$ are not closed because the sequence $f_n(x)=-frac{1}{n}$ when $nrightarrow infty$ will converge to $f(x)=0$, which is not in either of them.

I still have no idea as for $C,D$ being closed or not, and which of the subgroups is open.



Thanks.






metric-spaces







share|cite|improve this question











$endgroup$




Looking at the metric space C[-1,1] with the max metric.



Show that only one of the following subgroups is open.

How many of them are closed?



$A = {f in C[-1,1] : f(x)<1 ,,,forall x in [-1,0);, f(x)<0 ,,, forall x in [0,1]}$
$B = {f in C[-1,1] : f(x)<1 ,,,forall x in [-1,0];, f(x)<0 ,,, forall x in (0,1]}$
$C = {f in C[-1,1] : f(x) leq 1 ,,,forall x in [-1,0);, f(x) leq 0 ,,, forall x in [0,1]}$
$D = {f in C[-1,1] : f(x) leq 1 ,,,forall x in [-1,0];, f(x) leq 0 ,,, forall x in (0,1]}$



I think $A,B$ are not closed because the sequence $f_n(x)=-frac{1}{n}$ when $nrightarrow infty$ will converge to $f(x)=0$, which is not in either of them.

I still have no idea as for $C,D$ being closed or not, and which of the subgroups is open.



Thanks.






metric-spaces




metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 15:47







shahaf finder

















asked Dec 7 '18 at 15:29









shahaf findershahaf finder

262




262








  • 1




    $begingroup$
    Hint: Consider functions with $f(0) = 0$.
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 2:18












  • $begingroup$
    Thanks! I can see how this helps with groups being open / not open. But I'm still now sure about C,D being closed or not
    $endgroup$
    – shahaf finder
    Dec 8 '18 at 16:41








  • 1




    $begingroup$
    if $y_n to y$ and for all $n, y_n le 1$. Is it possible that $y > 1$?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 18:43






  • 1




    $begingroup$
    Let $f_n to f$ with $f_n in C$ for all $n$. What can you prove about $f$?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 19:19






  • 1




    $begingroup$
    In this case $C[-1,1]$ is first-countable, so identifying limit points by sequences is sufficient. But it is not necessary for this argument. I just used sequences as an example to get you thinking about it. You could use more general limits, or turn the argument around and show that every function not in $C$ (resp. $D$) has a neighborhood that misses it.
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 19:35














  • 1




    $begingroup$
    Hint: Consider functions with $f(0) = 0$.
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 2:18












  • $begingroup$
    Thanks! I can see how this helps with groups being open / not open. But I'm still now sure about C,D being closed or not
    $endgroup$
    – shahaf finder
    Dec 8 '18 at 16:41








  • 1




    $begingroup$
    if $y_n to y$ and for all $n, y_n le 1$. Is it possible that $y > 1$?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 18:43






  • 1




    $begingroup$
    Let $f_n to f$ with $f_n in C$ for all $n$. What can you prove about $f$?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 19:19






  • 1




    $begingroup$
    In this case $C[-1,1]$ is first-countable, so identifying limit points by sequences is sufficient. But it is not necessary for this argument. I just used sequences as an example to get you thinking about it. You could use more general limits, or turn the argument around and show that every function not in $C$ (resp. $D$) has a neighborhood that misses it.
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 19:35








1




1




$begingroup$
Hint: Consider functions with $f(0) = 0$.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 2:18






$begingroup$
Hint: Consider functions with $f(0) = 0$.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 2:18














$begingroup$
Thanks! I can see how this helps with groups being open / not open. But I'm still now sure about C,D being closed or not
$endgroup$
– shahaf finder
Dec 8 '18 at 16:41






$begingroup$
Thanks! I can see how this helps with groups being open / not open. But I'm still now sure about C,D being closed or not
$endgroup$
– shahaf finder
Dec 8 '18 at 16:41






1




1




$begingroup$
if $y_n to y$ and for all $n, y_n le 1$. Is it possible that $y > 1$?
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:43




$begingroup$
if $y_n to y$ and for all $n, y_n le 1$. Is it possible that $y > 1$?
$endgroup$
– Paul Sinclair
Dec 8 '18 at 18:43




1




1




$begingroup$
Let $f_n to f$ with $f_n in C$ for all $n$. What can you prove about $f$?
$endgroup$
– Paul Sinclair
Dec 8 '18 at 19:19




$begingroup$
Let $f_n to f$ with $f_n in C$ for all $n$. What can you prove about $f$?
$endgroup$
– Paul Sinclair
Dec 8 '18 at 19:19




1




1




$begingroup$
In this case $C[-1,1]$ is first-countable, so identifying limit points by sequences is sufficient. But it is not necessary for this argument. I just used sequences as an example to get you thinking about it. You could use more general limits, or turn the argument around and show that every function not in $C$ (resp. $D$) has a neighborhood that misses it.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 19:35




$begingroup$
In this case $C[-1,1]$ is first-countable, so identifying limit points by sequences is sufficient. But it is not necessary for this argument. I just used sequences as an example to get you thinking about it. You could use more general limits, or turn the argument around and show that every function not in $C$ (resp. $D$) has a neighborhood that misses it.
$endgroup$
– Paul Sinclair
Dec 8 '18 at 19:35










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