Open and closed subgroups of continous functions with max metric
$begingroup$
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
asked Dec 7 '18 at 15:29
shahaf findershahaf finder
262
$endgroup$
|
show 3 more comments
$begingroup$
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
asked Dec 7 '18 at 15:29
shahaf findershahaf finder
262
$endgroup$
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
|
show 3 more comments
$begingroup$
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
asked Dec 7 '18 at 15:29
shahaf findershahaf finder
262
$endgroup$
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
asked Dec 7 '18 at 15:29
shahaf findershahaf finder
262
asked Dec 7 '18 at 15:29
shahaf findershahaf finder
262
edited Dec 7 '18 at 15:47
shahaf finder
asked Dec 7 '18 at 15:29
shahaf findershahaf finder
262
asked Dec 7 '18 at 15:29
shahaf findershahaf finder
262
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
|
show 3 more comments
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
|
show 3 more comments
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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