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 $n\rightarrow \infty$ 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.


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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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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