Mixing
Which sets are “clompact�
Clash Royale CLAN TAG#URR8PPP
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This is an exercise taken verbatim from Abbott's Understanding Analysis:
Let’s call a set clompact if it has the property that every
closed cover (i.e., a cover consisting of closed sets) admits a finite subcover.
Describe all of the clompact subsets of $mathbf R$.
I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover
$$ 0cupbigcup_1^inftyleft[frac1n+1,frac1nright] $$
for $[0,1]$ does not have a finite subcover. Similar constructions of closed covers show that $[a,b]$, $(a,b]$, $[b,a)$ and $(a,b)$ are not clompact as well. In addition, $mathbf R$ itself and any unbounded interval is also not clompact.
Is anyone able to help in solving this problem? Any assistance is appreciated.
real-analysis analysis
asked 2 hours ago
YiFan
1,2181210
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up vote
1
down vote
favorite
This is an exercise taken verbatim from Abbott's Understanding Analysis:
Let’s call a set clompact if it has the property that every
closed cover (i.e., a cover consisting of closed sets) admits a finite subcover.
Describe all of the clompact subsets of $mathbf R$.
I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover
$$ 0cupbigcup_1^inftyleft[frac1n+1,frac1nright] $$
for $[0,1]$ does not have a finite subcover. Similar constructions of closed covers show that $[a,b]$, $(a,b]$, $[b,a)$ and $(a,b)$ are not clompact as well. In addition, $mathbf R$ itself and any unbounded interval is also not clompact.
Is anyone able to help in solving this problem? Any assistance is appreciated.
real-analysis analysis
asked 2 hours ago
YiFan
1,2181210
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is an exercise taken verbatim from Abbott's Understanding Analysis:
Let’s call a set clompact if it has the property that every
closed cover (i.e., a cover consisting of closed sets) admits a finite subcover.
Describe all of the clompact subsets of $mathbf R$.
I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover
$$ 0cupbigcup_1^inftyleft[frac1n+1,frac1nright] $$
for $[0,1]$ does not have a finite subcover. Similar constructions of closed covers show that $[a,b]$, $(a,b]$, $[b,a)$ and $(a,b)$ are not clompact as well. In addition, $mathbf R$ itself and any unbounded interval is also not clompact.
Is anyone able to help in solving this problem? Any assistance is appreciated.
real-analysis analysis
asked 2 hours ago
YiFan
1,2181210
This is an exercise taken verbatim from Abbott's Understanding Analysis:
Let’s call a set clompact if it has the property that every
closed cover (i.e., a cover consisting of closed sets) admits a finite subcover.
Describe all of the clompact subsets of $mathbf R$.
I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover
$$ 0cupbigcup_1^inftyleft[frac1n+1,frac1nright] $$
for $[0,1]$ does not have a finite subcover. Similar constructions of closed covers show that $[a,b]$, $(a,b]$, $[b,a)$ and $(a,b)$ are not clompact as well. In addition, $mathbf R$ itself and any unbounded interval is also not clompact.
Is anyone able to help in solving this problem? Any assistance is appreciated.
real-analysis analysis
real-analysis analysis
asked 2 hours ago
YiFan
1,2181210
asked 2 hours ago
YiFan
1,2181210
asked 2 hours ago
YiFan
1,2181210
asked 2 hours ago
YiFan
1,2181210
asked 2 hours ago
YiFan
1,2181210
1,2181210
add a comment |Â
add a comment |Â
1 Answer
1
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6
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Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.
answered 2 hours ago
Aweygan
12.7k21441
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.
answered 2 hours ago
Aweygan
12.7k21441
add a comment |Â
up vote
6
down vote
accepted
Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.
answered 2 hours ago
Aweygan
12.7k21441
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.
answered 2 hours ago
Aweygan
12.7k21441
Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.
answered 2 hours ago
Aweygan
12.7k21441
answered 2 hours ago
Aweygan
12.7k21441
answered 2 hours ago
Aweygan
12.7k21441
answered 2 hours ago
Aweygan
12.7k21441
12.7k21441
add a comment |Â
add a comment |Â
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