Mixing
$U$ is open then find a sequence of closed sets $ I_{n+1} subset I_{n}$
Let $U = ]0,1[$, then I am trying to find a sequence of closed sets $(I_n)_{n in mathbb{N}}$ of $mathbb{R}$ such that :
$$ I_{n+1} subset I_n text{ and } cap_{n in mathbb{N}} I_n subset U$$
And such that there isn’t an $ r in mathbb{N}$ such that :
$$forall N > r, I_N subset U$$
I didn’t succeed in finding such a sequence of closed sets.
The main problem is that the intersection of closed sets is still a closed set hence the second condition makes it hard.
So for exemple : $I_n =[1/2, 1+1/n]$ don’t work since $I_{infty} = [1/2,1]$ wich is closed and hence not a subset of $U$.
calculus real-analysis general-topology borel-sets
asked Nov 20 '18 at 12:58
Interesting problems
13310
add a comment |
Let $U = ]0,1[$, then I am trying to find a sequence of closed sets $(I_n)_{n in mathbb{N}}$ of $mathbb{R}$ such that :
$$ I_{n+1} subset I_n text{ and } cap_{n in mathbb{N}} I_n subset U$$
And such that there isn’t an $ r in mathbb{N}$ such that :
$$forall N > r, I_N subset U$$
I didn’t succeed in finding such a sequence of closed sets.
The main problem is that the intersection of closed sets is still a closed set hence the second condition makes it hard.
So for exemple : $I_n =[1/2, 1+1/n]$ don’t work since $I_{infty} = [1/2,1]$ wich is closed and hence not a subset of $U$.
calculus real-analysis general-topology borel-sets
asked Nov 20 '18 at 12:58
Interesting problems
13310
add a comment |
Let $U = ]0,1[$, then I am trying to find a sequence of closed sets $(I_n)_{n in mathbb{N}}$ of $mathbb{R}$ such that :
$$ I_{n+1} subset I_n text{ and } cap_{n in mathbb{N}} I_n subset U$$
And such that there isn’t an $ r in mathbb{N}$ such that :
$$forall N > r, I_N subset U$$
I didn’t succeed in finding such a sequence of closed sets.
The main problem is that the intersection of closed sets is still a closed set hence the second condition makes it hard.
So for exemple : $I_n =[1/2, 1+1/n]$ don’t work since $I_{infty} = [1/2,1]$ wich is closed and hence not a subset of $U$.
calculus real-analysis general-topology borel-sets
asked Nov 20 '18 at 12:58
Interesting problems
13310
Let $U = ]0,1[$, then I am trying to find a sequence of closed sets $(I_n)_{n in mathbb{N}}$ of $mathbb{R}$ such that :
$$ I_{n+1} subset I_n text{ and } cap_{n in mathbb{N}} I_n subset U$$
And such that there isn’t an $ r in mathbb{N}$ such that :
$$forall N > r, I_N subset U$$
I didn’t succeed in finding such a sequence of closed sets.
The main problem is that the intersection of closed sets is still a closed set hence the second condition makes it hard.
So for exemple : $I_n =[1/2, 1+1/n]$ don’t work since $I_{infty} = [1/2,1]$ wich is closed and hence not a subset of $U$.
calculus real-analysis general-topology borel-sets
calculus real-analysis general-topology borel-sets
asked Nov 20 '18 at 12:58
Interesting problems
13310
asked Nov 20 '18 at 12:58
Interesting problems
13310
asked Nov 20 '18 at 12:58
Interesting problems
13310
asked Nov 20 '18 at 12:58
Interesting problems
13310
asked Nov 20 '18 at 12:58
Interesting problems
13310
13310
add a comment |
add a comment |
1 Answer
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Such a sequence exists.
First note that since $mathbb{R} backslash U$ is not compact then by Borel- Lebesgue it is possible to find a sequence that fulfills the condition of the problem.
Simply take :
$$I_n = [n, +infty)$$
Then we clearly have :
$$bigcap_{i = 1}^{infty} I_i = emptyset subset U$$
Hence such a sequence exists.
answered Nov 20 '18 at 17:08
Thinking
91916
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Such a sequence exists.
First note that since $mathbb{R} backslash U$ is not compact then by Borel- Lebesgue it is possible to find a sequence that fulfills the condition of the problem.
Simply take :
$$I_n = [n, +infty)$$
Then we clearly have :
$$bigcap_{i = 1}^{infty} I_i = emptyset subset U$$
Hence such a sequence exists.
answered Nov 20 '18 at 17:08
Thinking
91916
add a comment |
Such a sequence exists.
First note that since $mathbb{R} backslash U$ is not compact then by Borel- Lebesgue it is possible to find a sequence that fulfills the condition of the problem.
Simply take :
$$I_n = [n, +infty)$$
Then we clearly have :
$$bigcap_{i = 1}^{infty} I_i = emptyset subset U$$
Hence such a sequence exists.
answered Nov 20 '18 at 17:08
Thinking
91916
add a comment |
Such a sequence exists.
First note that since $mathbb{R} backslash U$ is not compact then by Borel- Lebesgue it is possible to find a sequence that fulfills the condition of the problem.
Simply take :
$$I_n = [n, +infty)$$
Then we clearly have :
$$bigcap_{i = 1}^{infty} I_i = emptyset subset U$$
Hence such a sequence exists.
answered Nov 20 '18 at 17:08
Thinking
91916
Such a sequence exists.
First note that since $mathbb{R} backslash U$ is not compact then by Borel- Lebesgue it is possible to find a sequence that fulfills the condition of the problem.
Simply take :
$$I_n = [n, +infty)$$
Then we clearly have :
$$bigcap_{i = 1}^{infty} I_i = emptyset subset U$$
Hence such a sequence exists.
answered Nov 20 '18 at 17:08
Thinking
91916
edited Nov 20 '18 at 17:22
answered Nov 20 '18 at 17:08
Thinking
91916
answered Nov 20 '18 at 17:08
Thinking
91916
answered Nov 20 '18 at 17:08
Thinking
91916
91916
add a comment |
add a comment |
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