Do the following subsets of $mathbb{R}$ form a complete space?
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Consider the following sets
$$mathbb{R}setminus mathbb{Q}, mathbb{Z}, [0,1), [0,infty).$$
For a subset of complete space to be complete, it must be the case that it is also closed. The set of irrational numbers, $[0,1)$ and $[0,infty)$ are neither open or closed and so they cannot form a complete space. However, $mathbb{Z}$ forms a complete space with the usual metric since any Cauchy Sequence in $mathbb{Z}$ will become eventually constant and thus convergent.
Is this answer correct?
real-analysis
asked Nov 25 '18 at 13:42
Hello_WorldHello_World
4,11621731
$endgroup$
add a comment |
$begingroup$
Consider the following sets
$$mathbb{R}setminus mathbb{Q}, mathbb{Z}, [0,1), [0,infty).$$
For a subset of complete space to be complete, it must be the case that it is also closed. The set of irrational numbers, $[0,1)$ and $[0,infty)$ are neither open or closed and so they cannot form a complete space. However, $mathbb{Z}$ forms a complete space with the usual metric since any Cauchy Sequence in $mathbb{Z}$ will become eventually constant and thus convergent.
Is this answer correct?
real-analysis
asked Nov 25 '18 at 13:42
Hello_WorldHello_World
4,11621731
$endgroup$
$begingroup$
But $[0,infty)$ is closed and $mathbb{R} $ $mathbb{Q}$ is not closed.
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– John_Wick
Nov 25 '18 at 14:18
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So $[0,infty)$ is complete and $mathbb{R}setminus mathbb{Q}$ is not complete, right?
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– Hello_World
Nov 25 '18 at 14:47
$begingroup$
Yes that is correct
$endgroup$
– John_Wick
Nov 25 '18 at 19:35
add a comment |
$begingroup$
Consider the following sets
$$mathbb{R}setminus mathbb{Q}, mathbb{Z}, [0,1), [0,infty).$$
For a subset of complete space to be complete, it must be the case that it is also closed. The set of irrational numbers, $[0,1)$ and $[0,infty)$ are neither open or closed and so they cannot form a complete space. However, $mathbb{Z}$ forms a complete space with the usual metric since any Cauchy Sequence in $mathbb{Z}$ will become eventually constant and thus convergent.
Is this answer correct?
real-analysis
asked Nov 25 '18 at 13:42
Hello_WorldHello_World
4,11621731
$endgroup$
Consider the following sets
$$mathbb{R}setminus mathbb{Q}, mathbb{Z}, [0,1), [0,infty).$$
For a subset of complete space to be complete, it must be the case that it is also closed. The set of irrational numbers, $[0,1)$ and $[0,infty)$ are neither open or closed and so they cannot form a complete space. However, $mathbb{Z}$ forms a complete space with the usual metric since any Cauchy Sequence in $mathbb{Z}$ will become eventually constant and thus convergent.
Is this answer correct?
real-analysis
real-analysis
asked Nov 25 '18 at 13:42
Hello_WorldHello_World
4,11621731
asked Nov 25 '18 at 13:42
Hello_WorldHello_World
4,11621731
asked Nov 25 '18 at 13:42
Hello_WorldHello_World
4,11621731
asked Nov 25 '18 at 13:42
Hello_WorldHello_World
4,11621731
asked Nov 25 '18 at 13:42
Hello_WorldHello_World
4,11621731
4,11621731
$begingroup$
But $[0,infty)$ is closed and $mathbb{R} $ $mathbb{Q}$ is not closed.
$endgroup$
– John_Wick
Nov 25 '18 at 14:18
$begingroup$
So $[0,infty)$ is complete and $mathbb{R}setminus mathbb{Q}$ is not complete, right?
$endgroup$
– Hello_World
Nov 25 '18 at 14:47
$begingroup$
Yes that is correct
$endgroup$
– John_Wick
Nov 25 '18 at 19:35
add a comment |
$begingroup$
But $[0,infty)$ is closed and $mathbb{R} $ $mathbb{Q}$ is not closed.
$endgroup$
– John_Wick
Nov 25 '18 at 14:18
$begingroup$
So $[0,infty)$ is complete and $mathbb{R}setminus mathbb{Q}$ is not complete, right?
$endgroup$
– Hello_World
Nov 25 '18 at 14:47
$begingroup$
Yes that is correct
$endgroup$
– John_Wick
Nov 25 '18 at 19:35
$begingroup$
But $[0,infty)$ is closed and $mathbb{R} $ $mathbb{Q}$ is not closed.
$endgroup$
– John_Wick
Nov 25 '18 at 14:18
$begingroup$
But $[0,infty)$ is closed and $mathbb{R} $ $mathbb{Q}$ is not closed.
$endgroup$
– John_Wick
Nov 25 '18 at 14:18
$begingroup$
So $[0,infty)$ is complete and $mathbb{R}setminus mathbb{Q}$ is not complete, right?
$endgroup$
– Hello_World
Nov 25 '18 at 14:47
$begingroup$
So $[0,infty)$ is complete and $mathbb{R}setminus mathbb{Q}$ is not complete, right?
$endgroup$
– Hello_World
Nov 25 '18 at 14:47
$begingroup$
Yes that is correct
$endgroup$
– John_Wick
Nov 25 '18 at 19:35
$begingroup$
Yes that is correct
$endgroup$
– John_Wick
Nov 25 '18 at 19:35
add a comment |
1 Answer
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$begingroup$
You got three out of four. From Wikipedia on complete metric spaces, a subspace of a complete metric space is complete if and only if it is closed. $Bbb{R}$ of course is complete, so the complete subspaces in your list are just the closed ones, which are $Bbb{Z}$ and $[0, infty)$.
Your direct argument for $Bbb{Z}$ is good, and I think you can easily also give examples of Cauchy sequences in $Bbb{R} setminus Bbb{Q}$ and $[0, 1)$ which do not converge in those subspaces. You might try proving directly that $[0, infty)$ is complete.
answered Nov 25 '18 at 15:20
Hew WolffHew Wolff
2,245716
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1 Answer
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$begingroup$
You got three out of four. From Wikipedia on complete metric spaces, a subspace of a complete metric space is complete if and only if it is closed. $Bbb{R}$ of course is complete, so the complete subspaces in your list are just the closed ones, which are $Bbb{Z}$ and $[0, infty)$.
Your direct argument for $Bbb{Z}$ is good, and I think you can easily also give examples of Cauchy sequences in $Bbb{R} setminus Bbb{Q}$ and $[0, 1)$ which do not converge in those subspaces. You might try proving directly that $[0, infty)$ is complete.
answered Nov 25 '18 at 15:20
Hew WolffHew Wolff
2,245716
$endgroup$
add a comment |
$begingroup$
You got three out of four. From Wikipedia on complete metric spaces, a subspace of a complete metric space is complete if and only if it is closed. $Bbb{R}$ of course is complete, so the complete subspaces in your list are just the closed ones, which are $Bbb{Z}$ and $[0, infty)$.
Your direct argument for $Bbb{Z}$ is good, and I think you can easily also give examples of Cauchy sequences in $Bbb{R} setminus Bbb{Q}$ and $[0, 1)$ which do not converge in those subspaces. You might try proving directly that $[0, infty)$ is complete.
answered Nov 25 '18 at 15:20
Hew WolffHew Wolff
2,245716
$endgroup$
add a comment |
$begingroup$
You got three out of four. From Wikipedia on complete metric spaces, a subspace of a complete metric space is complete if and only if it is closed. $Bbb{R}$ of course is complete, so the complete subspaces in your list are just the closed ones, which are $Bbb{Z}$ and $[0, infty)$.
Your direct argument for $Bbb{Z}$ is good, and I think you can easily also give examples of Cauchy sequences in $Bbb{R} setminus Bbb{Q}$ and $[0, 1)$ which do not converge in those subspaces. You might try proving directly that $[0, infty)$ is complete.
answered Nov 25 '18 at 15:20
Hew WolffHew Wolff
2,245716
$endgroup$
You got three out of four. From Wikipedia on complete metric spaces, a subspace of a complete metric space is complete if and only if it is closed. $Bbb{R}$ of course is complete, so the complete subspaces in your list are just the closed ones, which are $Bbb{Z}$ and $[0, infty)$.
Your direct argument for $Bbb{Z}$ is good, and I think you can easily also give examples of Cauchy sequences in $Bbb{R} setminus Bbb{Q}$ and $[0, 1)$ which do not converge in those subspaces. You might try proving directly that $[0, infty)$ is complete.
answered Nov 25 '18 at 15:20
Hew WolffHew Wolff
2,245716
answered Nov 25 '18 at 15:20
Hew WolffHew Wolff
2,245716
answered Nov 25 '18 at 15:20
Hew WolffHew Wolff
2,245716
answered Nov 25 '18 at 15:20
Hew WolffHew Wolff
2,245716
2,245716
add a comment |
add a comment |
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$begingroup$
But $[0,infty)$ is closed and $mathbb{R} $ $mathbb{Q}$ is not closed.
$endgroup$
– John_Wick
Nov 25 '18 at 14:18
$begingroup$
So $[0,infty)$ is complete and $mathbb{R}setminus mathbb{Q}$ is not complete, right?
$endgroup$
– Hello_World
Nov 25 '18 at 14:47
$begingroup$
Yes that is correct
$endgroup$
– John_Wick
Nov 25 '18 at 19:35