Properties of Cantor set
$[0,1]$ is not homeomorphic to $[0,1]×[0,1]$ but $C$ is homeomorphic to $C times C$ where $C$ is the Cantor set.
I know both the proof. I am asking which property of $C$ is the reason of this absurdity!
general-topology cantor-set
edited Dec 9 at 17:11
Martin Sleziak
44.7k7115270
asked Dec 9 at 15:53
Santanu Debnath
1529
|
show 3 more comments
$[0,1]$ is not homeomorphic to $[0,1]×[0,1]$ but $C$ is homeomorphic to $C times C$ where $C$ is the Cantor set.
I know both the proof. I am asking which property of $C$ is the reason of this absurdity!
general-topology cantor-set
edited Dec 9 at 17:11
Martin Sleziak
44.7k7115270
asked Dec 9 at 15:53
Santanu Debnath
1529
If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
– Jack D'Aurizio
Dec 9 at 15:56
Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
– Santanu Debnath
Dec 9 at 15:59
I hope that is not true for finite set
– Santanu Debnath
Dec 9 at 16:01
No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
– Henno Brandsma
Dec 9 at 16:15
@JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
– Henno Brandsma
Dec 9 at 16:22
|
show 3 more comments
$[0,1]$ is not homeomorphic to $[0,1]×[0,1]$ but $C$ is homeomorphic to $C times C$ where $C$ is the Cantor set.
I know both the proof. I am asking which property of $C$ is the reason of this absurdity!
general-topology cantor-set
edited Dec 9 at 17:11
Martin Sleziak
44.7k7115270
asked Dec 9 at 15:53
Santanu Debnath
1529
$[0,1]$ is not homeomorphic to $[0,1]×[0,1]$ but $C$ is homeomorphic to $C times C$ where $C$ is the Cantor set.
I know both the proof. I am asking which property of $C$ is the reason of this absurdity!
general-topology cantor-set
general-topology cantor-set
edited Dec 9 at 17:11
Martin Sleziak
44.7k7115270
asked Dec 9 at 15:53
Santanu Debnath
1529
edited Dec 9 at 17:11
Martin Sleziak
44.7k7115270
asked Dec 9 at 15:53
Santanu Debnath
1529
edited Dec 9 at 17:11
Martin Sleziak
44.7k7115270
edited Dec 9 at 17:11
Martin Sleziak
44.7k7115270
edited Dec 9 at 17:11
Martin Sleziak
44.7k7115270
44.7k7115270
asked Dec 9 at 15:53
Santanu Debnath
1529
asked Dec 9 at 15:53
Santanu Debnath
1529
asked Dec 9 at 15:53
Santanu Debnath
1529
1529
If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
– Jack D'Aurizio
Dec 9 at 15:56
Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
– Santanu Debnath
Dec 9 at 15:59
I hope that is not true for finite set
– Santanu Debnath
Dec 9 at 16:01
No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
– Henno Brandsma
Dec 9 at 16:15
@JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
– Henno Brandsma
Dec 9 at 16:22
|
show 3 more comments
If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
– Jack D'Aurizio
Dec 9 at 15:56
Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
– Santanu Debnath
Dec 9 at 15:59
I hope that is not true for finite set
– Santanu Debnath
Dec 9 at 16:01
No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
– Henno Brandsma
Dec 9 at 16:15
@JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
– Henno Brandsma
Dec 9 at 16:22
If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
– Jack D'Aurizio
Dec 9 at 15:56
If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
– Jack D'Aurizio
Dec 9 at 15:56
Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
– Santanu Debnath
Dec 9 at 15:59
Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
– Santanu Debnath
Dec 9 at 15:59
I hope that is not true for finite set
– Santanu Debnath
Dec 9 at 16:01
I hope that is not true for finite set
– Santanu Debnath
Dec 9 at 16:01
No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
– Henno Brandsma
Dec 9 at 16:15
No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
– Henno Brandsma
Dec 9 at 16:15
@JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
– Henno Brandsma
Dec 9 at 16:22
@JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
– Henno Brandsma
Dec 9 at 16:22
|
show 3 more comments
1 Answer
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oldest
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$C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.
If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.
Other spaces with unique charaterisations also have such preservations by finite products:
$mathbb{Q}$: the unique countable metric space without isolated points.
$mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).
$Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.
And some more exist too. All of the above are homeomorphic to their squares.
answered Dec 9 at 16:21
Henno Brandsma
105k346114
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
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votes
$C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.
If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.
Other spaces with unique charaterisations also have such preservations by finite products:
$mathbb{Q}$: the unique countable metric space without isolated points.
$mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).
$Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.
And some more exist too. All of the above are homeomorphic to their squares.
answered Dec 9 at 16:21
Henno Brandsma
105k346114
add a comment |
$C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.
If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.
Other spaces with unique charaterisations also have such preservations by finite products:
$mathbb{Q}$: the unique countable metric space without isolated points.
$mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).
$Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.
And some more exist too. All of the above are homeomorphic to their squares.
answered Dec 9 at 16:21
Henno Brandsma
105k346114
add a comment |
$C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.
If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.
Other spaces with unique charaterisations also have such preservations by finite products:
$mathbb{Q}$: the unique countable metric space without isolated points.
$mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).
$Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.
And some more exist too. All of the above are homeomorphic to their squares.
answered Dec 9 at 16:21
Henno Brandsma
105k346114
$C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.
If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.
Other spaces with unique charaterisations also have such preservations by finite products:
$mathbb{Q}$: the unique countable metric space without isolated points.
$mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).
$Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.
And some more exist too. All of the above are homeomorphic to their squares.
answered Dec 9 at 16:21
Henno Brandsma
105k346114
edited Dec 9 at 16:28
answered Dec 9 at 16:21
Henno Brandsma
105k346114
answered Dec 9 at 16:21
Henno Brandsma
105k346114
answered Dec 9 at 16:21
Henno Brandsma
105k346114
105k346114
add a comment |
add a comment |
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If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
– Jack D'Aurizio
Dec 9 at 15:56
Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
– Santanu Debnath
Dec 9 at 15:59
I hope that is not true for finite set
– Santanu Debnath
Dec 9 at 16:01
No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
– Henno Brandsma
Dec 9 at 16:15
@JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
– Henno Brandsma
Dec 9 at 16:22