Every perfect set has cardinality $2^{aleph_0}$?
$begingroup$
It is well known that perfect sets in $mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:
Every perfect set in $mathbb{R}^n$ has cardinality $2^{aleph_0}$.
This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?
I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...
general-topology descriptive-set-theory
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
66.2k8160252
asked Dec 31 '18 at 0:58
ColescuColescu
3,28911137
$endgroup$
add a comment |
$begingroup$
It is well known that perfect sets in $mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:
Every perfect set in $mathbb{R}^n$ has cardinality $2^{aleph_0}$.
This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?
I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...
general-topology descriptive-set-theory
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
66.2k8160252
asked Dec 31 '18 at 0:58
ColescuColescu
3,28911137
$endgroup$
2
$begingroup$
The proof is by showing there is a "Cantor set" in any such perfect sets
$endgroup$
– Paul Plummer
Dec 31 '18 at 1:07
$begingroup$
In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
$endgroup$
– spaceisdarkgreen
Dec 31 '18 at 1:19
add a comment |
$begingroup$
It is well known that perfect sets in $mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:
Every perfect set in $mathbb{R}^n$ has cardinality $2^{aleph_0}$.
This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?
I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...
general-topology descriptive-set-theory
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
66.2k8160252
asked Dec 31 '18 at 0:58
ColescuColescu
3,28911137
$endgroup$
It is well known that perfect sets in $mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:
Every perfect set in $mathbb{R}^n$ has cardinality $2^{aleph_0}$.
This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?
I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...
general-topology descriptive-set-theory
general-topology descriptive-set-theory
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
66.2k8160252
asked Dec 31 '18 at 0:58
ColescuColescu
3,28911137
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
66.2k8160252
asked Dec 31 '18 at 0:58
ColescuColescu
3,28911137
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
66.2k8160252
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
66.2k8160252
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
66.2k8160252
66.2k8160252
asked Dec 31 '18 at 0:58
ColescuColescu
3,28911137
asked Dec 31 '18 at 0:58
ColescuColescu
3,28911137
asked Dec 31 '18 at 0:58
ColescuColescu
3,28911137
3,28911137
2
$begingroup$
The proof is by showing there is a "Cantor set" in any such perfect sets
$endgroup$
– Paul Plummer
Dec 31 '18 at 1:07
$begingroup$
In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
$endgroup$
– spaceisdarkgreen
Dec 31 '18 at 1:19
add a comment |
2
$begingroup$
The proof is by showing there is a "Cantor set" in any such perfect sets
$endgroup$
– Paul Plummer
Dec 31 '18 at 1:07
$begingroup$
In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
$endgroup$
– spaceisdarkgreen
Dec 31 '18 at 1:19
2
2
$begingroup$
The proof is by showing there is a "Cantor set" in any such perfect sets
$endgroup$
– Paul Plummer
Dec 31 '18 at 1:07
$begingroup$
The proof is by showing there is a "Cantor set" in any such perfect sets
$endgroup$
– Paul Plummer
Dec 31 '18 at 1:07
$begingroup$
In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
$endgroup$
– spaceisdarkgreen
Dec 31 '18 at 1:19
$begingroup$
In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
$endgroup$
– spaceisdarkgreen
Dec 31 '18 at 1:19
add a comment |
1 Answer
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$begingroup$
Yes, it does not require the continuum hypothesis to prove.
Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.
Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.
Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).
It might seem like we used the axiom of choice here, in two places:
Choosing $l(X)$ and $r(X)$, for a perfect set $X$.
Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.
However, we don't actually need AC here:
If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...
There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
129k10155296
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add a comment |
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$begingroup$
Yes, it does not require the continuum hypothesis to prove.
Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.
Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.
Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).
It might seem like we used the axiom of choice here, in two places:
Choosing $l(X)$ and $r(X)$, for a perfect set $X$.
Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.
However, we don't actually need AC here:
If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...
There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
129k10155296
$endgroup$
add a comment |
$begingroup$
Yes, it does not require the continuum hypothesis to prove.
Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.
Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.
Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).
It might seem like we used the axiom of choice here, in two places:
Choosing $l(X)$ and $r(X)$, for a perfect set $X$.
Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.
However, we don't actually need AC here:
If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...
There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
129k10155296
$endgroup$
add a comment |
$begingroup$
Yes, it does not require the continuum hypothesis to prove.
Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.
Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.
Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).
It might seem like we used the axiom of choice here, in two places:
Choosing $l(X)$ and $r(X)$, for a perfect set $X$.
Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.
However, we don't actually need AC here:
If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...
There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
129k10155296
$endgroup$
Yes, it does not require the continuum hypothesis to prove.
Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.
Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.
Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).
It might seem like we used the axiom of choice here, in two places:
Choosing $l(X)$ and $r(X)$, for a perfect set $X$.
Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.
However, we don't actually need AC here:
If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...
There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
129k10155296
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
129k10155296
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
129k10155296
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
129k10155296
129k10155296
add a comment |
add a comment |
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$begingroup$
The proof is by showing there is a "Cantor set" in any such perfect sets
$endgroup$
– Paul Plummer
Dec 31 '18 at 1:07
$begingroup$
In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
$endgroup$
– spaceisdarkgreen
Dec 31 '18 at 1:19