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
65.8k8160251
asked Dec 31 '18 at 0:58
ColescuColescu
3,26711136
$endgroup$
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
65.8k8160251
asked Dec 31 '18 at 0:58
ColescuColescu
3,26711136
$endgroup$
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
65.8k8160251
asked Dec 31 '18 at 0:58
ColescuColescu
3,26711136
$endgroup$
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
65.8k8160251
asked Dec 31 '18 at 0:58
ColescuColescu
3,26711136
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
65.8k8160251
asked Dec 31 '18 at 0:58
ColescuColescu
3,26711136
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
65.8k8160251
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
65.8k8160251
edited Dec 31 '18 at 18:57
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Dec 31 '18 at 0:58
ColescuColescu
3,26711136
asked Dec 31 '18 at 0:58
ColescuColescu
3,26711136
asked Dec 31 '18 at 0:58
ColescuColescu
3,26711136
3,26711136
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
1
active
oldest
votes
$begingroup$
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
127k10151292
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057339%2fevery-perfect-set-has-cardinality-2-aleph-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
127k10151292
$endgroup$
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
127k10151292
$endgroup$
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
127k10151292
$endgroup$
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
127k10151292
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
127k10151292
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
127k10151292
answered Dec 31 '18 at 1:08
Noah SchweberNoah Schweber
127k10151292
127k10151292
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057339%2fevery-perfect-set-has-cardinality-2-aleph-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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