Ultraproducts with a non-principal ultrafilter is aleph-one compact
$begingroup$
I came across the following statement:
Given an $omega$-sequence of $mathcal{L}$-structures, $(mathcal{M}_i : i < omega)$, and a non-principal ultrafilter $mathcal{U}$ on $omega$, the ultraproduct of the $mathcal{L}$-structures,
$$ mathcal{M} = prod_{mathcal{U}} M_i $$
$aleph_1$-compact, i.e. for any countable (non-empty) collection of non-empty definable sets in $mathcal{M}$ that has the Finite Intersection Property, the collection has a non-empty intersection.
Perhaps the given statement was incomplete (it was in the middle of a lecture, as an aside), but in its proof, it is assumed that the collection is nested, and the rest of the proof inherently relies on this property.
I wonder if the "nestedness" of the collection is actually a necessary condition. I set out for a proof, but I am unable to arrive at a conclusion. Here is an outline of my attempt:
Let $(F_n)_{n < omega}$ be the collection given by the assumption, and let $phi_n$ be the $mathcal{L}$-formula that defines $F_n$, for each $n < omega$. Then for any $N < omega$, by the Finite Intersection Property, we can find a realization $[a_N] in bigcap_{n = 0}^{N} F_n$ such that
$$mathcal{M} models bigwedge_{n=0}^{N} phi_n ([a_N]).$$
By Łoś, we have
$$ left{ i in I : mathcal{M}_i models bigwedge_{n = 0}^{N} phi_n(a_N(i)) right} in mathcal{U}. $$
That is where I am stuck at. I reckon that my goal is to perhaps show that
$$left{ i in I : mathcal{M}_i models bigwedge_{n < omega} phi_n(a'(i)) right} in mathcal{U}$$
and complete the proof by Łoś, but I am not entirely certain if that is what I should be pursuing.
model-theory
asked Dec 5 '18 at 20:33
JaporizedJaporized
132
$endgroup$
add a comment |
$begingroup$
I came across the following statement:
Given an $omega$-sequence of $mathcal{L}$-structures, $(mathcal{M}_i : i < omega)$, and a non-principal ultrafilter $mathcal{U}$ on $omega$, the ultraproduct of the $mathcal{L}$-structures,
$$ mathcal{M} = prod_{mathcal{U}} M_i $$
$aleph_1$-compact, i.e. for any countable (non-empty) collection of non-empty definable sets in $mathcal{M}$ that has the Finite Intersection Property, the collection has a non-empty intersection.
Perhaps the given statement was incomplete (it was in the middle of a lecture, as an aside), but in its proof, it is assumed that the collection is nested, and the rest of the proof inherently relies on this property.
I wonder if the "nestedness" of the collection is actually a necessary condition. I set out for a proof, but I am unable to arrive at a conclusion. Here is an outline of my attempt:
Let $(F_n)_{n < omega}$ be the collection given by the assumption, and let $phi_n$ be the $mathcal{L}$-formula that defines $F_n$, for each $n < omega$. Then for any $N < omega$, by the Finite Intersection Property, we can find a realization $[a_N] in bigcap_{n = 0}^{N} F_n$ such that
$$mathcal{M} models bigwedge_{n=0}^{N} phi_n ([a_N]).$$
By Łoś, we have
$$ left{ i in I : mathcal{M}_i models bigwedge_{n = 0}^{N} phi_n(a_N(i)) right} in mathcal{U}. $$
That is where I am stuck at. I reckon that my goal is to perhaps show that
$$left{ i in I : mathcal{M}_i models bigwedge_{n < omega} phi_n(a'(i)) right} in mathcal{U}$$
and complete the proof by Łoś, but I am not entirely certain if that is what I should be pursuing.
model-theory
asked Dec 5 '18 at 20:33
JaporizedJaporized
132
$endgroup$
add a comment |
$begingroup$
I came across the following statement:
Given an $omega$-sequence of $mathcal{L}$-structures, $(mathcal{M}_i : i < omega)$, and a non-principal ultrafilter $mathcal{U}$ on $omega$, the ultraproduct of the $mathcal{L}$-structures,
$$ mathcal{M} = prod_{mathcal{U}} M_i $$
$aleph_1$-compact, i.e. for any countable (non-empty) collection of non-empty definable sets in $mathcal{M}$ that has the Finite Intersection Property, the collection has a non-empty intersection.
Perhaps the given statement was incomplete (it was in the middle of a lecture, as an aside), but in its proof, it is assumed that the collection is nested, and the rest of the proof inherently relies on this property.
I wonder if the "nestedness" of the collection is actually a necessary condition. I set out for a proof, but I am unable to arrive at a conclusion. Here is an outline of my attempt:
Let $(F_n)_{n < omega}$ be the collection given by the assumption, and let $phi_n$ be the $mathcal{L}$-formula that defines $F_n$, for each $n < omega$. Then for any $N < omega$, by the Finite Intersection Property, we can find a realization $[a_N] in bigcap_{n = 0}^{N} F_n$ such that
$$mathcal{M} models bigwedge_{n=0}^{N} phi_n ([a_N]).$$
By Łoś, we have
$$ left{ i in I : mathcal{M}_i models bigwedge_{n = 0}^{N} phi_n(a_N(i)) right} in mathcal{U}. $$
That is where I am stuck at. I reckon that my goal is to perhaps show that
$$left{ i in I : mathcal{M}_i models bigwedge_{n < omega} phi_n(a'(i)) right} in mathcal{U}$$
and complete the proof by Łoś, but I am not entirely certain if that is what I should be pursuing.
model-theory
asked Dec 5 '18 at 20:33
JaporizedJaporized
132
$endgroup$
I came across the following statement:
Given an $omega$-sequence of $mathcal{L}$-structures, $(mathcal{M}_i : i < omega)$, and a non-principal ultrafilter $mathcal{U}$ on $omega$, the ultraproduct of the $mathcal{L}$-structures,
$$ mathcal{M} = prod_{mathcal{U}} M_i $$
$aleph_1$-compact, i.e. for any countable (non-empty) collection of non-empty definable sets in $mathcal{M}$ that has the Finite Intersection Property, the collection has a non-empty intersection.
Perhaps the given statement was incomplete (it was in the middle of a lecture, as an aside), but in its proof, it is assumed that the collection is nested, and the rest of the proof inherently relies on this property.
I wonder if the "nestedness" of the collection is actually a necessary condition. I set out for a proof, but I am unable to arrive at a conclusion. Here is an outline of my attempt:
Let $(F_n)_{n < omega}$ be the collection given by the assumption, and let $phi_n$ be the $mathcal{L}$-formula that defines $F_n$, for each $n < omega$. Then for any $N < omega$, by the Finite Intersection Property, we can find a realization $[a_N] in bigcap_{n = 0}^{N} F_n$ such that
$$mathcal{M} models bigwedge_{n=0}^{N} phi_n ([a_N]).$$
By Łoś, we have
$$ left{ i in I : mathcal{M}_i models bigwedge_{n = 0}^{N} phi_n(a_N(i)) right} in mathcal{U}. $$
That is where I am stuck at. I reckon that my goal is to perhaps show that
$$left{ i in I : mathcal{M}_i models bigwedge_{n < omega} phi_n(a'(i)) right} in mathcal{U}$$
and complete the proof by Łoś, but I am not entirely certain if that is what I should be pursuing.
model-theory
model-theory
asked Dec 5 '18 at 20:33
JaporizedJaporized
132
asked Dec 5 '18 at 20:33
JaporizedJaporized
132
asked Dec 5 '18 at 20:33
JaporizedJaporized
132
asked Dec 5 '18 at 20:33
JaporizedJaporized
132
asked Dec 5 '18 at 20:33
JaporizedJaporized
132
132
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's not a necessary assumption. If $(D_i)_{iinomega}$ is a family of sets with the finite intersection property, then let $E_i=bigcap_{jle i}D_i$; we have $E_0supseteq E_1supseteq ...$ and each $E_i$ is nonempty by the finite intserction property of the $D_i$s. Moreover, if the $D_i$s were definable, so are the $E_i$s, and anything in the intersection of the $E_i$s is in the intersection of the $D_i$s. So we can go from an arbitrary collection to a nested collection, find something in the intersection of the elements of the nested collection, and say that it's also in the intersection of the elements of the original collection.
answered Dec 5 '18 at 20:46
Noah SchweberNoah Schweber
123k10150285
$endgroup$
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%2f3027612%2fultraproducts-with-a-non-principal-ultrafilter-is-aleph-one-compact%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
$begingroup$
It's not a necessary assumption. If $(D_i)_{iinomega}$ is a family of sets with the finite intersection property, then let $E_i=bigcap_{jle i}D_i$; we have $E_0supseteq E_1supseteq ...$ and each $E_i$ is nonempty by the finite intserction property of the $D_i$s. Moreover, if the $D_i$s were definable, so are the $E_i$s, and anything in the intersection of the $E_i$s is in the intersection of the $D_i$s. So we can go from an arbitrary collection to a nested collection, find something in the intersection of the elements of the nested collection, and say that it's also in the intersection of the elements of the original collection.
answered Dec 5 '18 at 20:46
Noah SchweberNoah Schweber
123k10150285
$endgroup$
add a comment |
$begingroup$
It's not a necessary assumption. If $(D_i)_{iinomega}$ is a family of sets with the finite intersection property, then let $E_i=bigcap_{jle i}D_i$; we have $E_0supseteq E_1supseteq ...$ and each $E_i$ is nonempty by the finite intserction property of the $D_i$s. Moreover, if the $D_i$s were definable, so are the $E_i$s, and anything in the intersection of the $E_i$s is in the intersection of the $D_i$s. So we can go from an arbitrary collection to a nested collection, find something in the intersection of the elements of the nested collection, and say that it's also in the intersection of the elements of the original collection.
answered Dec 5 '18 at 20:46
Noah SchweberNoah Schweber
123k10150285
$endgroup$
add a comment |
$begingroup$
It's not a necessary assumption. If $(D_i)_{iinomega}$ is a family of sets with the finite intersection property, then let $E_i=bigcap_{jle i}D_i$; we have $E_0supseteq E_1supseteq ...$ and each $E_i$ is nonempty by the finite intserction property of the $D_i$s. Moreover, if the $D_i$s were definable, so are the $E_i$s, and anything in the intersection of the $E_i$s is in the intersection of the $D_i$s. So we can go from an arbitrary collection to a nested collection, find something in the intersection of the elements of the nested collection, and say that it's also in the intersection of the elements of the original collection.
answered Dec 5 '18 at 20:46
Noah SchweberNoah Schweber
123k10150285
$endgroup$
It's not a necessary assumption. If $(D_i)_{iinomega}$ is a family of sets with the finite intersection property, then let $E_i=bigcap_{jle i}D_i$; we have $E_0supseteq E_1supseteq ...$ and each $E_i$ is nonempty by the finite intserction property of the $D_i$s. Moreover, if the $D_i$s were definable, so are the $E_i$s, and anything in the intersection of the $E_i$s is in the intersection of the $D_i$s. So we can go from an arbitrary collection to a nested collection, find something in the intersection of the elements of the nested collection, and say that it's also in the intersection of the elements of the original collection.
answered Dec 5 '18 at 20:46
Noah SchweberNoah Schweber
123k10150285
answered Dec 5 '18 at 20:46
Noah SchweberNoah Schweber
123k10150285
answered Dec 5 '18 at 20:46
Noah SchweberNoah Schweber
123k10150285
answered Dec 5 '18 at 20:46
Noah SchweberNoah Schweber
123k10150285
123k10150285
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%2f3027612%2fultraproducts-with-a-non-principal-ultrafilter-is-aleph-one-compact%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
