About the unionset axiomConfusion regarding one formulation of the Axiom of Choice.What does it mean for a set of sentences $mathcalT$ to “secure” a set of sentences $Delta$?Is the Axiom of Choice needed here?Universal Specification axiomNaive set theory really need axiom of power?are $a$'s, $1$'s in a family $a, a, a, 1, 2, 2, 4$ themselves sets?Is it okay to say 'an element of a family'?Set Theory: Equivalency of Axiom of Choice and Choice FunctionHow does Axiom of Choice imply Axiom of Dependent Choice?Sets Without a Minimal Element (Axiom of Foundation)
Golf game boilerplate
Can a Gentile theist be saved?
Giant Toughroad SLR 2 for 200 miles in two days, will it make it?
Word describing multiple paths to the same abstract outcome
Can the harmonic series explain the origin of the major scale?
Invariance of results when scaling explanatory variables in logistic regression, is there a proof?
A workplace installs custom certificates on personal devices, can this be used to decrypt HTTPS traffic?
Does "Dominei" mean something?
Can I use my Chinese passport to enter China after I acquired another citizenship?
How will losing mobility of one hand affect my career as a programmer?
Bob has never been a M before
What is the opposite of 'gravitas'?
Could solar power be utilized and substitute coal in the 19th century?
Should my PhD thesis be submitted under my legal name?
My boss asked me to take a one-day class, then signs it up as a day off
In Star Trek IV, why did the Bounty go back to a time when whales were already rare?
Installing PowerShell on 32-bit Kali OS fails
Greatest common substring
Why are all the doors on Ferenginar (the Ferengi home world) far shorter than the average Ferengi?
Did US corporations pay demonstrators in the German demonstrations against article 13?
Female=gender counterpart?
Can I create an upright 7-foot × 5-foot wall with the Minor Illusion spell?
Why are on-board computers allowed to change controls without notifying the pilots?
Indicating multiple different modes of speech (fantasy language or telepathy)
About the unionset axiom
Confusion regarding one formulation of the Axiom of Choice.What does it mean for a set of sentences $mathcalT$ to “secure” a set of sentences $Delta$?Is the Axiom of Choice needed here?Universal Specification axiomNaive set theory really need axiom of power?are $a$'s, $1$'s in a family $a, a, a, 1, 2, 2, 4$ themselves sets?Is it okay to say 'an element of a family'?Set Theory: Equivalency of Axiom of Choice and Choice FunctionHow does Axiom of Choice imply Axiom of Dependent Choice?Sets Without a Minimal Element (Axiom of Foundation)
$begingroup$
I found this formulation of the unionset axiom:
For each set $mathcal E$ there exists a set $B$ whose members are the members of the members of $mathcal E$, i.e., so that it satisfies the equivalence $$tin Biff (exists Xinmathcal E)(tin X)$$
An then:
The unionset operation is obviously most useful when $mathcal E$ is a family of sets, i.e., a set all of whose members are also sets.
So the last statement implies that the elements of $mathcal E$ are not necessarily sets. But doesn't the first quotation imply that all elements of $mathcal E$ are sets? Otherwise how can $t$ be a member of an element of $mathcal E$?
elementary-set-theory logic
edited Mar 17 at 0:08
Andrés E. Caicedo
65.8k8160251
asked Mar 16 at 23:05
logiclogic
948
$endgroup$
add a comment |
$begingroup$
I found this formulation of the unionset axiom:
For each set $mathcal E$ there exists a set $B$ whose members are the members of the members of $mathcal E$, i.e., so that it satisfies the equivalence $$tin Biff (exists Xinmathcal E)(tin X)$$
An then:
The unionset operation is obviously most useful when $mathcal E$ is a family of sets, i.e., a set all of whose members are also sets.
So the last statement implies that the elements of $mathcal E$ are not necessarily sets. But doesn't the first quotation imply that all elements of $mathcal E$ are sets? Otherwise how can $t$ be a member of an element of $mathcal E$?
elementary-set-theory logic
edited Mar 17 at 0:08
Andrés E. Caicedo
65.8k8160251
asked Mar 16 at 23:05
logiclogic
948
$endgroup$
1
$begingroup$
If none of the elements of E are sets, then B is the empty set.
$endgroup$
– user21793
Mar 16 at 23:30
1
$begingroup$
In ZF(C) everything is a set. However, in everyday math we often use other objects (e.g. numbers) that we don't regard as sets, and though in ZF(C) they are presented so, we don't want to take their elements..
$endgroup$
– Berci
Mar 16 at 23:33
$begingroup$
@Berci But it wasn't stated prior to the the unionset axiom that everything is a set. (I'm talking about "Notes on Set Theory" by Moschovakis. And if everything is a set, then I don't see why the author made the second comment I quoted (when he referred to the "special case" when $mathcal E$ is a set all of whose members are also sets).
$endgroup$
– logic
Mar 16 at 23:40
add a comment |
$begingroup$
I found this formulation of the unionset axiom:
For each set $mathcal E$ there exists a set $B$ whose members are the members of the members of $mathcal E$, i.e., so that it satisfies the equivalence $$tin Biff (exists Xinmathcal E)(tin X)$$
An then:
The unionset operation is obviously most useful when $mathcal E$ is a family of sets, i.e., a set all of whose members are also sets.
So the last statement implies that the elements of $mathcal E$ are not necessarily sets. But doesn't the first quotation imply that all elements of $mathcal E$ are sets? Otherwise how can $t$ be a member of an element of $mathcal E$?
elementary-set-theory logic
edited Mar 17 at 0:08
Andrés E. Caicedo
65.8k8160251
asked Mar 16 at 23:05
logiclogic
948
$endgroup$
I found this formulation of the unionset axiom:
For each set $mathcal E$ there exists a set $B$ whose members are the members of the members of $mathcal E$, i.e., so that it satisfies the equivalence $$tin Biff (exists Xinmathcal E)(tin X)$$
An then:
The unionset operation is obviously most useful when $mathcal E$ is a family of sets, i.e., a set all of whose members are also sets.
So the last statement implies that the elements of $mathcal E$ are not necessarily sets. But doesn't the first quotation imply that all elements of $mathcal E$ are sets? Otherwise how can $t$ be a member of an element of $mathcal E$?
elementary-set-theory logic
elementary-set-theory logic
edited Mar 17 at 0:08
Andrés E. Caicedo
65.8k8160251
asked Mar 16 at 23:05
logiclogic
948
edited Mar 17 at 0:08
Andrés E. Caicedo
65.8k8160251
asked Mar 16 at 23:05
logiclogic
948
edited Mar 17 at 0:08
Andrés E. Caicedo
65.8k8160251
edited Mar 17 at 0:08
Andrés E. Caicedo
65.8k8160251
edited Mar 17 at 0:08
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Mar 16 at 23:05
logiclogic
948
asked Mar 16 at 23:05
logiclogic
948
asked Mar 16 at 23:05
logiclogic
948
948
1
$begingroup$
If none of the elements of E are sets, then B is the empty set.
$endgroup$
– user21793
Mar 16 at 23:30
1
$begingroup$
In ZF(C) everything is a set. However, in everyday math we often use other objects (e.g. numbers) that we don't regard as sets, and though in ZF(C) they are presented so, we don't want to take their elements..
$endgroup$
– Berci
Mar 16 at 23:33
$begingroup$
@Berci But it wasn't stated prior to the the unionset axiom that everything is a set. (I'm talking about "Notes on Set Theory" by Moschovakis. And if everything is a set, then I don't see why the author made the second comment I quoted (when he referred to the "special case" when $mathcal E$ is a set all of whose members are also sets).
$endgroup$
– logic
Mar 16 at 23:40
add a comment |
1
$begingroup$
If none of the elements of E are sets, then B is the empty set.
$endgroup$
– user21793
Mar 16 at 23:30
1
$begingroup$
In ZF(C) everything is a set. However, in everyday math we often use other objects (e.g. numbers) that we don't regard as sets, and though in ZF(C) they are presented so, we don't want to take their elements..
$endgroup$
– Berci
Mar 16 at 23:33
$begingroup$
@Berci But it wasn't stated prior to the the unionset axiom that everything is a set. (I'm talking about "Notes on Set Theory" by Moschovakis. And if everything is a set, then I don't see why the author made the second comment I quoted (when he referred to the "special case" when $mathcal E$ is a set all of whose members are also sets).
$endgroup$
– logic
Mar 16 at 23:40
1
1
$begingroup$
If none of the elements of E are sets, then B is the empty set.
$endgroup$
– user21793
Mar 16 at 23:30
$begingroup$
If none of the elements of E are sets, then B is the empty set.
$endgroup$
– user21793
Mar 16 at 23:30
1
1
$begingroup$
In ZF(C) everything is a set. However, in everyday math we often use other objects (e.g. numbers) that we don't regard as sets, and though in ZF(C) they are presented so, we don't want to take their elements..
$endgroup$
– Berci
Mar 16 at 23:33
$begingroup$
In ZF(C) everything is a set. However, in everyday math we often use other objects (e.g. numbers) that we don't regard as sets, and though in ZF(C) they are presented so, we don't want to take their elements..
$endgroup$
– Berci
Mar 16 at 23:33
$begingroup$
@Berci But it wasn't stated prior to the the unionset axiom that everything is a set. (I'm talking about "Notes on Set Theory" by Moschovakis. And if everything is a set, then I don't see why the author made the second comment I quoted (when he referred to the "special case" when $mathcal E$ is a set all of whose members are also sets).
$endgroup$
– logic
Mar 16 at 23:40
$begingroup$
@Berci But it wasn't stated prior to the the unionset axiom that everything is a set. (I'm talking about "Notes on Set Theory" by Moschovakis. And if everything is a set, then I don't see why the author made the second comment I quoted (when he referred to the "special case" when $mathcal E$ is a set all of whose members are also sets).
$endgroup$
– logic
Mar 16 at 23:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't have Moschovakis on hand, but if I recall correctly he's working in (essentially) a set theory with urelements (EDIT: also called atoms); this is a set theory where not everything is a set, but instead we have a collection of "urelements" which can be elements of sets but are not themselves sets.
- In some sense, in a model of set theory with urelements we build a set-theoretic universe on top of the collection of urelements. There is a lot of freedom here (e.g. should there be a "set of all urelements"?). The usual set theory ZFC (and ZF) does not allow urelements. However, we can whip up a "ZF(C) with urelements" without serious difficulty. Conversely, set theory with urelements isn't really very different (in most contexts, anyways) than set theory without urelements, so not much is lost by omitting them.
In a set theory with urelements, we can have a set some of whose elements are sets and others of which aren't. For example, suppose $a$ and $b$ are urelements. Then $$mathcalE=a, b, a,b, ,$$ is a perfectly valid set. The union construction can be applied to such an $mathcalE$, giving $$bigcupmathcalE=b, a,b, .$$ Note that the element $a$ of $mathcalE$ doesn't contribute anything to $bigcupmathcalE$: by definition $bigcupmathcalE$ is the set of all elements of elements of $mathcalE$, and $a$ - while an element of $mathcalE$ - has no elements of its own. More trivially, "$$" is an element of $mathcalE$ which doesn't contribute anything to the union.
- Note that there is no "typing" here: even if $a$ is an urelement, an expression like "$tin a$" makes perfect grammatical sense (it's just false).
Note that if $mathcalE$ is any set then $bigcupmathcalE=bigcupmathcalE'$ where $mathcalE'$ is the subset of $mathcalE$ gotten by removing all urelements (= all non-sets); so there's no real reason to consider taking the union of a set which isn't a family of sets. And in any reasonable set theory with urelements, being an urelement is a definable property (so we can form $mathcalE'$ from $mathcalE$ via the separation/subset axiom). There are two obvious ways to guarantee this:
We could have a predicate naming the urelements.
We could have a constant symbol $emptyset$ naming the emptyset; then an object $x$ of our universe is an urelement iff $xnot=emptysetwedgeforall y(ynotin x)$.
Note that these are ultimately equivalent.
answered Mar 16 at 23:59
Noah SchweberNoah Schweber
127k10151293
$endgroup$
$begingroup$
I guess "urelements" are the same thing as "atoms", in which case you indeed recall correctly that "he's working in a set theory with urelements" (he calls them atoms).
$endgroup$
– logic
Mar 17 at 0:27
$begingroup$
@logic Yup, that's right - edited.
$endgroup$
– Noah Schweber
Mar 17 at 2:17
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%2f3150959%2fabout-the-unionset-axiom%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$
I don't have Moschovakis on hand, but if I recall correctly he's working in (essentially) a set theory with urelements (EDIT: also called atoms); this is a set theory where not everything is a set, but instead we have a collection of "urelements" which can be elements of sets but are not themselves sets.
- In some sense, in a model of set theory with urelements we build a set-theoretic universe on top of the collection of urelements. There is a lot of freedom here (e.g. should there be a "set of all urelements"?). The usual set theory ZFC (and ZF) does not allow urelements. However, we can whip up a "ZF(C) with urelements" without serious difficulty. Conversely, set theory with urelements isn't really very different (in most contexts, anyways) than set theory without urelements, so not much is lost by omitting them.
In a set theory with urelements, we can have a set some of whose elements are sets and others of which aren't. For example, suppose $a$ and $b$ are urelements. Then $$mathcalE=a, b, a,b, ,$$ is a perfectly valid set. The union construction can be applied to such an $mathcalE$, giving $$bigcupmathcalE=b, a,b, .$$ Note that the element $a$ of $mathcalE$ doesn't contribute anything to $bigcupmathcalE$: by definition $bigcupmathcalE$ is the set of all elements of elements of $mathcalE$, and $a$ - while an element of $mathcalE$ - has no elements of its own. More trivially, "$$" is an element of $mathcalE$ which doesn't contribute anything to the union.
- Note that there is no "typing" here: even if $a$ is an urelement, an expression like "$tin a$" makes perfect grammatical sense (it's just false).
Note that if $mathcalE$ is any set then $bigcupmathcalE=bigcupmathcalE'$ where $mathcalE'$ is the subset of $mathcalE$ gotten by removing all urelements (= all non-sets); so there's no real reason to consider taking the union of a set which isn't a family of sets. And in any reasonable set theory with urelements, being an urelement is a definable property (so we can form $mathcalE'$ from $mathcalE$ via the separation/subset axiom). There are two obvious ways to guarantee this:
We could have a predicate naming the urelements.
We could have a constant symbol $emptyset$ naming the emptyset; then an object $x$ of our universe is an urelement iff $xnot=emptysetwedgeforall y(ynotin x)$.
Note that these are ultimately equivalent.
answered Mar 16 at 23:59
Noah SchweberNoah Schweber
127k10151293
$endgroup$
$begingroup$
I guess "urelements" are the same thing as "atoms", in which case you indeed recall correctly that "he's working in a set theory with urelements" (he calls them atoms).
$endgroup$
– logic
Mar 17 at 0:27
$begingroup$
@logic Yup, that's right - edited.
$endgroup$
– Noah Schweber
Mar 17 at 2:17
add a comment |
$begingroup$
I don't have Moschovakis on hand, but if I recall correctly he's working in (essentially) a set theory with urelements (EDIT: also called atoms); this is a set theory where not everything is a set, but instead we have a collection of "urelements" which can be elements of sets but are not themselves sets.
- In some sense, in a model of set theory with urelements we build a set-theoretic universe on top of the collection of urelements. There is a lot of freedom here (e.g. should there be a "set of all urelements"?). The usual set theory ZFC (and ZF) does not allow urelements. However, we can whip up a "ZF(C) with urelements" without serious difficulty. Conversely, set theory with urelements isn't really very different (in most contexts, anyways) than set theory without urelements, so not much is lost by omitting them.
In a set theory with urelements, we can have a set some of whose elements are sets and others of which aren't. For example, suppose $a$ and $b$ are urelements. Then $$mathcalE=a, b, a,b, ,$$ is a perfectly valid set. The union construction can be applied to such an $mathcalE$, giving $$bigcupmathcalE=b, a,b, .$$ Note that the element $a$ of $mathcalE$ doesn't contribute anything to $bigcupmathcalE$: by definition $bigcupmathcalE$ is the set of all elements of elements of $mathcalE$, and $a$ - while an element of $mathcalE$ - has no elements of its own. More trivially, "$$" is an element of $mathcalE$ which doesn't contribute anything to the union.
- Note that there is no "typing" here: even if $a$ is an urelement, an expression like "$tin a$" makes perfect grammatical sense (it's just false).
Note that if $mathcalE$ is any set then $bigcupmathcalE=bigcupmathcalE'$ where $mathcalE'$ is the subset of $mathcalE$ gotten by removing all urelements (= all non-sets); so there's no real reason to consider taking the union of a set which isn't a family of sets. And in any reasonable set theory with urelements, being an urelement is a definable property (so we can form $mathcalE'$ from $mathcalE$ via the separation/subset axiom). There are two obvious ways to guarantee this:
We could have a predicate naming the urelements.
We could have a constant symbol $emptyset$ naming the emptyset; then an object $x$ of our universe is an urelement iff $xnot=emptysetwedgeforall y(ynotin x)$.
Note that these are ultimately equivalent.
answered Mar 16 at 23:59
Noah SchweberNoah Schweber
127k10151293
$endgroup$
$begingroup$
I guess "urelements" are the same thing as "atoms", in which case you indeed recall correctly that "he's working in a set theory with urelements" (he calls them atoms).
$endgroup$
– logic
Mar 17 at 0:27
$begingroup$
@logic Yup, that's right - edited.
$endgroup$
– Noah Schweber
Mar 17 at 2:17
add a comment |
$begingroup$
I don't have Moschovakis on hand, but if I recall correctly he's working in (essentially) a set theory with urelements (EDIT: also called atoms); this is a set theory where not everything is a set, but instead we have a collection of "urelements" which can be elements of sets but are not themselves sets.
- In some sense, in a model of set theory with urelements we build a set-theoretic universe on top of the collection of urelements. There is a lot of freedom here (e.g. should there be a "set of all urelements"?). The usual set theory ZFC (and ZF) does not allow urelements. However, we can whip up a "ZF(C) with urelements" without serious difficulty. Conversely, set theory with urelements isn't really very different (in most contexts, anyways) than set theory without urelements, so not much is lost by omitting them.
In a set theory with urelements, we can have a set some of whose elements are sets and others of which aren't. For example, suppose $a$ and $b$ are urelements. Then $$mathcalE=a, b, a,b, ,$$ is a perfectly valid set. The union construction can be applied to such an $mathcalE$, giving $$bigcupmathcalE=b, a,b, .$$ Note that the element $a$ of $mathcalE$ doesn't contribute anything to $bigcupmathcalE$: by definition $bigcupmathcalE$ is the set of all elements of elements of $mathcalE$, and $a$ - while an element of $mathcalE$ - has no elements of its own. More trivially, "$$" is an element of $mathcalE$ which doesn't contribute anything to the union.
- Note that there is no "typing" here: even if $a$ is an urelement, an expression like "$tin a$" makes perfect grammatical sense (it's just false).
Note that if $mathcalE$ is any set then $bigcupmathcalE=bigcupmathcalE'$ where $mathcalE'$ is the subset of $mathcalE$ gotten by removing all urelements (= all non-sets); so there's no real reason to consider taking the union of a set which isn't a family of sets. And in any reasonable set theory with urelements, being an urelement is a definable property (so we can form $mathcalE'$ from $mathcalE$ via the separation/subset axiom). There are two obvious ways to guarantee this:
We could have a predicate naming the urelements.
We could have a constant symbol $emptyset$ naming the emptyset; then an object $x$ of our universe is an urelement iff $xnot=emptysetwedgeforall y(ynotin x)$.
Note that these are ultimately equivalent.
answered Mar 16 at 23:59
Noah SchweberNoah Schweber
127k10151293
$endgroup$
I don't have Moschovakis on hand, but if I recall correctly he's working in (essentially) a set theory with urelements (EDIT: also called atoms); this is a set theory where not everything is a set, but instead we have a collection of "urelements" which can be elements of sets but are not themselves sets.
- In some sense, in a model of set theory with urelements we build a set-theoretic universe on top of the collection of urelements. There is a lot of freedom here (e.g. should there be a "set of all urelements"?). The usual set theory ZFC (and ZF) does not allow urelements. However, we can whip up a "ZF(C) with urelements" without serious difficulty. Conversely, set theory with urelements isn't really very different (in most contexts, anyways) than set theory without urelements, so not much is lost by omitting them.
In a set theory with urelements, we can have a set some of whose elements are sets and others of which aren't. For example, suppose $a$ and $b$ are urelements. Then $$mathcalE=a, b, a,b, ,$$ is a perfectly valid set. The union construction can be applied to such an $mathcalE$, giving $$bigcupmathcalE=b, a,b, .$$ Note that the element $a$ of $mathcalE$ doesn't contribute anything to $bigcupmathcalE$: by definition $bigcupmathcalE$ is the set of all elements of elements of $mathcalE$, and $a$ - while an element of $mathcalE$ - has no elements of its own. More trivially, "$$" is an element of $mathcalE$ which doesn't contribute anything to the union.
- Note that there is no "typing" here: even if $a$ is an urelement, an expression like "$tin a$" makes perfect grammatical sense (it's just false).
Note that if $mathcalE$ is any set then $bigcupmathcalE=bigcupmathcalE'$ where $mathcalE'$ is the subset of $mathcalE$ gotten by removing all urelements (= all non-sets); so there's no real reason to consider taking the union of a set which isn't a family of sets. And in any reasonable set theory with urelements, being an urelement is a definable property (so we can form $mathcalE'$ from $mathcalE$ via the separation/subset axiom). There are two obvious ways to guarantee this:
We could have a predicate naming the urelements.
We could have a constant symbol $emptyset$ naming the emptyset; then an object $x$ of our universe is an urelement iff $xnot=emptysetwedgeforall y(ynotin x)$.
Note that these are ultimately equivalent.
answered Mar 16 at 23:59
Noah SchweberNoah Schweber
127k10151293
edited Mar 17 at 2:17
answered Mar 16 at 23:59
Noah SchweberNoah Schweber
127k10151293
answered Mar 16 at 23:59
Noah SchweberNoah Schweber
127k10151293
answered Mar 16 at 23:59
Noah SchweberNoah Schweber
127k10151293
127k10151293
$begingroup$
I guess "urelements" are the same thing as "atoms", in which case you indeed recall correctly that "he's working in a set theory with urelements" (he calls them atoms).
$endgroup$
– logic
Mar 17 at 0:27
$begingroup$
@logic Yup, that's right - edited.
$endgroup$
– Noah Schweber
Mar 17 at 2:17
add a comment |
$begingroup$
I guess "urelements" are the same thing as "atoms", in which case you indeed recall correctly that "he's working in a set theory with urelements" (he calls them atoms).
$endgroup$
– logic
Mar 17 at 0:27
$begingroup$
@logic Yup, that's right - edited.
$endgroup$
– Noah Schweber
Mar 17 at 2:17
$begingroup$
I guess "urelements" are the same thing as "atoms", in which case you indeed recall correctly that "he's working in a set theory with urelements" (he calls them atoms).
$endgroup$
– logic
Mar 17 at 0:27
$begingroup$
I guess "urelements" are the same thing as "atoms", in which case you indeed recall correctly that "he's working in a set theory with urelements" (he calls them atoms).
$endgroup$
– logic
Mar 17 at 0:27
$begingroup$
@logic Yup, that's right - edited.
$endgroup$
– Noah Schweber
Mar 17 at 2:17
$begingroup$
@logic Yup, that's right - edited.
$endgroup$
– Noah Schweber
Mar 17 at 2:17
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%2f3150959%2fabout-the-unionset-axiom%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
1
$begingroup$
If none of the elements of E are sets, then B is the empty set.
$endgroup$
– user21793
Mar 16 at 23:30
1
$begingroup$
In ZF(C) everything is a set. However, in everyday math we often use other objects (e.g. numbers) that we don't regard as sets, and though in ZF(C) they are presented so, we don't want to take their elements..
$endgroup$
– Berci
Mar 16 at 23:33
$begingroup$
@Berci But it wasn't stated prior to the the unionset axiom that everything is a set. (I'm talking about "Notes on Set Theory" by Moschovakis. And if everything is a set, then I don't see why the author made the second comment I quoted (when he referred to the "special case" when $mathcal E$ is a set all of whose members are also sets).
$endgroup$
– logic
Mar 16 at 23:40