The usual definition of a finite number $n$ in the ZFC set theory is $n\in\mathbb{N}$, which is equivalent to &quot;$n$ is 0 or a successor ordinal, and so are all its elements&quot;. But this is not obviously equivalent to the intuition of a finite number, which would rather be a group of sticks that one can draw on a paper or a wall. And by the compactness theorem, we can consistently add a new constant symbol $c$ to ZFC, together with the axioms $c\in\mathbb{N}$ and $0 &lt; c, 1 &lt; c, 2 &lt; c, ...$ We call numbers such as $c$ non standard. I am particularly interested in proving the termination in finite time of certain algorithms or computer programs. And I am now concerned with the following type of reasoning : assume by contradiction that for all $n\in\mathbb{N}$, the program runs for more than $n$ steps, and derive a contradiction. Conclude that there exists an $n\in\mathbb{N}$ such as the program terminates in less than $n$ steps. What if this number $n$ was non standard ? It would prove nothing regarding the intended meaning of termination of a program. Is there a meta-theoretical property of ZFC, telling that every definable $n\in\mathbb{N}$ is standard ? Or telling that the proved existing $n$ above is standard ?
123639
This is a bit of a philosophical question. According to "Set Theory" by Jech, the set $\omega$ of natural numbers is defined as the least nonzero limit ordinal. After thinking about this definition recently I found no reason why this set should neccessarily be restricted to contain only the numbers obtained by applying the successor operation to zero, what is the intuitive notion of natural number.   I did not even find a way to formulate the statement that all elements of $\omega$ are of this form.   To be more technical, is it in fact possible to consistently add an element $x$ to the theory together with axioms that assert $x\neq0,x\neq 1,\ldots$ ?
41131
Sorry but I don't think I can know, since it's a definition. Please tell me. I don't think that $0=\emptyset\,$ since I distinguish between empty set and the value $0$. Do all sets, even the empty set, have infitinite emptiness e.g. do all sets including the empty set contain infinitely many empty sets?
71031
Below, the  of second-order logic is used.    My question is about a second-order analogue of $ZFC$ other than the usual "second-order $ZFC$." Rather than define the latter, I'll just state that its (set-sized) models are exactly the $V_\kappa$s with $\kappa$ strongly inaccessible (or if you prefer, the uncountable Grothendieck universes).  Specifically, I'm interested in the second-order theory consisting of the usual (first-order) axioms of Pairing, Extensionality, Union, Choice, Infinity, and Powerset, and the Separation and Replacement schemes ranging over second-order formulas. I'll call this "$ZFC_2^{scheme}$."    Specifically, for each formula $\varphi(x, y_1,...,y_k)$ of second-order logic with only the displayed variables, each of which is first-order, we have a corresponding Separation and Replacement instance. We do not allow second-order parameters here, so formulas with second-order free variables don't yield Separation/Replacement instances.   The passage from $ZFC$ to $ZFC_2^{scheme}$ has a nice uniformity: it's an example of a more general operation $\mathcal{ZFC}$ which spits out a $ZFC$-analogue given a logic at least as strong as first-order logic.    This is the reason for the parameter restriction above: I want $ZFC_2^{scheme}$ to look just like $\mathcal{ZFC}(\mathcal{L})$ for any other logic $\mathcal{L}$, and not take into account the particular feature that we now have a new kind of variable at our disposal - and in particular, take as "syntax-free" an approach to logics as we can get away with. (This is a choice of arguable value, but it's the one I'm making for now.)   It's easy to show that every model of $ZFC_2^{scheme}$ is well-founded, so to understand its models we only need to look at transitive sets. A natural hope at this point is that $ZFC_2^{scheme}$ is just second-order $ZFC$ in disguise, that is, that for a transitive set $M$ we have $M\models ZFC_2^{scheme}$ iff $M=V_\kappa$ for some strongly inaccessible $\kappa$. However, this isn't at all obvious to me (although it is easy to show the right-to-left direction).     Question. What are the (set-sized) models of $ZFC_2^{scheme}$?   (Note by contrast that the "arithmetic analogues" $PA_2^{scheme}$ and second-order $PA$ are equivalent since each pins down $\mathbb{N}$ up to isomorphism - although that leads to .)  (I'm happy to drop Choice if that would help.)  EDIT: As Hanul Jeon pointed out below,  shows that consistently $ZFC_2^{scheme}$ (there called "$ZFC_2^{def}$") has countable models, so we have a partial answer. However, I don't see at the moment a way to get an outright $ZFC$ proof that $ZFC_2^{scheme}$ is strictly weaker than second-order $ZFC$.
86488
For a finite set $S$ there are $\vert S\vert!$ orderings of its elements.  What is the cardinality of all orderings of $\mathbb{N}$?  What would $$\vert \mathbb{N}\vert!$$ mean?  Is it $\prod_{n\in\mathbb{N}} \aleph_0=\aleph_0^{\aleph_0}=\aleph_1$?   says the "equivalence classes of well-orderings of the natural numbers" has cardinality $\aleph_1$.  But I can't put together in my mind the connection between well-orderings of $\mathbb{N}$ and all orderings of $\mathbb{N}$.  (Are there non well-ordered orderings of $\mathbb{N}$?)  This leads to wondering what the cardinality of all orderings of $\mathbb{R}$ is.  (Actually my thought process went in the other direction, from $\mathbb{R}$ to $\mathbb{N}$.)  I think knowing the answer to the above question will teach how to determine what $$\vert \mathbb{R}\vert !$$ means, if anything.  I will be happy if you assume the continuum hypothesis to be true for the purposes of this question, but any level of completeness is welcome.
35224
I have seen that some authors are defing the number field as a subfield of $ \mathbb C$ which is a finite extension of the rational numbers $ \mathbb Q $, while some others without referering to complex numbers $ \mathbb C$ . I think we don't need $ K$ to be a subfield of $ \mathbb C$ in the definition. So, my question is the follwing:     Is it neceserily to define $K$ as a subfield of $ \mathbb C$ or not ?   And if no why ? Is it true that if we omit this in the definition, that then $K$ will turn out to be a subfield of $ \mathbb C$?   I came up with this question, when I saw that in order to define infinite primes in a number field then these are determined by the embeddings $ K \to \mathbb C $  Any idea would be really appreciated.  Thank you in advance.
76482
The Cantor set, as showed in books and , is defined in terms of $C_k$, the finite Cantor set of level $k$:  $$ \mathcal{C} = \bigcap_{k=1}^\infty C_k $$  But after intersection only the "last" remains, so, why not to define it by a limit?    $$\mathcal{C} = \lim_{k \to \infty} C_k$$  It is perhaps a naive intuition, but I not see a good justification.    (adding here a note after first answer, only for comment the comments)  NOTE: if it is not only a  question of choice of notation, but also about context and semantics. Can I tell an engineer that the intersection is a kind of specification, something like a project to explain "what I need", and the limit is a "what  I get", the end result?&nbsp; &nbsp; ...Or perhaps the inverse, as suggested by @HansLundmark (thanks the comment! also thnaks @SangchulLee!). I am supposing that $C_k$ is a "decreasing sequence",  $C_1 \supseteq C_2 \supseteq C_3 \supseteq \dotsb$, so using  we can say  &nbsp;"it's natural to define the limit as the intersection: $C_n \to \mathcal{C}$ as $n\to\infty$",  &nbsp;where $\mathcal{C}$ is defined by intersection.  About comment of @LordShark: its is possible to use  "limits" notation in the context of set sequences without "develop a theory" for it?  @HansLundmark's link is a satisfactory answer for it?
120369
Is it possible for there to be a model of ZFC with the property that, for every set $S$ in the model, there is a unary predicate in the language of ZFC such that $S$ is the is the only set satisfying the predicate?  I'm pretty sure I've been told the answer is "yes", but I am never able to find a reference when I want one. So this question (assuming it is answered in the affirmative) is to be that reference!
51180
From what I understood, cardinal numbers are defined as:  $\aleph_0$ = the cardinality of $\mathbb{N}$ $\aleph_{n+1}$ = is the least cardinal number greater than $\aleph_n$  The continuum hypothesis states that the cardinality of $\mathbb{R}$ is  $\aleph_1$. I more or less understand this cannot be proven (does not follow from ZFC).  However, can it be proven dat there in fact is a 'least cardinal number' greater than $\aleph_0$?  Is it possible that there are sets with cardinality $\aleph_{1/2}$, $\aleph_{1/3}$, et cetera, so to speak?  Or expressed differently: is it possible that for every set A with cardinality CA > $\aleph_0$, there exists another set B with cardinality CB where CA > CB > $\aleph_0$?
93844
Consider the formula F =∀x ∃y ∃z (¬(x=y) ∧ ¬P(x,y) ∧ ¬(x=z) ∧ P(x,z)). I have to find a universe $U$ which cardinality is given by n, while n &gt;= 3. Furthermore I should find a predicate $P: U^{2} \rightarrow \{0,1\} $ which satisfies the given formula F. The universe should be defined in terms of n, meaning that if I would tell you n = 5 everything should work. I tried several  things but don't understand how to define this universe in terms of n.
14558
We all know that $|\mathbb{N}| = \aleph_0$. Since $|\{-1\} \cup \mathbb{N}| = \aleph_0$ as well, I guess you could say that $\aleph_0 + 1 = \aleph_0$.  You can go on to derive that $\aleph_0 + \aleph_0 = \aleph_0$, and by induction $n \times \aleph_0 = \aleph_0$ (assuming that $n$ is finite, anyway).  Now, here's a thing: What is $\aleph_0 - \aleph_0$?  Well, $|\mathbb{P}| = \aleph_0$, and $|\mathbb{P} \backslash \mathbb{N}| = \aleph_0$, so perhaps $\aleph_0 - \aleph_0 = \aleph_0$?  No, wait. Consider the set $S = \{n \in \mathbb{N}: n &gt; 5\}$. Now we have $|S| = \aleph_0$, and yet $|\mathbb{N} \backslash S| = 5$. So maybe $\aleph_0 - \aleph_0 = 5$?  But that is absurd, since we can redefine $S$ to make the result any finite number we wish. Or we can define $S$ such that the result is countably infinite.  Does this mean that the notion of $\aleph_0 - \aleph_0$ simply has no definite answer? Or am I just being too simplistic here?
96383
In  (to ) it has been noted:     One more point to make about the paradoxical decomposition to more parts than elements [...]   currently we do not know of any model of +$\lnot$ where such decomposition does not exist.    Namely, as far as we know, in all models where choice fails there is some set which can be partitioned into more parts than elements.   (Just for clarification: From the context of the above references the comparison "more parts than" is obviously meant in terms of the  of the suitable  being strictly larger than the cardinality of the suitable initial set itself.)  Now, I find the negation or denial of such "paradoxial partitioning" interesting; i.e. the statement (proposition, "$\lnot$PP"):  "Each partition of each given set has cardinality less than, or at most equal to, the cardinality of the given set."  And I like to further explore how this suggested "proposition $\lnot$PP" (being considered along with ZF, of course) relates to "the standard set theory including Axiom of Choice", . Therefore  My questions:      Are there any models of ZFC known (or could there be any such models, in principle) in which there is some set which can be partitioned into more parts than elements ?  And:  Are there any models of ZF known (or could there be any such models, in principle) in which there is no set which can be partitioned into more parts than elements, and which is not also a model of ZFC ?  (And just for reference: Is there a conventional or concise way of expressing the suggested "proposition $\lnot$PP" in terms of standard notation, such as used in the sources linked above? Has it perhaps been discussed already, by some other name? ...)
109777
This question came from  and from  where it's noted that "integers have a finite count of digits", so that the "number" in the title is not at all a number (not integer nor rational or real) . This statement seems to me justified because I suspect that if we admit an infinite count of digits we build a set of "numbers" that is not countable (but I've not a proof). I'm wrong?    Reading comments and answers I'm a bit confused. So I add a more specific question: Is the string in the title a number in some model of Peano Axioms?
46782
I am wondering what the difference between a definition and an axiom.  Isn't an axiom something what we define to be true?  For example, one of the axioms of Peano Arithmetic states that $\forall n:0\neq S(n)$, or in English, that zero isn't the successor of any natural number. Why can't we define 0 to be the natural number such that it isn't the successor of any natural number?  In general, why is an axiom not a definition?   Are there cases where we can formulate definitions as axioms or vice versa?
92059
$\Bbb Z$ - integers $\Bbb N$ - natural numbers (starting from 1) $\Bbb R$ - real numbers    I believe the answer is the set of real numbers ($\Bbb R$), seeing as $b$ will not equal $0$ as the set of natural numbers start from $1$. Thoughts?
106244
Let $\Omega=\mathbb{N}$. For each $E\subset\Omega$ let $N_n(E)$ be the cardinality of the set $E\cap \{1,2,\ldots,n\}$. Define $C=\left\{E: \lim_{n\rightarrow \infty} \frac{N_n(E)}{n} \text{ exists}\right\}$. Show that $C$ isn't a field.  I already know that it isn't closed to finite union of non-disjoint set but I can't see why. I saw a post that somebody said &quot;the set of natural numbers whose first digit is 1 doesn't have this limit&quot;. I'm in trouble to see why this limit wouldn't exists in finite unions. Thanks in advance.
81480
Infinity, is clearly not . But there is some dissent on whether or not zero is finite. I have seen authors use "finite" to indicate the value of $0$ is excluded as well as infinity.  Is there any rule about this?
117063
The usual way to define the "size" of an infinite set is through cardinality, so that e.g. the sets $\{1, 2, 3, 4, \ldots\}$ and $\{0, 1, 2, 3, 4, \ldots\}$ have the same cardinality. However, is this the only way to define a useful "size" of an infinite set? Could one conceivably define a "size" where the two example sets have different sizes?