New answers tagged

2

The idea of viewing a circle as a (regular?) infinite-sided polygon goes back at least as far as Nicholas of Cusa (sometimes referred to as Cusanus). The idea was picked up by Kepler who used it in area calculations, many years before the idea was developed formally in a mathematically adequate form. The torch was picked up by Leibniz who most likely ...


0

The integral really ought to read: $$\int T_{n}(x) \, dx = \frac{1}{2} \, \left( \frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1} \right)+C$$ and we can rewrite this as $$\int T_{n}(x) \, dx = \frac{1}{2} \, \left( \frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)-1}{n-1} \right)+C'$$ Note now that L'Hopital's Rule gives us: ...


0

Preliminaries Firstly there is no such thing as choosing a completely random point in the plane. Why so? If we assume that there is ever such a thing, then notice that for it to be completely random, every point in the plane must be equally likely to be chosen. How likely? It cannot be absolutely zero likelihood, otherwise no point can ever be chosen. ...


0

The C++ differs from C in that part as it doesn't define limits of what the compliation or execution environment is required to handle. Consequently there are valid C++ programs that cannot be compiled by any implementation. Say you for example in your main nests $2^{2^{64}}$ curly brackets, the standard allows for this but there is no computer available ...


0

The set of programs is countable: Set of all finite strings ( A subset such as set of all valid programs will be countable as well) The set of all valid programs is not finite: If it is simply add a blank remark at the end of the largest program in the set....so the set of all valid programs has no largest program hence infinite. Alternatively Since C++ is ...


6

Cantor had Jewish roots, which is probably why he was familiar with the Hebrew alphabet. But it's unlikely to be the reason de jure or de facto for the choice. From Georg Cantor: His Mathematics and Philosophy of the Infinite By Joseph Warren Dauben: Not wishing to invent a new symbol himself, he chose the aleph, the first letter of the Hebrew alphabet. ...


2

According to the book "Set theory, logic and their limitations" by Moshé Machover, aleph is the first letter of the word "einsoph", which is the Hebrew word for infinity, and is also used in cabbalistic traditions as a word for God. Given Cantor's interest in the connection between the infinite and the divine, this seems like the reason for his choice of ...


8

BrianO's answer is spot-on, but it seems to me you may not be too familiar with models and consistency proofs, so I'll try to provide a more complete explanation. If anything it may better steer you towards what you need to study, as admittedly I'm about to gloss over a lot of material. Why do we need the axiom of infinity? Because we know (and can prove) ...


1

The point is once you collect all the natural numbers into a set, you can now treat that set as an atomic object like any other and you can do all the things you can do with a set to it. So, for example, you can make a set that has the set of natural numbers as an element, you can construct the powerset of natural numbers, you can make functionals ...


7

The existence of each natural number follows from the other axioms of set theory, but if you drop the Axiom of Infinity (AxInfinity), the resulting theory ZFC-AxInfinity has a (transitive) model consisting of the hereditarily finite sets, which contains no infinite sets. The axioms of ZFC-AxInfinity provide no way to gather all the natural numbers into a ...


0

$\mathbb R[X]$ has the same cardinality as $\mathbb R$ itself. One fairly simple way to see this is to know that there are bijections $f: \mathbb R \to \mathcal P(\mathbb N)$ and $g: \mathbb N\times\mathbb N \to \mathbb N$. Then $$h(a_0+a_1X+\cdots a_n X^n) = \{g(p,q)\mid p\in f(a_q)\}$$ defines an injection $h:\mathbb R[X]\to \mathcal P(\mathbb N)$, and ...


1

I'm going to assume that by the limit definition of $e$, you mean $$ e=\lim_{n \rightarrow \infty} {\left(1+\frac{1}{n}\right)^n} \quad \text{where } n\in \mathbb{N} $$ In a way (but not all ways!), the $\infty$ can be thought of as the number of objects $n$ will pass by on the way to $e$. The number of elements in $\mathbb{N}$, called the cardinality of ...


2

Zero. The value of the integral doesn't depend on $x$ at all.


0

There are different kinds of infinity. When talking about limits you see if you consult the definition of $\lim_{x\to\infty}$ you'll see that there's actually no infinity involved. This is because the concept of infinity is a bit confusing and here's a occasion where you can easily avoid an actual infinity. This is a case of infinity that could be called ...


1

I would say that yes, $0$ is correct. These two lines only intersect in one point, and so, based on the usual meaning of "between" I would say either the area between the curves is infinite, or zero. Both seem pretty reasonable conclusions, and it just depends on what one considers inside vs outside. I think that since in most cases we are assuming that the ...


1

To disprove a statement all you need is one counterexample. As you observed you have for sure $\mbox{ span } \{ x_1,.., x_k\} \subset \mbox{ span } \{ y_1,.., y_l\} $. So this part cannot fail. This means that you must come up with a single example such that $\mbox{ span } \{ y_1,.., y_l\}$ is not inside $\mbox{ span } \{ x_1,.., x_k\}$. And as you said ...


0

With $t:=\frac1x$, The limit is the same as $$\lim_{t\to0^+}\frac{(1+t)^{1/3}-(1-t)^{1/3}}{t}.$$ By L'Hospital, $$\lim_{t\to0^+}\frac{(1+t)^{1/3}-(1-t)^{1/3}}{t}=\frac13\lim_{t\to0^+}\left((1+t)^{-2/3}+(1-t)^{-2/3}\right).$$ Alternatively, with the generalized binomial theorem, ...


1

You can use the identity $$a^3−b^3=(a−b)(a^2+ab+b^2)$$ with $a=(x^3+x^2)^{1/3}$ and $b=(x^3-x^2)^{1/3}$. So, you want to multiply (numerator and denominator) with $a^2+ab+b^2$. This will give you ...


0

$$ \lim_{x\to\infty} (x(1+1/x)^{1/3} - x(1-1/x)^{1/3})) $$ and then expand $(1\pm\epsilon)^{1/3}\approx 1\pm\epsilon/3...$ for $\epsilon\to 0\ .$


0

In an infinitesimal-enriched number system, the operator of taking limit as $x\to\infty$ is decomposed into two stages: (1) evaluating at an infinite value of $x$, say $x=H$; and (2) taking the standard part, or shadow. In the example you considered, $\frac{1}{H}$ is infinitesimal, and its shadow is zero. So the answer to your question is, it depends. ...


1

It is possible to do arithmetic with infinite natural numbers if one chooses the right theory. I am interested in a theory I call Modular Arithmetic (MA). MA has the same axioms as first order Peano arithmetic (PA) except $\forall x Sx \neq 0$ is replaced with $\exists x Sx=0$. MA has finite models: the rings $\mathbb{Z}/n\mathbb{Z}$. MA is a combination of ...


8

With your definition of "infinite set" (which is Dedekind's definition, not the usual one), no axioms beyond ZF are needed to prove that $\aleph_0$ is the smallest infinite cardinal. Let $A$ be an infinite set, and let $\phi:A\to A$ be an injection which is not a bijection. Choose an element $a\in A\setminus\phi(A).$ Then ...


6

The "most natural axiom" to add is perhaps the axiom of countable choice which asserts the existence of a choice function for every countable family of non-empty sets. I'd even argue that the true "natural axiom" would be the principle of dependent choice, which is a slight (but significant) strengthening of countable choice which posits that recursive ...


5

You can indeed adopt the axiom "every infinite set has a countably infinite subset", i.e., "every set that is not equipotent to a natural number (=finite ordinal) has a subset equipotent to the set $\omega = \mathbb{N}$ of natural numbers". This can also be reformulated as "every D-finite set is finite", where a D[edekind]-finite set is one that does not ...


2

This answer may not be what you're looking for, but the fact is that in this proof, one rarely actually gives a bijection itself. Instead, the countability of the rationals is proved much more easily using some simpler theorems. Namely, a countable union of countable sets is countable. Now define $A_n$ as the set of rationals in reduced form with ...


1

Consider this sequence and see if it helps: 1 maps to 1/1 2 maps to 2/1 3 maps to 1/2 4 maps to 1/3 5 maps to 3/1 (As the 2/2 is already included in the list we are building) 6 maps to 4/1 7 maps to 3/2 8 maps to 2/3 9 maps to 1/4 10 maps to 1/5 11 maps to 5/1 (As the others are already done on this diagonal.) 12 maps to 6/1 13 maps to 5/2 14 ...


0

You can't have $f(x)\to \infty$ for some $x$. $f(x)$ is a fixed value and not something that can "tend" anywhere. When you say $f(x)=\lim_{n\to\infty} f_n(x)$ you say (I hope, at least implicitly) that the limit $\lim_{n\to\infty} f_n(x)$ exists (for all $x$). However, if for some $xn \Bbb R$, the sequence with $n$th term $f_n(x)$ is divergent, then $f(x)$ ...


3

Let me make a few remarks about the constructive aspects. The standard definition is the following: a set $X$ is finite if there is a natural number $n$ and a bijection between $X$ and $\{ i \in \mathbb{N} : i < n \}$. Some of the expected properties are true: The disjoint union of two finite sets is finite. The product of two finite sets is finite. The ...


0

I looked over the existing answers and they don't seem to address the concern of the OP as I understood it, so I will try to provide a separate answer. The point is that the $(\epsilon,N)$-type definition of convergence is a first-order property and therefore by the transfer principle is still satisfied over the hyperreals. More specifically, a Cauchy ...


1

At the elementary level of the sum of a typical infinite series such as $\sum_{n=1}^\infty \frac{1}{n^2}$ one can illustrate the idea of infinities smaller than the superscript $\infty$ in the sum by using the hyperreal framework. Here a choice of a positive nonstandard hyperinteger $H$ gives a hyperfinite sum $\sum_{n=1}^H \frac{1}{n^2}$ which is ...


1

Since you are not a mathematician, I will try to explain in simple terms why mathematicians consider that these two "infinities" are equal. Imagine you have a classroom and you want to know whether you have more students than chairs (or the opposite). Then you ask the students to seat down and the result is clear: if some students are still standing, there ...


2

They are the same size infinity, specifically they both are the same size as the whole real line. To see that they are the same, notice we can map the interval $[2,3]$ onto the interval $[2,4]$ using the function $f(x)=2x-2$ this hits everything in $[2,4]$ since any $c$ in $[2,4]$ is hit by $(c+2)/2$which is in $[2,3]$ and only one thing hits $c$. (In this ...


7

Your argument requires further elaboration before it can be considered a "mathematical argument": In particular, you need to explain what a "infinitely large natural number" is. You then need to explain, in your new mathematical system, what the technical definition of "containing itself" means. In the standard sense, it means that there exists some ...


32

There's a few things I can think of which might fit the bill: We could work in a non-$\omega$ model of ZFC. In such a model, there are sets the model thinks are finite, but which are actually infinite; so there's a distinction between "internally infinite" and "externally infinite." (A similar thing goes on in non-standard analysis.) Although their ...


25

Well, there are a few notions of "infinite" sets that aren't equivalent in $\mathsf{ZF}.$ One sort is called Dedekind-infinite ("D-infinite", for short) which is a set with a countably infinite subset, or equivalently, a set which has a proper subset of the same cardinality. So, a set is D-finite if and only if the Pigeonhole Principle holds on that set. The ...


2

::choosing a nonstandard $H$ and calculating $f(H)$ and then taking standard part This is utterly meaningless. You cannot pretend to calculating $f(H)$ prior you `calculate' $H$ (whatever that means anyway).


2

Is arithmetic with infinite numbers fictitious? It depends on your definition of "arithmetic with infinite numbers" and "fictitious". The meaning of fictitious that this question was written to oppose, was in reference to certain descriptions of how Robinson's nonstandard analysis is used for calculus. Those descriptions don't have any obvious ...


1

" I do not understand how all distances $d(x_m,x_n)$ will be less than all positive real numbers." This is the key to your misunderstanding. All distances can't be. But for any real number no matter who small we can find an infinite number of distances smaller. (So we could say all real numbers are bigger than an infinite number of the distances...) ...


17

The subject line currently reads “How can a Cauchy Sequence converge to an irrational number?”. If we construe that literally, then one easy way a Cauchy sequence (lower-case initial "s") can converge to $\pi$ is that every term of the Cauchy sequence is $\pi$. Thus: $x_1=\pi, x_2=\pi, x_3=\pi,\ldots\,{}$. I suspect you meant “How can a Cauchy sequence of ...


7

It's not the case that "all distances $d(x_m, x_n)$ will be smaller than every real number". That is not even true "eventually", in the sense that for some $N$, it's true for all $m, n > N$. That would imply that all such distances are $0$. Furthermore, of course there is no "smallest [positive] real number". The point is that for any positive real ...


1

first of all it is not that all the distances are less than every positive number. given a positive number , one can find a stage after which any two terms are the given number close to each other. The idea of a sequence converging to some point do not necessarily imply that the distance between two terms of the sequence become zero. What it says is that, ...


4

If $n >m \ge 1$ you have $d(x_n,x_m) < 10^{n-1}$. Choose $\epsilon>0$ and $N$ such that $10^{N-1} < \epsilon$, then if $n,m \ge N$ we have $d(x_n,x_m) < \epsilon$. Note that the statement is for any $\epsilon>0$ there is some $N$ such that blah, blah, blah, and not that there is some $N$ such that for all $\epsilon>0$ we have blah, ...


3

I think your confusion comes from "for every positive real number $\epsilon$". In fact you have to fix $\epsilon$ first when finding $N$.


2

The exact content of the cited K - Shelah result is that there is a concrete formula $A(x)$ in the set theoretic language such that ZFC (not ZF!) proves that 1st there is unique $x$ satisfying $A(x)$ and 2nd every $x$ satisfying $A(x)$ is a ctbly saturated elementary extension of the reals. In brief, there is a definable ctbly saturated elementary extension ...


2

The series $1 - \frac12 + \frac13 - \frac14 + - \ldots$ is not absolutely convergent. It is a theorem that if the series converges, but not absolutely, we can rearrange the sequence of the terms to make the series converge to whatever value we like. The second series in the question is just a particular rearrangement of the series $1 - \frac12 + \frac13 - ...


3

They are conditionally convergent series. The definition is a series that converges but if you take the absolute value of all the terms the series does not converge. If you take the absolute value of all the terms in the first, you get the harmonic series which diverges like $\log N$. When summing a conditionally convergent series the result is not the ...



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