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3

Well, for one thing, there are different kinds of infinity, which lets us distinguish between the sizes of infinite sets. One cool thing is that this plays a central role in the proof of the Banach-Tarski theorem. Geometrically, $\Bbb R^n\cup\{\infty\}$ is homeomorphic to $S^n$, or the $n$-dimensional hypersphere. This also allows for geometry on the ...

2

Your professor was talking about the distinction between countable & uncountable sets, established by Cantor's diagonal argument. See also Aleph number, Beth number, equinumerosity, & Cantor's contributions to set theory.

0

Well we need some technique here and its a fairly standard technique. We need to understand that if $x \in (0, \pi/2)$ then $$\dfrac{\sin x}{x} > \cos x = 1 - 2\sin^{2}(x/2) > 1 - 2(x/2)^{2} = 1 - \dfrac{x^{2}}{2}$$ Putting $x = 1/n$ we get $$n\sin(1/n) > 1 - \frac{1}{2n^{2}}$$ and hence $$(n + 2)^{2}\sin(1/n) > n^{2}\sin(1/n) > n\left(1 - ... 2 Hint:$$ {(n+2)^2}\sin\left( \frac{1}{n}\right) > M \implies {(n+2)^2}\left( \frac{1}{n}\right)>M,$$since$$ \sin\left( \frac{1}{n}\right) \leq \frac{1}{n}$$1 Let us write it as$$\lim_{x\rightarrow 0}(\frac{1}{x}+2)^{2}\sin[x]$$which is the continuous version of the problem with x=\frac{1}{n}. We expand \sin[x] by$$x-\frac{x^{3}}{6}+\frac{x^{5}}{120}\cdots$$The left hand side is 2+\frac{1}{x^{2}}+2\frac{1}{x}. We know 2\sin[x] and 2\frac{\sin[x]}{x} is bounded. But from the above expression it is ... 3$$ \lim_{n\rightarrow \infty}(n+2)^2\sin\ \frac{1}{n} = \lim_{n\rightarrow \infty}(n+2)\frac{\sin\ \frac{1}{n}}{\frac{1}{n+2}} = \lim_{n\rightarrow \infty}(n+2) =\infty$$since$$ \lim_{n\rightarrow \infty} \frac{\sin\ \frac{1}{n}}{ \frac{1}{n+2}} =1$$11 The set you are building is very specific. It indeed has the same size as the reals. There are natural ways in which we can identify it with certain sets of reals. These sets are "simple", in a technical sense. It is an early theorem in descriptive set theory that all these simple sets either are countable or have the same size as the reals. (In technical ... 6 I'm not sure I understand your notation, but I think your argument boils down to the following: \mathbb{Z} is countable. \mathbb{Z}^2 is countable. \mathbb{Z}^3 is countable... ... \mathbb{Z}^{\mathbb{N}} (the set of all countably infinite sequences of integers) has the same cardinality as \mathbb{R}. Therefore, every subset of \mathbb{R} is ... 2 Let's say that you want to divide 1 pie among an infinity of your friends. Each friend will get 1/Infinity of the pie or, as you say 0. After you give out the pie, you rethink matters and say to your friends to give the pie back to you. Each friend will give you his portion of the pie that is a 0. So you end up with 0 pies. Where has the pie gone? 1 The answer is simple. Mathematics works with precise definitions. To say that two sets have the same cardinality is to say that these two sets satisfy that certain definition. To say that they don't have the same cardinality means that they don't satisfy the definition. That is all. We define two sets A and B to have the same cardinality when there is ... 1 The reason we know that there are more real numbers than there are natural numbers, in terms of cardinality, is the following: Assume we have a complete, enumerated list of all the real numbers (which would allow a bijection between \Bbb N and \Bbb R). Let's get a sample of that list:$$\begin{align}&.123789714361\ldots\tag{1}\\ ...

2

The standard result here is that if $S$ is an arbitrary set, then the cardinality of $S$ is strictly smaller then the cardinality of the power set $\mathcal P(S)$, i.e., there is no surjection $\varphi: S \twoheadrightarrow \mathcal P(S)$. The proof is Cantor's classic diagonal argument: suppose $\varphi$ were any such map, and consider the set $B = \{s \in ... 3 For the particular results that the cardinality of the real numbers is greater than the cardinality of the integers, there are many proofs. Perhaps the most famous one is Cantor's diagonal argument for that fact, which actually exploits a typographical fact about the representation of real numbers as strings of digits. There are other proofs that involve ... 1 The classic proof that the cardinality of the set of real numbers is greater than that of the set of integers is Cantor's diagonalization argument. In general, two infinite sets are considered the same "size" if there exists a one-to-one correspondence between them. There does not exist a one-to-one correspondence between the set of integers and the set of ... 3 The function$g(u) =(f(u))^2 +2\cdot \cos u $is increasing and bounded hence there exists$\lim_{u\to \infty } g(u) $hence$\lim_{u\to \infty } f(u) $cannot exist. 1 A given term$(4t)^i/i!$occurs for$k=1,\dots, i/2,$so you have the sum break up into two pieces (for odd and even$i$) The even part looks like$\sum_{i=1}^\infty (4t)^i/(2(i-1)!),$the odd part is very slightly more complicated, but both are easy to evaluate. 3 Regarding the second question, "Is there any first order theory which is consistent by finitary proofs but inconsistent with infinitary proofs?" The answer is yes, when particular additional infinitary inference rules are used. Lord_Farin's answer shows that infinitary inference rules are necessary to make this happen. It is well known that the set$X$of ... 5 Yes, of course. We might need to "go one level higher" so to speak, and base our discussion in some given model of set theory (instead of on the metalanguage, where the behaviour of infinite objects is more shaky), but there are no real problems. Barring the introduction of new, infinitary proof rules, the answer is no. For, we have: Let$\Sigma\vdash ...

0

This is a question for surreal numbers. Surreal numbers are a really amazing thing invented by John Conway that include numbers like 0 and 3/4, but also things like "twice the square root of infinity, all plus an infinitesimal". This question depends on the values of the infinite and infinitesimal, but the way it works is this. The number ω is defined as the ...

4

These are not isomorphic: the second has bridges, and the first has none. --o----o----o----o-- | | | | --o----o----o----o-- o o o /|\ /|\ /|\ / | \ / | \ / | \ --o | o----o | o----o | o-- \ | / \ | / ...

4

Yes and no. Think about the implications. Is $0\cdot\infty=1$ ? Because normally, if $\displaystyle\frac ab=c$ , then $a=bc$ But this isn't really the case here, is it ? Because, since all limits of the form $\displaystyle{\lim_{n\to\infty}\frac kn}$ are $0$, for all finite numbers k, then the product $0\cdot\infty$ becomes meaningless. Sometimes it can even ...

15

That notation is used as a shorthand for "As $x$ approaches infinity, the denominator blows up without bound, and hence since the numerator is constant, the value of the function approaches zero (i.e. gets arbitrarily close to zero), and hence its limit is zero." $\dfrac 1\infty$ does NOT literally mean "divide $1$ by $\infty$! So literally, it is ...

1

One of the most frequently considered infinite-dimensional spaces is $\ell^2(\mathbb R)$, which is the set of all infinite sequences $\mathbf x = (x_1,x_2,x_3,\ldots)$ of real numbers for which $x_1^2+x_2^2+x_3^2+\cdots<\infty$. One sort of "basis" used for that space is $\{(1,0,0,0,\ldots), (0,1,0,0,\ldots), (0,0,1,0,\ldots),\ldots)$. This is an ...

0

There is one issue that has not been raised in the fine answers given earlier. The issue is implicit in the OP's phrasing and it is worth making it explicit. Namely, the OP is assuming that, just as $0.9$ or $0.99$ or $0.999$ denote terminating decimals with a finite number of 9s, so also $0.999\ldots$ denotes a terminating decimal with an infinite number ...

2

Question 1. Is it necessary to talk about uncountability of R after a session on Hilbert's hotel paradox? No, not on the first day—an introduction to the concept that additional rooms can be made available by just re-arranging people in a hotel with no vacancy is enough to chew on for one day. I might even split this up into two days. On day ...

3

Real numbers are best avoided until properly treated, which requires a fair dose of analysis. Uncountability can be demonstrated on other sets. I had, on several occasions already, explained Hilbert's Hotel to kids of various ages. After the hotel seems never to fill up, I bring in an infinite amount of families to the hotel. Each family has a child. Each ...

2

I see no reason you need to get into uncountable sets if you just want to play around with weird things that can happen with infinite sets. As for your second question, I remember seeing the argument for uncountability of $\mathbb R$ in elementary school and it stuck pretty well then, and I don't consider myself to have been especially precocious, so I see ...

4

It is definitely not necessary to talk about the uncountability of $\Bbb R$, at any point. Or even mention the real numbers. Most people who didn't fiddle with mathematics for a bit cannot even grasp what it means to be a real number and I often find that when trying to start and explain someone what is a real number, they make some effort to understand, but ...

0

I believe it should be pointed out that there is a very significant and important similarity between $0.\dot 9$ and $\frac1\infty$. It may seem easy to accept that $0.\dot9 = 1$ because calculations like $0.\dot9-0.0\dot9$ seem obvious, but you are none the less manipulating two expressions, each of which represents a number with a countably infinite number ...

0

The limit is undefined. Since (-1)^(2n+1)=-1 & (-1)^(2n+1)=1, raising to n where approaches infinity means that you can't define whether the result is +1 or -1.

2

Let $A_n = n$ and $B_n = 1/n$. Then the sequence $C_n = A_nB_n$ is constant, so certainly has a limit. But $A_n$ diverges to infinity and $B_n$ has a limit. However, if $A_n$ diverges to infinity and $B_n$ has a nonzero limit, then certainly their product will diverge. To prove this, consider that if $B_n$ has a nonzero limit $B$, then beyond some $n$ all ...

0

The problem people tend to have with this concept is that we sometimes assume our number system is the exact representation of the numbers themselves. We are, in fact, dealing with a system similar to $\mathbb Z_{10}$, where we roll over to 0 again in a particular position after reaching 9. There isn't a representation for 10 in this system, but instead a ...

2

It's important to note that $0.\overline{9}$ doesn't have a limit: it's just a single number. However, we like to define decimals so that they can be used to represent real numbers. And, in particular, we would like $0.\overline{9}$ to be equal to the limit of the sequence $$0.9, 0.99, 0.999, 0.9999, \cdots$$ It is this sequence that "tries" to reach 1, ...

5

Even though your proof is perfectly valid, here is another one; maybe this convinces you more. The sum of the terms of an infinite geometric series $a_n$ with first term $a$ and ratio $|r|<1$ is $$S_{\infty}=\frac{a}{1-r}$$ Hence, $$0.999\cdots=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\cdots=\frac{\frac{9}{10}}{1-\frac{1}{10}}=1$$

0

An infinitely small part of infinity can indeed be a real number. For example, if $H$ is an infinite hypernatural, then $\epsilon=\frac{1}{H}$ is infinitesimal, and then an epsilonth part of $H$ will be exactly... 1. By the way, this does have real-life applications. The infinitesimal approach is a more accessible way of teaching calculus (see Elementary ...

6

Often traditional reasoning from arithmetic breaks down when you try to think about infinity, and percentages are no exception. Examples: You agree to pay me $B(n)/n$ dollars per year, where $B(n)$ is a function giving your total wealth after $n$ years (thanks!). So as time goes by, the portion of your income you're paying me is $1/n$ (or $100/n$ percent). ...

0

It all depends on what you'd like to mean with "order of infinity". If you mean "cardinality", then $\mathbb R$ and $\mathbb R^n$ are indistinguishable. That is, there is a bijection from $\mathbb R$ to $\mathbb R^n$. Of course, this answer could be somewhat unsatisfactory. Heuristically, an element of $\mathbb R$ has one "degree of freedom", whereas an ...

-2

C∪{∞} has infinity in its definition. Infinity is undefined therefore complex infinity is undefined. If infinity were definable then it would be finite.

0

Another sense in which $10$ is closer to $\infty$ than $1$ is to $\infty$ is that $\frac 1{10} \lt \frac 11$. This comes into play if you do Taylor series "around infinity" by making use of the variable $\frac 1x$. The powers of $\frac 1x$ get small more quickly at $x=10$ than at $x=1$.

3

Okay let's try to make rigorous sense of this question: First you need to define what you mean by a 10-adic number (10 is not prime!). The most sensible thing to consider would be, in this context, $R = \varprojlim \mathbb{Z}/10^n\mathbb{Z}$ with the obvious maps. $R$ is profinite and has a natural topology; it is even a topological ring. An open ...

0

Geometrically speaking...(sketch) I would say no.. since if we need to speak about a distance to $\infty$ then it must be a point in our space..from the context I assume we want to work with $\mathbb{R}$ and infinity. Naturally, lets take $\hat{\mathbb{R}}$ of the one point compactification of the reals. This is the same as the circle $S^1$. Fixing one ...

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