We can thus consider the set of all sets and subsets of an infinite set as having a cardinality equal to the cardinality of the base set (in this case, the Natural numbers set N). But, what about that 11 correspondence between the set of all Real numbers to the set of all Natural numbers? If we can divide the set of all Natural numbers up into an infinite number of subsets whos cadinality is equal to the parent set then how can we fail to find a corresponding element in the set of Natural numbers to the set of Real numbers? I have shown that I can divide the Natural numbers into an infinite set of infinite sets. Further more, each subset of infinite cardinality can contain completely unique elements from any other division of the parent set. So, before I have even made a dent in exhausting the elements of the Natural numbers, I can divide off a subset that could itself be divided off into a collection of infinite sets just to cover the elements between 0 and 1 in the Real number set. If there aren't enough, I can divide that sub sub sub set again and still have an infinily divisible set to cover the correspondance with. I would never lack for a unique element from the Natural numbers set to match up with an element to the Real numbers set.
I know that the numbers in the Natural number set would become absolutly enormous by the time we covered the interval from 0 to 1 in the Reals but, so what, I still have plenty to move to the next interval. In fact I still have the same number of elements in the Natural numbers that I had before I used up an infinitely small fraction of them to cover all previous intervals. I am not going to run out of Natural numbers by placing them in a 1 to 1 correspondence with any other set!
]]>Proof:
Let $H=[a_{H},b_{H}]$ and $K=[a_{K},b_{K}]$ where $H$ and $K$ are mutually exclusive.
Since $H$ and $K$ are both closed point sets they each have a left most point and a right most point (by the definition of a closed point set).
If there is a point $p \in H$ between any two points $m$ and $n$ of $K$, then by axiom 1.6, there is a point $q$ between both ($m$ and $q$) and ($q$ and $n$). Thus $H \cup K$ is not a closed interval.
If $H<K$ then by axiom 1.6 there exists some point $p_x$between the right most point of $H$
and the left most point of $K$ where $p_x \notin H \cup K$ and again, no colosed interval exists.
By symetrical arguement for the case where $K<H$ there is a point between the the left most point and right most point of $K \cup H$ that is not a member.
$\therefore$ the union of two mutually exclusive closed point sets in not a closed interval.
█
Proof:
Let $f$ be a function with domain $D$.
Let $S = [a_{S},b_{S}] \in D$.
Let there be a point $p \in S$.
Let there be a point set $Q = \{mm = (x,f(x)) = (x,f(p))\}$ such that $Q \in S$.
The set $Q$ is bounded by $S$.
If $Q$ is a closed point set then $Q$ has a left most point and a right most point.
Since $Q$ is bounded by $S$, then the left most point of $Q$ can not be less then the left most point of $S$.
Thus the left most point of $Q$, $a_{Q} \geq a_{S}$.
By symetrical arguement the same can be said for the right most point of $Q$.
This shows that if $a_{Q} = a_{S}$ and $b_{Q} = b_{S}$ then $Q$ is closed.
For the case of $a_{Q} > a_{S}$ or $b_{Q} < b_{S}$, then since $f$ is continuous there are no
points $f(x) < a_{Q}$ where $f(x) \in Q$ thus $a_{Q}$ is the left most point of $Q$.
By symetrical arguement, $b_{Q}$ has no points greater within $Q$.
$\therefore Q$ is a closed point set.
█
Proof:
Let $S=\{xx \in \mathbb{R}, 0<x<1\}$.
Since $S$ is made up of all $x \in \mathbb{R}$ between 0 and 1, $S$is an infinite set, meaning $S$ is not finite (by the definition of infinite).
$S$has a first point to the right equal to $1$where $1 \notin S$. (This is by the completeness axiom and the definition of an open set).
Let there be a point $p \in S$of the form $p = {110^{n} \over 10^{n}}$where $n \in \mathbb{N}$. (note: It is considered evident that for any number $n$, $p$ then can be written in the form $0.999...9_{n}$ where the decimal expansion is equal in length to $n$).
Let there then be a point $q \in S$ where $q = {110^{n+1} \over 10^{n+1}}$.
It is then demonstrable that $p<q$where $p \in S$ and $q \in S$.
As $S \in \mathbb{R}$ and is an infinite set, by the axiom that between any two points there is always another point in an infinite set, then for any value of $n, p<q<1$.
As $n \to \infty$, the length of $p$'s decimal expantion also $\to \infty$.
Since $q > p$, $q = 0.999...$ as $n \to \infty$.
Since $1$ is the first point to the right of $S$ and since $1 \notin S$ and $q \in S$, $q \neq 1$.
$\therefore 0.999... \neq 1$.
Q.E.D.
You need to pick one of our four definitions for continuity and probably stick with it throughout the proof. The proof I'm thinking of uses the definition involving open intervals. Here are some hints to get you started:
Let $S=(a,b)$ be an open interval containing $h(p)$. Since $f(p)=h(p)=g(p)$, $f(p)$ and $g(p)$ are also contained in $S$. We are assuming that $f$ and $g$ are continuous. What does our definition of continuity involving open sets tell us in this case? You should say something about two open intervals on the $x$axis. Now, using the two intervals that you have associated to $f$ and $g$, you need to construct an open interval $T_h$ on the $x$axis such that if $t\in T_h$, then $h(t)\in S$. At some point, you'll need to encounter a statement like: $a<f(t)\leq h(t)\leq g(t)<b$.
]]>http://jeromyanglim.blogspot.com/2010/10/gettingstartedwithwriting.html
]]>Proof:
Let set $S = \{x  x \in \mathbb{R}$ and $0 < x < 1\}$.
Let $S$ converge to $1$ by definition of convergence to a point.
By construction and the definition of an open point set, $1 \notin S$.
By the definition of a limit point, $1$ is a limit point of $S$.
Let there be a point $p \in S$ such that there is also a point $b$ where $p < b < 1$.
(an example of such a set of points is where $p = 0.8$ and $b=0.9$, $0.8 < 0.9 < 1$).
By definition of a limit point, $b \in S$ and $b$ is between $p$ and $1$.
By the axiom, for any two points $p$ and $q$ there is a point between them.
By definition of first point to the right, $1$ is the first point to the right of $S$ and greater then every point in $S$.
By the definition of infinite, $S$ is infinite meaning not finite and there are always $n + 1$ points in $S$ where $n \in \mathbb{N}$ showing that $S$ does not meet the definition of finite.
Let $M$ be an open interval $(a, b)$ containing $1$ and $p$ by the definition of a limit point.
For all values of $p < 1$, $b \in M$.
For any value of $p < 0.999...$, $b > p$.
For some $p \leq 0.999...$, $b > p \in S$.
By the definition of infinite there is a value of $b = 0.999...$ for some $M$ where $b \in S$.
Since $b \in S$ and $1 \notin S$, then $b \neq 1$.
$\therefore 0.999... \neq 1$.
█
If $p_1, p_2, p_3, \ldots$ is a nondecreasing sequence and there is a point, $x$, to the right of each point of the sequence, then the sequence converges to some point.
Jeff claimed (please correct me if I am wrong) that the word every should have been used (as it is in the Completeness Axiom) instead of each. I claimed that it didn't matter which word was used and that the meaning was the same. I've just thought about this a little more and figured that I might as well share my thoughts.
The quantifier on $x$ indicates that we are talking about a single $x$. This single $x$ is to right of each point in the sequence. That is, if you pick a random $p_i$ in the sequence, $p_i<x$. This implies that $x$ is to the right of every point in the sequence. The $x$ is a "one size fits all" point. So, I maintain it doesn't matter whether we use each or every.
Contrast the statement of the theorem with the following statement:
For each point in the sequence, there is a point $x$ that is to the right.
In this case, the dependence of the quantifiers is reversed and we no longer have a "one size fits all" $x$. Each individual point $p_i$ in the sequence has a point $x_i$ to the right, where I am using a subscript on $x$ to emphasize that this point depends on $p_i$. The point to the right of $p_i$ may be different than the point to the right of $p_j$ (assuming $i\neq j$). This is a very different situation than that in Theorem 1.60.
]]>All this can be said without using "infinity," potential or actual, or the pseudoconcept of "the number of natural numbers." One simply identifies a systematic way of representing rational numbers, and the rest follows.
I think Will's difficulty with infinities is (as I said elsewhere) ontological, not mathematical. The Zeno argument that best highlights the difficulty is not Achilles v Tortoise, but the much balder "stade" argument (from the Greek adverb for "standing" or "motionless", with a suggestion of punning on "stadion", the noun for "measure"). Take an arrow shot from a bow. It appears to move smoothly through the air. But at each instant of time, it has an exact (even, theoretically, calculable) position, and an exact measure from end to end. At each instant, then, the arrow simply IS as it is, in a fixed place at a fixed time. But time is simply a sequence of instants. If in no instant does the arrow move, then how can it move over any sequence of instants?
Unlike the Achilles, this does not involve us in summing smaller and smaller fractions of an interval, a mathematical task to which the concept of "limit" seems wholely adequate. Here we are summing genuine zeroes, which always sum to naught.
Where, then, is the motion? Zeno, of course, was arguing precisely that in reality there is none, and the fact that it appears to us to move shows that our perceptual world is not the real world. But there are other ways to approach the paradox. One (not so easy to defend) is to argue that there cannot be an instant of time. But however you pick it up, this paradox is about how things are in the world, not about mathematical entities.
]]>What we can talk about are numbers that grow without bounds as long as they grow on the right side of the decimal point. $0.999999999...$ Is not the same number as $0.111111111...$ . This is pretty easy to see, right? Here we are dealing with infinitesimals. As these numbers grow it is not so much the number getting bigger that is interesting, and they are getting bigger though not growing towards Infinity but the difference between the number $0.999...99$ and $0.999...999$ gets smaller without bounds as there is another $0.000...0009$ to add. That is, as the length of the number on the right side of the decimal point increases without bounds the added component that is represented by the next digit to the right is smaller and smaller with no limit to how small. Here is the odd part though, as a number grows without bounds it is indistinguishable from another number that grows without bounds if there is a sequence of numbers that starts at the decimal point (in the left most position) and grows without bounds that exactly matches the left most unbounded sequence in another number.
That might not have been very well said so I will try to explain what I mean with an example. If I were to take a sequence of numbers like $0.999888$ and compare it to another number that was close but not the same like $0.999777$, I could easily show they were different simply by subtracting one from the other and show the result is nonzero. Now if I, instead, take a function that creates numbers for some value $n$ where the size of that sequence can grow without bounds say as $n$ starts at $1$ then $n=2$ then $3$ and so on forever, I might get a series of numbers like $f(1) = 0.98, f(2) = 0.9988, f(3) = 0.999888, … f(n) = 0.9999...8888...$ where the length of the number grows without bounds. Then I could compare it to a function that generated a series like $0.91, 0.9911, 0.999111, … 0.9999...1111...$ . I can see that they aren’t the same number and the functions are not equal but once I let $n$ grow without bounds, even though we can’t really talk about numbers that grow without bounds to the left of the decimal point, I can also show that as $n =>$ Infinity these two function become more and more equal (not more equal, they are equal or not equal, but less and less different) and the only thing we can talk about is the first sequence of numbers that grow without bounds, the $9$s, in this case. It would not matter if the two functions generated sequences that were even more different either, like $f(n) = 0.998, 0.999988, 0.999999888, ...$ compared to $f(n) = 0.33, 0.93333, 0.99333333, 0.9933333333, …$ (here the leading $9$s only increase when $n=$a prime number but the $3$s appear in pairs with each Whole number increase of $n$). As long as some sequence no matter how small the proportion at the beginning of the number compared to the numbers at the end became a sequence that grew without bounds and was the same as another number‘s leading boundless sequence. All values after the boundless sequence become infinitesimal in value and disappear.
Below is a table of elementary operations that are given as undefined for Infinity. More about operations with Infinity can be found at these Web addresses: http://en.wikipedia.org/wiki/User:Caue.cm.rego/infinity , http://www.mathsisfun.com/numbers/infinity.html , http://www.mathworks.com/help/toolbox/mupad/stdlib/undefined.html .
Undefined Operations on Infinity 

0 × ∞ 
0 × ∞ 
∞ + ∞ 
∞  ∞ 
∞ / ∞ 
∞^{0} 
1^{∞} 
What is odd about this table is the fact that while we can not subtract Infinity from Infinity. Apparently it is ok, though, to add or subtract one boundless number from another: $a= 0.999... , b =  0.999..., a + b = 0$. Is this really ok? What if we derived $a = 0.999...$ In one way (by some function $A())$ and $b = 0.999...$ By a completely different method (some function $B()$)? Do we really know they are equal but with opposite signs? Infinity has no value, it is just the property of being unbounded. If we knew by the nature of the two functions that somewhere, at any point we want to give $n$ a finite value and check the results, the ending sequence of digits would be different between those two functions. Subtracting one from the other we would get a nonzero answer. Such a set of functions could look like the ones we were looking at above: $0.999888  0.999111 = 0.000777$. On the other hand, what if we knew they were boundless. What if you tried to do the same thing in that case: $0.999...888...  0.999...111...= 0.000...777...$ . This number is zero, right? The string of numbers to the immediate right of the decimal point is a boundlessly increasing number of $0$s Those $0.000...777...$s are lost. We can’t get them back because we can’t truly know their value. Are they infinitely small or not even there or is there no difference? The above table says we can’t even try to know because the operation of subtracting Infinity from Infinity is undefined. I can easily see the wisdom of that choice but again, what about subtracting $0.999...$ From $0.999...$ To get $0$? How do we know it is zero if Infinity has no value? We can not truely say there are the same number of $9$s in one sequence or the other, all we can say is that they are both unbounded in their sequence.
]]>0 
1 
0  0 
0  1 
1  0 
1  1 
0  0  0 
0  0  1 
0  1  0 
0  1  1 
1  0  0 
1  0  1 
1  1  0 
1  1  1 
Here are some charts that shows all permutations of the binary number system limited to one, two and three digits in length. Thus the number of possible combinations are $2^1, 2^2$ and $2^3$ respectively. This demonstrates that the number of permutations for a binary number of $n$ length is $2^n$. It is easy to see here that the number of possible combinations of $1$s and $0$s is much greater then the number of digits in the number; 1 digit numbers have 2 possible combinations, 2 digit numbers have 4 possible combinations, 3 digit numbers have 8 possible combinations, etc., etc., etc. In fact, as the number of digits $n$ grows bigger the possible combinations grow exponentially by $n$. Eventually, as $n$ approaches infinity the difference between the number of digits in a binary number and the number of combinations those digits can represent is far, far greater then $n$. It should be $2^nn$. For example: $21=1, 42=2, 83=5$. Even the difference grows faster then $n$. However, when $n$ actually is infinity, that is, a binary number is infinite in length, all bets are off and they all cross the finish line neckandneck. Everything is infinite. The size of the number versus the number of numbers becomes square. For a chart that looks something like:

This chart has been drawn in a square configuration, the number of digits is represented horizontally and the number of combinations is on the vertical, to illustrate the above fact that there are the same number of digits to the left and right as there are up and down, both reaching to Infinity.
Let it be noted here that Infinity is not a number. There is no value attached to any quantity of infinity. Infinity is more a class denoting the uncountability of a value. As such, any portion of that value is equally uncountable unless the portion that is taken from it has been specified. For example: An infinite value  1. Now, of course, the 1 is countable but the remainder is still just as uncountable as the starting value. In fact the starting value has not been affected at all.
What about for another number system? Say, a decimal system.

 Here the number of permutations versus the length of the number is $10^n$ for a number that is $n$ digits long. Compared to the binary system; for a number of $n$ length, the $10^n$ values for a decimal system are a lot more then the $2^n$ values for a binary number of the same length. In this case as $n$ approaches Infinity, the number of permutations or values representable is astronomically more then for a binary digit of the same length. I mean, like dude, take $10^{100}$ versus $2^{100$ and compare the difference then try $10^{500}$ versus $2^{500}$ and $10^{1000}$ versus $2^{1000}$ and compare their differences. What is the trend here? Well, $10^{100}$ is represented on my calculator as $10^{100}$ (We know, however, this number is 101 digits long) but $2^{100}$ is $1,267,650,600,228,229,401,496,703,205,376$ (31 digits long) which in terms of scientific notation equals a bit less then $1.3 x 10^{30}$. That’s a large difference, right? But these things have to be read in context or by comparison to something else. Now try $10^{500}$ versus $2^{500}$. Now we have a number we can sink our teeth into. $2^{500} = 3.3 x 10^{150}$. That is a big number but the decimal number is still much bigger. In fact, the difference has grown even greater. What about the last set of numbers I mentioned earlier? $10^{1000}$ versus $2^{1000}$. $2^{1000}$ equals $1.1 x 10^{301}$. That has the biggest difference yet. We can notice that the proportion of the exponents remain much the same but the differences between the two numbers is growing by that proportion. Will the value of a decimal number therefore reach Infinity before the value of a binary number as n goes to Infinity? Will the number of permutations so far out strip the number of digits between the decimal number and the binary number that the decimal number’s value reaches Infinity before the binary number’s digits n reaches Infinity? It is clearly approaching Infinity faster, isn’t it? The fact remains that all values that head towards the Magical Realm Infinity will reach it at the same time, even though order is maintained. The Cops on the road to Infinity are very insistent about maintaining order. However, once the Realm of Infinity has been reached Chaos is the rule. It’s kind of like how once you square a negative number you loose the negative property and no amount of square rooting can get it back. Oh yeah, I remember now. We made up a number that allows us to do that. We just imagined there was such a number and POOF! i. WHOA! 
I don't know about you but doesn't there seem to be something strange about a concept that at once has no value and a universal value of equality for all members?
]]>I feel like I need to refer to definition 1.27: A sequence is a function with its domain $\mathbb{N}$and its range $\in \mathbb{R}$ and definition 1.28: $p_1, p_2, p_3, ...$ converges to x means if $S$is an open interval containing $x$ then there is a positive integer $N$such that if $n$ is a positive integer and $n\geq N$ then $p_n$$\in S$. Thus, all I really need to do is show that if $x$ is $\in\mathbb{R}$ then $c\cdot x$ is also $\in\mathbb{R}$. It makes sence but I don't see how this necessarily follows.
I can create an open interval $S=(a,b)$ containing $c\cdot x$ but how do I prove there is a $c\cdot p_n$ within that interval if $c\cdot p_n$ may not be a Real number?
Any ideas?
WillTheSerious
]]>One can see that carrying this operation out forever will lead to an infinite string of 9s. We’ll just forget about the 1
because numbers disappear in Infinity. Infinity is a magical realm. Numbers go in but they never come out. Like
the Roach Hotel of Mathematics.
How about if we square .9 and square 1.
We can not do the operation out because we can’t see what happens at the right end of the string where it exists
in Infinity. So, let’s try it in the finite realm and see if we can tell what is happening in Infinity.
.9 x .9 = .81 No 9s in the resulting string of numbers and .81 is obviously not equal to .9.
How about:
$.99^{2} = .81 + .081 + .0891 = .8 + .17 + .010 +.0001 = .9801$ One 9 out of 4 digits but it’s a leading 9.
(I wanted to do this out to show all values are present and accounted for).
Ok, next problem: (Assume the same process of calculation for all that follows)
.999^2 = .998001 More leading 9s. However there are still less 9s then other numbers.
Again:
.9999^2 = .99980001 Wow, I see a pattern here. Now let’s assume this process goes on forever.
There will be just as many 0s as 9s and the 9s will be 1 less then the original
number of nines. Also, there are always 2 more digits that are neither 9s nor
0s then making the string of digits of non9s two longer then there are 9s.
Now here’s the thing about magic realms. All travelers arrive at once. None who leaves for a magic realm can get there before another as long as they left at the same time. There is no closer or farther away. Not only will the length of the string of numbers reach infinity but the 9s will get there at the same time and so will the 0s. You can’t get half way to infinity. You are either there or not. So, there are not ½  2 to Infinity of 9s. And, as I already stated about the magic realms including Infinity, once you go in, like magic, you disappear. Only the 9s can be found because they are behind in the journey to infinity. Although, they arrived there at the same time as the 1s, they are still last and push all the numbers ahead of them through the gates where they are lost forever, like magic. The nines have a foot hold in the finite realm and unless another traveler on its way to Infinity, and I might point out that they have to also be on the same path, starts later then the 9s, they will remain in the Finite realm. There is no indication that anyone else wants to travel this path to Infinity so the 9s are all that are left. Does that make sense? Magic!
Wait, wait, How do I know the 9s as well as the other numbers actually make it to Infinity? It’s a long long long way. They can’t make it on their own. Don’t I or someone have to keep doing to calculations never endingly. I got it! Let’s not and say we did. Even still. Any teacher would know I was lying. “You never did the calculations out!”
I would say, “Well… Ah, I did do the calculations, teach.”
“No, I don’t thinks so. I don’t know how you got those 9s to Infinity but I know you didn’t do all the calculations. Now did you?”
“Well… No, not all of them. But I did most of ‘um. Really, I did nearly all of ‘um.”
“I am unconvinced. You couldn’t have done that many in the time you had.”
“Ah… um, at least half of them. I did them ‘till the 8 got to infinity, then I had to go to bed. That‘s 1 more then half.”
“Son, if you did that many calculations then you would know that the 8 as well as the 1 never get to Infinity.”
I defend myself, “Yeah they do. They are gone, aren’t they? They disappeared once they got to Infinity.”
The teacher is really angry now, “No no no no, you didn’t do all the calculations and they can’t just disappear. What did you do with the 8 and the 1. There’s only one of each so you must have stashed them somewhere. I think you need to go into the corner and think about where those missing numbers are.”
What could I do but sit in the corner. I didn’t know where they went. I still don’t.
You should not look to resources outside the context of this course for help. That is, you should not be consulting the web, other texts, other faculty, or students outside of our course. On the other hand, you may use each other, the course notes, me, and your own intuition.
The spirit of this "rule" is to discourage "looking up answers." I want the emphasis to be on us creating mathematics. However, I am perfectly okay with people looking up things like "how do I prove something by contradiction?" or "what's the contrapositive?" or "what's the definition of a function?" I hope you get the idea and if I need to be more explicit, just ask.
]]>