I would like to show an idea on how to make real numbers infinitely countable. It is quite simple, too simple for me to believe it has been overlooked. So my question is, what have I overlooked?    So the "countable" list would start like this {0.1, 0.2, 1.1, 2.1, 1.2, 0.3, 0.4, 1.3, 2.2, 3.1 ...}, the way Cantor diagonally counted the list of rational numbers and potentially covering every real number there is.  And for the numbers after the decimal point, you go down a column and count normally after the decimal point but reverse the number.
123408
The proof that the set of real numbers is uncountably infinite is often concluded with a contradiction. In the following argument I use a similar proof by contradiction to show that the set of rational numbers can be shown to be uncontably infinite:  Let us assume that the set of rational numbers is countable. This implies that we can write all the rational numbers between 0 and 1 in a list. Let us say, it looks something like this:     $ 0.\color{red}0000000001000000...\\ 0.0\color{red}10000001110000...\\ 0.10\color{red}000001000000... \\ ....$   ans so on. Now, create a new number that picks $n^{th}$ digit from $n^{th}$ number and  changes it to $0$ if it is $1$ and vice versa.     $0.101.... $   This number, say $x$, is between 0 and 1, and so it must appear in the list. Suppose it appears at the $n^{th}$ position in the list. However, based on how we constructed $x$, we know that the $n^{th}$ digit of $x$ is different from the $n^{th}$ digit of the $n^{th}$ number in the list. So, $x$ should be different from the $n^{th}$ number in the list at the $n^{th}$ digit and hence, this is a contradiction. We can conclude that such a list of the rational numbers does not exist and hence, the set of rational numbers is uncontably infinite.  I know that there is another mapping between the set of rational numbers and natural numbers that can show that there is one-to-one mapping between the two sets (). However, the above proof in that sense is not really conclusive and that the set of real numbers might also be countably infinite and it is just that we are not yet aware of the mapping.  Please provide your comments if the above proof is somehow flawed for rational numbers or if I'm missing something.
26202
I just prove $$S_1=\{0.4,0.8\}$$ $$S_2=\{0.04,0.08,0.44,0.48,0.84,0.88\}$$ ... So I can calculate the number of elements in $S_n = 2 \times 3^{n-1}$  I just prove it is countably infinite.
22224
The Cantor set is closed, so its complement is open.  So the complement can be written as a countable union of disjoint open intervals.  Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?
53718
There are two questions:        Image of a countable set of real numbers under any continuous function is countable?      My claim is yes. Let $X$ is countable $\implies X=\{x_1,x_2,\ldots,\}$. Now $f(X)=\{f(x_1),f(x_2),\ldots,\}$ which can be atmost countable. Now my question is "What is the role of continuity here?"        Image of a uncountable set of real numbers under any non-constant continuous function is uncountable?      I feel this is true. But unable to proceed. Please provide me a hint.
76872
I am trying to show that for any real number a, there exist infinitely many rational numbers m/n with $ |a - m/n| &lt; 1 /n^{2} $. I've tried to attempt the question by assuming there are finite rational numbers and finding a contradiction.
120519
I realize the title of my question will probably cause the raising of some eyebrows, so let me explain. Not sure whether to file this under "math" or "philosophy". This also might be able to be explained away due to some possible misconceptions I have about irrational/transcendental numbers, so please correct me where I'm wrong, since I'm quite certain I am...but this is nagging me, so here it is:  Ok, so imagine a non-repeating decimal with an infinite number of digits in it. The numbers are random, but for the sake of argument, let's say the number starts with "0.2368210236..." and continues from there.  So I look at the first decimal digit - 2 - and think to myself "in the rest of this decimal, there are an infinite number of times that '2' shows up."  So I look at the first TWO decimal digits - 23 - and again, think to myself "due to the randomness and infinity of this decimal, there is statistically a 100% certainty that the numbers "23" show up, in that order, somewhere in the rest of this decimal.  Continuing this line of thought, I reach the conclusion that if I choose ANY number of digits (variable: x) that start off this infinite decimal number, then a) there is statistically a 100% certainty that particular sequence shows up somewhere else in the decimal, and b) that it shows up an infinite number of times.  If such is the case, then what happens as x approaches infinity? Would that not mean that, given enough length in the decimal, there would eventually come a point where the decimal starts repeating itself, ad infinitum? Would that not mean that what was originally thought of as an irrational number is actually rational? Would this line of thinking not apply to ALL irrational numbers?  I guess what I'm asking is this: could it be that there ARE no irrational numbers, just rational numbers whose scope and vastness outstrip our ability to see their rationality?  Alright, mathphiles; have at me!
26129
Can someone point me to a proof that the set of irrational numbers is uncountable? I know how to show that the set $\mathbb{Q}$ of rational numbers is countable, but how would you show that the irrationals are uncountable?
68243
I was reading about countably infinite sets and the definition goes as, "A set is countably infinite if its elements can be put in one-to-one correspondence with the set of natural numbers"(Source: Google). Well yes, this make sense. As the set $\mathbb{N}$ is countably infinite and set which can it can create a bijection with, would end up having the same number of elements and thus they too would be countably infinite. But this would be true iff $\mathbb{N}$ is countably infinite. How would you prove that? Or is the idea of countably infinite base on some assumption that $\mathbb{N}$ is countably infinite?
13091
Ok, here is the problem statement:     Prove that if $S$ is any finite set of real numbers, then the union of $S$ and the integers is countably infinite.   This seems pretty obvious to me, knowing that 2 countable sets are countable.  But is there some step by step way to prove this?  Like do I need to prove bijectivity or something?  Thanks!
46533
Can any one tell me how to prove that the set of rational numbers are countable? Prove give me a prove?  Thanks.
97682
Can the Cantor diagonal argument be use to check countability of natural numbers? I know how it sounds, but anyway.  According to the fundamental theorem of arithmetic, any natural number can be expressed as an unique product of primes.  if we denote primes $2,3,5,\ldots$ as $P_1,P_2,P_3,\ldots$ and include $1$ as part of the system, we will get,  $$D = [1, P_1, P_2, P_3,\ldots]$$  we can write any natural number as series of $[d]$ that goes on forever, as we can keep on appending $1$ to any product of primes forever.  for example $6$ is $2 \cdot 3 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1.....$  Now, we use the Cantor diagonalization, and enumerate all possible combinations of D.  $$1 - D_{11}, D_{12}, D_{13}, D_{14}, ...$$ $$...$$ $$n - D_{n1}, D_{n2}, D_{n3}, D_{n4}, ...$$ $$...$$  Can we show that there is another number that is not in the list? If we assume that there is a function, $\operatorname{Next}(D)$, that gives as the next element in $[D]$.  $$P_1 = \operatorname{Next}(1)$$ $$P{n+1} = \operatorname{Next}(P_n)$$  since there is always the next prime, we can construct,   $$m = \operatorname{Next}(D_{11}) \cdot \operatorname{Next}(D_{22}) \cdot \operatorname{Next}(D_{33}) \cdot \operatorname{Next}(D_{44})...$$  This is analogous to adding $1$ to diagonal elements applied to real numbers.  From the same diagonalization argument this number, '$m$' cannot be in the list. It is a contradiction, and natural numbers aren't countable...  Can someone point on a rigurous discussion of such a reasoning?  Thanks!
12509
Is the cofinite topology on a countable set first countable? I know it is not first countable on R but I don't know the case of a countable set.
110211
Is every open subset of $\mathbb{R}$ uncountable?  I was crafting a proof for the theorem that states every open subset of $\mathbb{R}$ can be written as the union of a countable number of disjoint intervals when this question came up.  I feel like the answer is yes, but I'm not sure how to go about proving it or whether there is a crazy construction (like the Cantor Set)  that creates a countable, open subset of $\mathbb{R}$. Any ideas?
14119
Consider a common proof of the uncountability of $(0,1]$, as presented  for example:  We assume that the reals can be arranged in a sequence $x_k$, represent every number in $x_k$ by its nonterminating decimal expansion, use the diagonal argument to create a sequence of decimal digits that differs from the nonterminating decimal expansion of each number in $x_k$ and then claim that this sequence of digits so generated must be the decimal expansion of some number not in $x_k$.  My problem is with this last step. Could it not be the case that the new sequence of digits is just the terminating decimal expansion of some number already in $x_k$?  Would it not be more correct to use the sequence $x_k$ to generate a sequence of all (both terminating and non-terminating) decimal expansion of all numbers in $[0,1)$ and then apply the diagonal argument?
23510
How many subsets of $\mathbb{N}$ have the same cardinality as $\mathbb{N}$?   I realize that any of the class of functions $f:x\to (n\cdot x)$ gives a bijection between $\mathbb{N}$ and the subset of $\mathbb{N}$ whose members equal multiples of n.  So, we have at least a countable infinity of sets which have the same cardinality of $\mathbb{N}$.  But, we could remove a single element from any countably infinity subset of the natural numbers and we still will end up with a countably infinite subset of $\mathbb{N}$.  So (the reasoning here doesn't seem quite right to me), do there exist uncountably many infinite subsets of $\mathbb{N}$ with the same cardinality of $\mathbb{N}$?    Also, is the class of all bijections $f: \mathbb{N} \to \mathbb{N}$ and a given countably infinite subset uncountably infinite or countably infinite?
59118
Let $S$ be an infinite subset of $\mathbb{R}$ with the property that the existence of $S$ can be proved within ZFC (and in particular the definition of $S$ does not invoke the negation of the continuum hypothesis), but the cardinality of $S$ is undecidable within ZFC.  Are there some concrete examples of such sets?  I.e. something more constructive than $S:=$ some subset of cardinality $\aleph_1$?  This question results from musing about intuitively obvious equivalents of $\neg CH$ in order to find evidence against the validity of $CH$. My (undoubtedly too optimistic) thinking was that if some such example of $S$ is concrete enough, it would be perhaps intuitively obvious whether it should be countable or continuum, despite this being undecidable formally.  My best guess for the moment is something along the lines of  $$S=\left\{\sum_n\frac{1}{a_n}:\; (a_n)\;\text{is a strictly increasing sequence of primes}\right\}$$  but surely the cardinality can be determined (though not quite sure how).
43900
Question 1: Why are irrational numbers uncountable and rationals countable?  I really struggle to understand this. I initially thought it had something to with the fact that between any two numbers there will always be a irrational number. But now it has come to my attention that the same holds for rational numbers. So I can't really see the difference between them. What is it that makes the difference? Why can we line up all rational numbers one-to-one with the natural numbers, but not the irrationals with the naturals?  Sorry for a perhaps basic question, but I really, really struggle with it. Thanks.