In the diagonal argument in which we prove $\mathbb{R}$ is uncountable, we take any function $f: \mathbb{N} \to \mathbb{R}$, suppose for the sake of contradiction that it is surjective, and consider any enumeration of $\mathbb{R}$, which will take the form: \begin{align*} f(0) &amp; = a_{00} . a_{01} a_{02} a_{03} \ldots \\ f(2) &amp; = a_{10} . a_{11} a_{12} a_{13} \ldots \\ f(3) &amp; = a_{20} . a_{21} a_{22} a_{23} \ldots \\ \vdots  \end{align*} The proof then constructs $x = b_0 b_1 b_2 \ldots$ where $b_i \neq a_{ii}$ for each $i$, so $x \not \in f(i)$ for any $i$, hence $x \not \in f(\mathbb{N})$. Here is the question I have: what about ending in an infinite string of $9$'s? I can require when constructing $f(n)$ for each $n$ that I represent a real number in its decimal expansion not terminating in $9's$ (so I write $0.5\overline{0}$ instead of $0.4\overline{9}$, for example), but what is to stop the $x$ I construct from ending in $9$'s? If that's the case, I can't go digit-by-digit and rule out each line in this enumeration.
37259
I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing?
58579
Let  $f : \mathbb{R}^2\rightarrow \mathbb{R}$ be a continuous map such that $f(x)=0$ only for finitely many values of $x$. Which of the following is true?   Either $f(x)\leq 0$ for all $x$ or $f(x)\geq 0$ for all $x$. the map $f$ is onto. the map $f$ is one one. None of the above.   What I have done so far is :   I would take polynomial in two variables.. This need not be like $f(x)\geq 0$ for all $x$ or $f(x)\geq 0$ for all $x$.So,first option is eliminated. The map is not one one assuming that $f$ has more than one zero. So, third option is wrong.   I could not think of an example in which it is not onto..  Only examples i am getting in my mind are polynomials and they are onto..  So, I am having trouble with surjectiveness of the function.  Please help me to clear this.  Thank you..
41993
Let $f:\mathbb{R}\to \mathbb{R}$ is a function such that for all real $x$ and $y$, $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$, then prove that $f$ must be one of the two following functions:   $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=0$ for all real $x$   OR   $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x$ for all real $x$   I got to the point where putting the two equations together, you get $f(x+y)f(x)= f(xy) + f(x)^2$ and plugging in $f(x)=x$ checks with it. So am I going in the right direction or am I just doing some guess work? Is there a more elegant way of doing it?   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!
86430
Problem: Consider the subfield $$E = \mathbb{Q} \big(\left\{\sqrt[n]{2} \mid n \in \mathbb{N} \setminus \left\{0\right\} \right\} \big)$$ of $\mathbb{R}$. Prove that $[E: \mathbb{Q}]$ is not finite.  Attempt: I proved that $E$ is algebraic over $\mathbb{Q}$. Because let $u \in E$. Then there is a $n \in \mathbb{N} \setminus \left\{0\right\}$ such that $u = \sqrt[n]{2}$. Then $f(x) = x^n - 2 \in \mathbb{Q}[x]$ is the minimal polynomial of $u$ over $\mathbb{Q}$ such that $f(u) = 0$. Hence $E$ is algebraic over $\mathbb{Q}$.  I now want to prove the degree of extension is not finite. I assume it is finite. Suppose then $[E: \mathbb{Q}] = k \in \mathbb{N}$. I don't know how to derive a contradiction from this.   I know that if $u$ is algebraic over $\mathbb{Q}$ and $f$ is the minimal polynomial of $u$ over $\mathbb{Q}$, then $[\mathbb{Q}(u) : \mathbb{Q}] = \deg(f)$.   Help/suggestions are appreciated.
74678
About a bijection between $\mathbb R$ and $\mathbb R \cup \{\infty\}$  If we'll take a line $f(x)=\begin{cases} ax &amp;, x\in\mathbb R\\ 0 &amp;, x=\infty  \end{cases}$ then we'll have a bijection.   But about this function: $g(x)=\begin{cases} x &amp;, x\in\mathbb R\setminus \{0\}\\ 0 &amp;, x=\infty  \end{cases}$ ? it appears to be a bijection and it's even simpler.
112884
I want to prove that the set of all real functions $\mathbb{R}^\mathbb{R}$ has a higher cardinality than the real numbers $\mathbb R$, by Cantor's diagonal argument.  I'm having difficulties with approaching this problem.  What I'm looking for is a hint in the right direction.  I've seen an example where it is shown that the power set $2^S$ of a countable set $S$ is uncountable, by the diagonal argument.
38749
I was just thinking about this recently, and I thought of a possible bijection between the natural numbers and the real numbers.  First, take the numbers between zero and one, exclusive.  The following sequence of real numbers is suggested so that we have bijection.    0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.02, ... , 0.09, 0.10, 0.11, ... , 0.99, 0.001, 0.002, ... , 0.999, 0.0001, etc.  Obviously, this includes repeats, but this set is countable.  Therefore, the set of all numbers between zero and one is a subset of the above countable set, and is thus countable.  Then we simply extend this to all real numbers and all the whole numbers themselves, and since the real numbers, as demonstrated above, between any two whole numbers is countable, the real numbers are the union of countably many countable sets, and thus the real numbers are countable.    Please help me with this.  I understand the diagonalization argument by Cantor, but I am curious specifically about this proof which I thought of and its strengths and flaws.  Thanks.
21174
Let $C(\mathbb{R})=\{ f:\mathbb{R}\to \mathbb{R} \colon \ f \text{ continuous}\}\subseteq \mathbb{R}^{\mathbb{R}} $. How to prove formally that $C(\mathbb{R}) \notin \mathcal{B}(\mathbb{R}^{\mathbb{R}})$ (the product sigma algebra)?  The classic proof said that   $A \in \mathcal{B}(\mathbb{R}^{\mathbb{R}})$ iff exists $J\subseteq \mathbb{R}$ with $|J|\leq \aleph_0$ and $B\in \mathcal{B}(\mathbb{R^J})$ such that $A=B \times \mathbb{R}^{\mathbb{R}\setminus J}=\{f \in \mathbb{R}^\mathbb{R} \colon  \ (f(j) \colon \ t \in J) \in B\}$. In simpler terms,  $A$ only can have countable restrictions but the set of continuous functions have uncountable restrictions since every function on $C(\mathbb{R})$ needs to be continuous in every point $x \in \mathbb{R}$.  The last part of this proof always didn't satisfied me at all, and I don't really know how to do this proof formally.   Any help or references will be appreciated.
12771
Recall that a countable set $S$ implies that there exists a bijection $\mathbb{N}\to S$. Now, I consider (0,1). I want to prove by contradiction that $(0,1)$ is not countable.  First, I assume the contrary that there exists a bijection $f$, and I can find an element in $S$, but not in the range of $f$. But I can't find such element. How can you construct such $f$?
35336
Math people:  It is my understanding that set theorists/logicians compare cardinalities of sets using injections rather than surjections.  Wikipedia defines countable sets in terms of injections.  Cantor's diagonal proof that the real numbers are uncountable involves showing that there is no surjection from $\mathbb{N}$ to $(0,1)$.  So do I need the Axiom of Choice to accept Cantor's Diagonal Proof?  I browsed the Similar Questions and I could not find an answer.  I apologize if this is a duplicate.  StEFAN (Stack Exchange FAN)
103017
This is different then the other versions of the questions here. I'm not supposed to use the diagonal argument. I'm looking to write a proof based on Cantor's theorem, and power sets.
70281
I am trying to prove that $D(\mathbb{N}$)={ $f\in \mathbb{N}^\mathbb{N}$ | f is a bijection such that $f(n)\neq n$ for all $n\in\mathbb{N}$} is uncountable. So, I was thinking of showing $D(\mathbb{N})$ ~ $P(\mathbb{N})$ using Cantor Bernstein Theorem.   For the one direction, is it found to say since $D(\mathbb{N})\subset \mathbb{N}^\mathbb{N}$ ~$P(\mathbb{N})$ then there is an injection from $D(\mathbb{N})$ to $P(\mathbb{N})$?   I am not too sure how to go about the other direction. Should I show equinumerous to functions from $\mathbb{N}^\mathbb{N}$ instead of using the power set?
2005
Let $f:\mathbb{N} \to \mathbb{Q}^+$ defined as follows : $$\begin{cases} f(0) = 0 \\ f(2n) = \dfrac{1}{f(n)+1} \\ f(2n+1) = f(n)+1 \end{cases}\,.$$ It is asked to prove the injectivity then the surjectivity of $f$.  After examining first values taken by $f$, I proved by induction that $f(2^k)=\dfrac{F_k}{F_{k+1}}$ where $(F_n)$ is the fibonacci sequence. Also one notes that if $m$ and $n$ have different parity then we can't have $f(m) = f(n)$ since $f(2\mathbb{N}) \subset (0,1)$ and $f(2\mathbb{N}+1) \subset [1,+\infty)$. Any ideas on how to finish the proof of injectivity and any thoughts about surjectivity are welcome. Thanks
118287
I'm a little stuck with this proof. Not sure where to go. I was thinking that I could first assume that $f$ is one to one and prove that it's onto, and then assume it's onto and prove that it's one to one... but I'm not sure what to do with the knowledge that $|A|=|B|=m\in\mathbb{Z}\geq 0$.  I've tried to start by defining what it means for $f$ to be one-to-one, and going from there...  \begin{align*} f\text{ is one-to-one}&amp;\Longrightarrow f=\{(a,b)\in A\times B\mid b=f(a)\wedge f^{-1}(b)=a\}\\ &amp;\Longrightarrow \dots\\ &amp;\Longrightarrow \text{$f$ is onto} \end{align*}  but not sure where to go from there... I have a feeling that my definition is too presumptuous to begin with, since I'm trying to show that $f$ is onto... not assume that it is.
77615
How would I show the following have a bijection. Which is one to one and onto ofcourse.  $\mathbb{Z}\rightarrow \mathbb{N}$  I know I need to find a function. But I do not see the pattern that makes integer go to natural. I already did this problem from natural to integer but now I want to see it done this way.  I have listed some numbers  $,..,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,..$  $1,2,3,4,5,6,7,8,9,10,11,12,13,14,.$
38882
Every real number in $[0,1]$ has a decimal expansion $0.d_1d_2d_3...$, so construct an infinite tree rooted at 0 where each node has branches leading to $\{0,1,2,3,4,5,6,7,8,9\} $, and let each path through the tree represent the successive decimal digits of a real number. To enumerate them, go through the tree breadth-first i.e. layer-by-layer.