Intuitively, it seems to me that $|\mathbb{R}^{2}| &gt; |\mathbb{R}|$. My intuition says something to the effect of: There exists a bijection between $\mathbb{R}$ and $\{(x,0) : x \in \mathbb{R}\}$, and the second set is a subset of $\mathbb{R}^{2}$.  But I remember reading somewhere that $|\mathbb{R}^{2}| &gt; |\mathbb{R}|$ is not true, though I may be mistaken.  I know only the very basics about countable sets which are infinite. I also know that $\mathbb{R}$ is uncountable.  To be clear, I am looking for:   Is $|\mathbb{R}^{2}| &gt; |\mathbb{R}|$? Why? Any quick resource/lecture notes to quickly brush up on related concepts.   The most relevant question I could find was:
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I am working on this exercise for an introductory Real Analysis course:     Show that |$\mathbb{R}$| = |$\mathbb{R}^2$|.   I know that $\mathbb{R}$ is uncountable. I also know that two sets $A$ and $B$ have the same cardinality if there is a bijection from $A$ onto $B$. So if I show that there exists a bijection from $\mathbb{R}$ onto $\mathbb{R}^2$ then I beleive that shows that |$\mathbb{R}$| = |$\mathbb{R}^2$|.   Let $x_i \in \mathbb{R}$, where each $x_i$ is expressed as an infinite decimal, written as $x_i = x_{i0}.x_{i1}x_{i2}x_{i3}...,$. Each $x_{i0}$ is an integer, and $x_{ik} \in \left \{ 0,1,2, 3, 4, 5, 6, 7, 8, 9 \right \}$. Then, let  $$f(x_i)=(x_{i0}.x_{i1}x_{i3}x_{i5}... ,x_{i0}.x_{i2}x_{i4}x_{i6}...)$$  What should I do to show that $f: \mathbb{R} \to \mathbb{R}^2$ is an injective function? Any suggestions or help with the question would be appreciated.
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Does separability imply a sort of Lindelöf property?  Since I can't prove this fact I'm beginning to think that my conjecture is false.  Intuitively, $\mathbb{R}$ has a countable subset $\mathbb{Q}$ which is used to form a countable basis for $\mathbb{R}$ with the usual topology and prove the Lindeöf property.
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Are there two closed sets $A,B\subset\mathbb{R}^2$ with the following properties?   $A\cap B=\emptyset$ $\forall \epsilon&gt;0$ there exist $a \in A$ and $b\in B$ such that $\|a-b\| &lt; \epsilon$
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Possible Duplicate:          As the topic,if $A\subset\mathbb{R^2}$ is countable, does $\mathbb{R^2}\setminus A$ path connected??? I know the answer is it is path connected but not sure how to prove it.
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I want to prove that $\mathbb{Q}$ is countable. So basically, I could find a bijection from $\mathbb{Q}$ to $\mathbb{N}$. But I have also recently proved that $\mathbb{Z}$ is countable, so is it equivalent to find a bijection from $\mathbb{Q}$ to $\mathbb{Z}$?
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I think it wants me to show that there's no bijection between the two sets.  I first tried to show that there's simply no bijection, then I realised that it doesn't work.  If I'm to show there's no bijective morphism that carries multiplication in nonzero real numbers to addition in real numbers, how am I supposed to do it? I'm thinking about doing something with 0, since it's the element of the first set but not the second set.  Thanks!
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I was wondering if it is possible to produce an explicit bijection $h\colon \mathbb{R} \rightarrow \mathbb{R}/\mathbb{Q}$. If we can produce an explicit injection $i\colon \mathbb{R} \rightarrow \mathbb{R}/\mathbb{Q}$, can the  be used constructively?  It is clear that the two sets have the same cardinality, so the existence of such a bijection is trivial. What I am really looking for is a nice-looking bijection, or a proof that no such nice-looking bijection exists, for a definition of "nice-looking" which I cannot quite figure out.  One problem which I think makes finding such a bijection difficult is that any natural injection of $\mathbb{R}/\mathbb{Q}$ (ie, those injections in which one representative is chosen from each coset) produces a non-measurable set, specifically a .  I hate to ask such a vague question, but I'm really not sure about whether the correct answer is constructive, or whether it is a proof that any such bijection is in some sense "very complicated."  As a final note, the motivation for this question came from , in which I was somewhat astonished to see such a clear, constructive bijection given between $\mathbb{R}$ and $\mathbb{R} \setminus S$, where $S$ is countable.
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Could any one give an example of a bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$?  Thank you.
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Statement: There are uncountably many surjections from $\mathbb N$ to $\mathbb N$.  Is the following proof valid:  The real numbers are uncountable.   There exists a bijection between the set of real numbers and the set of all functions from $\mathbb N$ to $\{0,1\}$.   Given a function, $f$, from $\mathbb N$ to $\{0,1\}$, construct a surjection from $\mathbb N$ to $\mathbb N$, denoted $g$, as follows:  We map the $n$th composite number to $n$, and the nth prime, to $f(n)$.   Clearly this is a surjection since the composite numbers are countable. Clearly different functions $f$ lead to different surjections $g$. Since there are uncountably many $f$s, there are uncountably many surjections $g$.
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The problem says:  List all the definable subsets of $\mathbb R^2$ in ($\mathbb R$,&lt;) with formulas.  And I found $x&lt;y$, $y&lt;x$ and $x=y$  are desired one. But, I want to say that any subset which could made up by them is also definable. For example, less than or equal to is also desired one, but I want to say that not just it is definable but it is definable since it is made by $x&lt;y$, $y&lt;x$ and $x=y$. I intuitively think that there're many definable sets like $\leq $ above. So, my question is that:   How can I express all of such definable sets by $x&lt;y$, $y&lt;x$ and $x=y$?
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I know that the set of functions $ f:\mathbb N \to \mathbb N$ is uncountable, but what if we consider only $f$ such that $f$ is increasing? I want to know if this set is countable D: and also the case of bijective and increasing (clearly if the firstone holds then this too). I think that it could be possible that it's countable
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Let $A$ be a countable subset of $\mathbb{R}^2$. Show that $\mathbb{R}^2-A$ is path connected.  These are my steps:   Let $x$ and $y$ be arbitrary points of $R^2$ Let $f^r:[0,1] \to \mathbb{R}^2$ be given by $f^r(t)=(1-t^r)x+t^ry$ for $r \in\mathbb{R}$.  Show that $f^r$ is a continuous path between $x$ and $y$ and that non of them intersect. Show that $f^r$ is bijective with $\mathbb{R}$ through $F(f^r)=r$ and conclude that there are uncountable non intersecting paths. Assume for the sake of contradiction that $A$ intersects all of the paths $f^r$ and deduce that $A$ must be uncountable (since it must intersect every $f^r$ at a different point).  Contradiction! ($A$ is countable) Hence there exist a paths $f^r$ for some $r \in\mathbb{R}$ which doesn't intersect $A$. $\mathbb{R}^2-A$ is path-connected.     Is this proof sound?
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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.
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I've heard that it is possible to present a bijection $\phi : \mathbb{R}^n \to \mathbb{R}^m$ due to Cantor that although not continuous show that those sets have the same cardinality independent of the dimension as a vector space. The problem is that I couldn't find about it anywhere. Can someone tell me where I can find about this bijection and how to show it's existance?  Thanks very much in advance!
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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)
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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?
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If I have $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f$ is onto and $f\circ f\circ f = f$, how can I prove that $f$ is bijective? I know that I only have to prove that it is 1-to-1 because I'm given the fact that it's onto, but how can I use the fact that $f\circ f\circ f=f$ to prove that it is 1-to-1?  Thanks!