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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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Can you use cardinal arithmetic? –  DonAntonio Jan 29 '13 at 20:29
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Related. –  Cameron Buie Jan 29 '13 at 20:32
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snip the index $i$ to simplify notation. –  user59671 Jan 29 '13 at 20:39
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Also, this post. –  David Mitra Jan 29 '13 at 21:11
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2 Answers 2

Edited to add: As Andres pointed out in a comment, I had this the wrong way round: to use Schroeder-Bernstein, we want to exhibit injections (not surjections) in each direction. I think I have it right now.

Using decimal (or binary, ternary,...) expansions like this is very messy.

  1. Your suggestion is obviously not an injection, because for every $(x,y)$ in the range, the integral parts of $x$ and $y$ are the same ($x_{i0}$ in your notation).
  2. There are any number of bijections between $\mathbb R$ and the interval $(0,1)$. I'm sure you can think of a few. So we can reduce the problem to finding a bijection between $(0,1)$ and $(0,1)\times(0,1)$. In this case, your approach looks promising. But unfortunately it is spoilt by the fact that $0.abc09999...$ is equal to $0.abc10000...$. You can make the notion of decimal expansion unambiguous by always choosing the non-terminating one, but then when you try to extract two such decimal expansions as you have done, there is no guarantee that they will both be non-terminating. For instance, $0.7170707070707070...$ and $0.7079797979797979...$ both unpack to $(0.777777...,0.1)$.

So the usual approach is to show the existence of an injection in each direction, which guarantees the existence of a bijection by the Schroeder–Bernstein theorem.

There is the obvious injection from $(0,1)$ to $(0,1)\times(0,1)$ given by $x \mapsto (x,0)$.

To go the other way, take $(x,y) \in (0,1)\times(0,1)$. We can express $x$ and $y$ as non-terminating decimal expansions $0.x_1x_2x_3\ldots$ and $0.y_1y_2y_3\ldots$ (non-terminating means that the number of non-zero digits is infinite). Now we map this to the number with decimal expansion $0.x_1y_1x_2y_2x_3y_3...$. This is not a bijection! But it is an injection, which is enough to prove that the cardinalities of $\mathbb R$ and $\mathbb R^2$ are equal.

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You are complicating things unnecessarily using surjections, as now you need the axiom of choice to ensure appropriate inverses which are injective, so that Bernstein-Schroeder can be invoked. You want injections in both directions. –  Andres Caicedo Jan 29 '13 at 20:52
    
@Andres: Oh, you're right. I'll fix it tomorrow, if I can. –  TonyK Jan 29 '13 at 20:58
    
@TonyK: Can you do $x\mapsto(x,0)$ for the injection $(0,1)\rightarrowtail(0,1)^2$, since $0\notin (0,1)$? –  5space Jan 12 at 17:52
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$$f(0.0090909090...)=(0.0999999,0)=(0.10000,0)=f(0.1000)$$ so $f$ is not a bijection.


we know there is a (continuous) surjective function $f:\mathbb{R}\to\mathbb{R}^2$ (or find any other surjective function). then by axiom of choice, there are some $A\subseteq\mathbb{R}$ and some bijection $g:A\to\mathbb{R^2}$. we have $$|\mathbb{R}^2|=|A|\leq|\mathbb{R}|\leq|\mathbb{R}^2|$$

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I don't understand the downvote. This is a valid objection! –  TonyK Jan 29 '13 at 20:54
    
they're right, $f(1)=(1,1)$ –  user59671 Jan 29 '13 at 21:03
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Yes, but $f(0.3039393939...) = f(0.3130303030...)$. –  TonyK Jan 29 '13 at 21:09
    
@CutieKrait: I was the one who downoted you. I have undone the downvote. However, your answer only points out that the function $f(x)$ from the questions isn't bijective. You don't actually help any further with hints on how to prove that the cardinality is the same. –  Thomas Jan 30 '13 at 2:30
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Those hints ruined your post! Now I've downvoted it. –  TonyK Jan 30 '13 at 20:51
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