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Is there an injection from the set of all real sequences to R?

user116498
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2 Answers2

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Yes. In fact there is an easy bijection, not that explicit, but it works.

Let $f: \mathbb{N}^\mathbb{N} \to \mathbb{R}$ be a bijection (there are many, for instance continued fractions). The set of real sequences is $\mathbb{R}^\mathbb{N}$, that corresponds, from the above bijection, to the set of sequences of "sequences of naturals" $(\mathbb{N}^\mathbb{N})^\mathbb{N}$.

But this last set is easily in bijective correspondence to $\mathbb{N}^{\mathbb{N}^2}$, the set of double-sequences of naturals. Finally, $\mathbb{N}^2$ has a bijection to $\mathbb{N}$, which you can chain again to obtain

$$ \mathbb{R}^\mathbb{N} \simeq (\mathbb{N}^\mathbb{N})^\mathbb{N} \simeq \mathbb{N}^{\mathbb{N}^2} \simeq \mathbb{N}^\mathbb{N} \simeq \mathbb{R} $$

rewritten
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  • I dont really understand the bijection by continued fractions from N^N→R could you elaborate a little? Thanks for he help :) – user116498 Dec 18 '13 at 13:34
  • Any non negative real number can be expressed uniquely as a continued fraction $$a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{a_4+\dots}}}}$$ where the $a_n$ are natural numbers. These coefficeints form a sequence of naturals, so this gives you an injection into $\mathbb{N}^\mathbb{N}$. – rewritten Dec 19 '13 at 11:36
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Yes. You can show there is a bijection with the following theorem.

If $a$ is an infinite cardinal number, and if $b$ is a cardinal number with $2 ≤ b ≤ 2^a$, then $b^a = 2^a$.

Also recall that $|\mathbb{R}| = \beth_1$ and $|\mathbb{N}|=\beth_0$. See also, beth numbers. Thus we may argue as follows. $$|\mathbb{R}^\mathbb{N}| = {\beth_1}^{\beth_0} = 2^{\beth_0} = \beth_1$$

We deduce that $|\mathbb{R}^\mathbb{N}|=|\mathbb{R}|$, or in other words that these sets can be put into bijective correspondence.

goblin GONE
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  • Thanks for the help, your solution went way over my head – user116498 Dec 18 '13 at 13:38
  • @user116498, do you know about cardinal numbers? – goblin GONE Dec 18 '13 at 13:40
  • Nope, sadly not only cardinality of power set – user116498 Dec 18 '13 at 13:57
  • @user116498, you should look them up. Here's a primer: the cardinal numbers measure the size of sets. So unsurprisingly, $0,1$ and $2$ are cardinal numbers (because you have have a set with $0$, $1$ and $2$ elements), as are all natural numbers. But it doesn't end there; we can define cardinal numbers that measure the sizes of infinite sets, too. The least infinite cardinal number is $\beth_0$, and this is the cardinality of $\mathbb{N}$. A much larger cardinal number is $\beth_1$, and this is the cardinality of $\mathbb{R}$. Anyway, you should read up about them if you have the time. – goblin GONE Dec 18 '13 at 14:02
  • Will do thanks :) – user116498 Dec 18 '13 at 14:24