prove that $|\{0,1\}^{\mathbb{N}}|\leq |\mathbb{R}|$

Asked Dec 15 '20 at 18:18

Active Dec 15 '20 at 20:02

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Could you give me a suggestion to create a correspondence?

Let $f:\{0,1\}^{\mathbb{N}}\rightarrow {\mathbb{R}}$

I must prove that f is injective but I don't know what correspondence I could use. Any suggestion?

edited Dec 15 '20 at 20:02

Eric Wofsey

asked Dec 15 '20 at 18:18

Kale_1729

You can use $f(x)=\sum_{k=0}^\infty 3^{-k}x(k)$. – Dec 15 '20 at 18:24
$f\to 0.f(0)f(1)f(2)f(3)...$ – Mark Dec 15 '20 at 18:24
@Gae.S. That $x$ would be a sequence that is in ${0,1}^{\mathbb{N}}$? – Kale_1729 Dec 15 '20 at 18:27
Seeing what the domain of $f$ is, $x$ would be an element of ${0,1}^{\Bbb N}$, yes. – Dec 15 '20 at 18:28
@Gae.S. So when proving that $x(k)=y(k)$, can I conclude that $x = y$? – Kale_1729 Dec 15 '20 at 18:32
It is true that $x=y$ if and only if, for all $k\in\Bbb N$, $x(k)=y(k)$. – Dec 15 '20 at 18:40
@Gae.S. I get it. One last doubt. Could you tell me by take that correspondence? – Kale_1729 Dec 15 '20 at 18:41
"Let $f\colon{0,1}^\Bbb N\to\Bbb R$ be $f(x)=0$, then $f$ is not injective. When you say "let ..." you are taking an arbitrary object satisfying your assumptions (which may or may not define a unique object, or be impossible to satisfy (e.g. "let $x$ be a natural number such that $x+1=0$")), so unless you specify what is $f$, you cannot prove it is injective, since there are such functions $f$ which are not injective. – Asaf Karagila Dec 15 '20 at 18:58

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