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Problem

Let $D$ be the set of all the functions $f\colon\mathbb N\to \mathbb N$ , where $\mathbb N$ is the set of natural numbers.

Let $E$ be the set of all functions $f\colon \mathbb N\to\{\,0,1\,\}$.

Prove both the sets have equal cardinality.

Attempt (failed)

$E$ is the set of all infinite binary sequences. By Cantor's second diagonal argument $E$ is uncountable. Since $E\subset D$, $D$ is also uncountable. If i prove there exists injections from $D$ to $E$ and also from $E$ to $D$, then Schroeder-Bernstein theorem can be used.

Attempt 2

Consider two functions $\alpha\colon D \to E$ and $\beta\colon E \to D $

Consider a function $f \in E$ is a function from $\mathbb N$ to $\{\,0,1\,\}$. $\beta(f) $ is a function that assigns to $n$ the $f(n)+1$ . This is a injective function .

I have difficulty finding $\alpha$.

Any hint or help will be appreciated.

Asaf Karagila
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Bluey
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3 Answers3

3

Sketch:

As you remark, we just need an injection of $D$ into $E$.

Consider $F\in D$, and let $p_i$ denote the $i^{th}$ prime. Define a function $\mathscr F\in E$ by letting $F(p_i^n)$ denote the $n^{th}$ digit in the binary expansion of $F(i)$. If $m$ is not a prime power just let $\mathscr F(m)=0$. Now just verify that the map $\Phi: D\to E$, $F\mapsto \mathscr F$ is injective.

lulu
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2

They are both equal to the cardinality of (0, 1)

dmtri
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1

In general, the set of all functions from a set $A$ to a set $B$ has cardinality $card(B)^{card(A)}$. So basically you nee to show that $\aleph_0^{\aleph_0}=2^{\aleph_0}$.

Nicky Hekster
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