Prove that the countable product of $ \ A \ $ has same cardinality to $ \ \mathbb{R} \ $

Question

Let $ \ A=\{0,1\} \ $. Prove that the countable product of $ \ A=\{0,1 \} \ $ has same cardinality to $ \ \mathbb{R} \ $.

Answer:

Let $ \ P(\mathbb{N}) \ $ be the power set of $ \ \mathbb{N} \ $.

Let $ \ \lambda: \prod_{i \in \mathbb{N}} A \to P(\mathbb{N}) \ $ (Cartesian product ) be the function defined as follows:

an element $ \ f \ $ in $ \ \prod_{i \in \mathbb{N}}A \ $ is a function $ \ f: \mathbb{N} \to \{0,1 \} \ $ .

So let $ \ \lambda(f)=\{ n \in \mathbb{N} |f(n)=0 \} \ $.

The proof will be complete if we show that $ \ \lambda \ $ is a bijective function.

I am unable to show $ \ \lambda \ $ is bijective .

Help me

Answer 1

If $\lambda(f) = \lambda(g)$ then $$f(n) = 0 \iff n \in \lambda(f) \iff n \in \lambda(g) \iff g(n) = 0$$ which also implies $f(n) = 1$ if and only if $g(n) = 1$ because $f $ and $g$ are functions to $\{0,1\}$. Therefore $f = g$. We conclude that $\lambda$ is injective.

Let $S \subseteq \mathbb{Z}^+$ be arbitrary and define $f: \mathbb{Z}^{+} \to \{0,1 \}$ as $$f(n) = \begin{cases} 0,& n \in S\\ 1,& n \notin S\\ \end{cases}$$

It is clear that $\lambda(f) = S$. Therefore $\lambda$ is surjective.

Answer 2

Consider the function

$$\chi\colon P(\mathbb N)\to\prod_{i\in \mathbb N}A$$

given by the function which sends a set $B\subset \mathbb N$ to the function $$\chi(B)\colon n\mapsto \begin{cases}0 \text{ , if }n\in B\\ 1\text{ , if }n\notin B\end{cases}$$

Then clearly $\chi$ is an inverse to $\lambda$, hence both are bijections.

Answer 3

Although it does not really matter, I suggest to define $\lambda(f) = \{ n \in \mathbb{Z}^{+} |f(n) = 1 \}$. If $f \ne g$, then there exists $n_0$ such that $f(n_0) \ne g(n_0)$. W.l.o.g. assume $f(n_0) = 0$, $g(n_0) = 1$. This implies that $\lambda(f) \ne \lambda(g)$ because $n_0 \notin \lambda(f)$, $n_0 \in \lambda(g)$. Hence $\lambda$ is injective. Now let $M \in P(\mathbb{Z}^+)$. Define $(x_i^M) \in \Pi_{i \in \mathbb{Z}^+} A$ by $x_i^M = 1$ if $i \in M$ and $x_i^M = 0$ if $i \notin M$. Then $\lambda((x_i^M) ) = M$, i.e. $\lambda$ is surjective.

Note that $\Pi_{i \in \mathbb{Z}^+} A$ is nothing else than the set of functions $f : \mathbb{Z}^+ \to A$. Then $\lambda(f) = \lambda^{-1}(1)$. Its inverse is $\mu : P(\mathbb{Z}^+) \to \Pi_{i \in \mathbb{Z}^+} A, \mu(M) = \chi_M$ = characteristic function of $M$.