Proposition. There is a bijective correspondence between $\mathscr P{(\mathbb{Z_+})}$, the power set of $\mathbb{Z_+}$, and $X^\omega$, the set of all infinite sequences of elements of $X$, where $X=\{0,1\}$.
(Source: Munkres Topology 7.3)
Proof
Let $f: \mathscr P{(\mathbb{Z_+})} \to X^\omega$.
Define $f(A) = (x_i)_{i\in\mathbb{Z_+}}$, for $A \in \mathscr P{(\mathbb{Z_+})}$, such that if $i \in A$, $x_i = 1$, otherwise $x_i = 0$.
We first show $f$ is an injection. Suppose $A,A' \in \mathscr P{(\mathbb{Z_+})}$ such that $A=A'$. It follows that $A,A' \subset \mathbb{Z_+}$.
Consider $f(A)$ and $f(A')$. Then $f(A) = x \in X^\omega$ and $f(A') = x^{'} \in X^\omega$. By definition of $X^\omega$, $x = (x_i)_{i\in\mathbb{Z_+}}$ and $x^{'} = (x^{'}_i)_{i\in\mathbb{Z_+}}$ for $x_i$, $x^{'}_i \in X$.
It follows from the definition of $f$, that $\forall j\in\mathbb{Z_+}$, if $j\in A$, then $x_j = 1$. Since $A = A'$ by assumption, $\forall j\in A$, $j\in A'$ and $x^{'}_j = 1$. Hence $\forall j\in \mathbb{Z_+} \cap A = \mathbb{Z_+} \cap A^{'}$, $x_j = 1 = x^{'}_j$. Further, $\forall j\in \mathbb{Z_+} – A = \mathbb{Z_+} – A^{'}$, $x_j = 0 = x^{'}_j$. Thus $\forall j\in\mathbb{Z_+}$, $x_j = x^{'}_j$. Hence, $x = x^{'}$. Therefore, $f$ is an injection.
To show $f$ is a surjection, arbitrarily choose $x \in X^\omega$. Then $x = (x_i)_{\mathbb{i\in Z_+}}$, such that each $x_i\in X = \{0,1\}$. Construct a set $A$, as follows.For each $i\in \mathbb{Z_+}$, if $x_i=1$, let $i$ be an element of $A$. Whether empty or not, $A \subset \mathbb{Z_+}$. Thus $A \in\mathscr{P(\mathbb{Z_+})}$. Hence $\forall x\in X^\omega$, $\exists A\in\mathscr{P(\mathbb{Z_+})}$, such that $x = f(A)$. Therefore, $f$ is a surjection.
We have show there exists a bijective correspondence between $\mathscr P{(\mathbb{Z_+})}$ and $X^\omega$.
Question
I would appreciate a verification of this proof. Also, is there a better way to notate the definition of $f$ above.
Best Answer
Your bijection is fine, but you’ve not shown that it’s injective: to do that you have to show that if $f(A)=f(A')$, then $A=A'$. Clearly you can’t do this by assuming that $A=A'$ to begin with. What you’ve shown is actually that $f$ is well-defined: if $A=A'$, then $f(A)=f(A')$. If you turn the argument around, starting with the assumption that $f(A)=f(A')$ and proving from that that $A=A'$, you’ll have shown that $f$ is injective. The argument can also be tightened up a bit:
Your proof that $f$ is surjective is fine, though again it can be tightened up a little: just note that if $A=\{i\in\Bbb Z_+:x_i=1\}$, then $f(A)=x$.