[proofplan] For each coordinate $n\in\omega$, we show that the set of Cohen conditions deciding the $n$th bit is dense and belongs to $M$. Genericity then gives a condition in $G$ whose domain contains $n$, so every coordinate receives at least one value. The filter property gives compatibility of any two conditions in $G$, and since Cohen conditions are functions, two conditions in $G$ cannot assign different values to the same coordinate. This produces a unique bit at each $n$, and the canonical interpretation of $\dot c$ identifies the resulting binary sequence with the characteristic function of $\dot c^G$. [/proofplan] [step:Meet the dense set of conditions deciding the coordinate $n$] Fix $n\in\omega$. Define \begin{align*} D_n=\{p\in \operatorname{Add}(\omega,1): n\in \operatorname{dom}(p)\}. \end{align*} Since $M$ is transitive and satisfies ZFC, every $n\in\omega$ belongs to $M$. The set $D_n$ belongs to $M$ by Separation in $M$, applied to the formula $n\in\operatorname{dom}(p)$ over the parameter set $\operatorname{Add}(\omega,1)\in M$ and the parameter $n\in M$. We verify that $D_n$ is dense in $\operatorname{Add}(\omega,1)$. Let $p\in \operatorname{Add}(\omega,1)$. If $n\in\operatorname{dom}(p)$, then $p\in D_n$. If $n\notin\operatorname{dom}(p)$, define $q:\operatorname{dom}(p)\cup\{n\}\to 2$ by requiring $q|_{\operatorname{dom}(p)}=p$ and $q(n)=0$. Then $q\in \operatorname{Add}(\omega,1)$, $q\le p$, and $q\in D_n$. Thus $D_n$ is dense. Since $G$ is $M$-generic and $D_n\in M$ is dense, $G\cap D_n\ne\varnothing$. Therefore there exists $p_n\in G$ such that $n\in\operatorname{dom}(p_n)$. [guided] Fix a coordinate $n\in\omega$. To prove that the Cohen real is a genuine total sequence, we first need some condition in the generic filter to say what the $n$th bit is. The relevant dense set is \begin{align*} D_n=\{p\in \operatorname{Add}(\omega,1): n\in \operatorname{dom}(p)\}. \end{align*} This set consists exactly of the finite binary conditions that already decide the value at coordinate $n$. We check density directly. Let $p\in \operatorname{Add}(\omega,1)$ be arbitrary. If $n\in\operatorname{dom}(p)$, then $p$ itself lies in $D_n$. If $n\notin\operatorname{dom}(p)$, we extend $p$ by assigning a value at $n$. Define $q:\operatorname{dom}(p)\cup\{n\}\to 2$ by $q|_{\operatorname{dom}(p)}=p$ and $q(n)=0$. The domain of $q$ is finite, $q$ maps its domain into $2$, so $q\in\operatorname{Add}(\omega,1)$. Since the order is extension, $q\supset p$ means $q\le p$. Also $n\in\operatorname{dom}(q)$, so $q\in D_n$. Hence below every condition there is a condition in $D_n$. Because $M$ is transitive, the fixed natural number $n\in\omega$ is an element of $M$. Because $M$ satisfies Separation and $\operatorname{Add}(\omega,1)\in M$, the subset of $\operatorname{Add}(\omega,1)$ defined inside $M$ by the condition $n\in\operatorname{dom}(p)$ belongs to $M$; that subset is precisely $D_n$. Since $G$ is $M$-generic, it meets every [dense subset](/page/Dense%20Subset) of $\operatorname{Add}(\omega,1)$ that belongs to $M$. Therefore $G\cap D_n\ne\varnothing$, so there is a condition $p_n\in G$ with $n\in\operatorname{dom}(p_n)$. This condition assigns some bit $p_n(n)\in 2$ to the coordinate $n$. [/guided] [/step] [step:Show that two conditions in $G$ assign the same value to $n$] Let $p,q\in G$ satisfy $n\in\operatorname{dom}(p)\cap\operatorname{dom}(q)$. Since $G$ is a filter in the reverse-inclusion ordering, there exists $r\in G$ such that $r\le p$ and $r\le q$. The order is extension, so $r\supset p$ and $r\supset q$. Therefore \begin{align*} p(n)=r(n)=q(n). \end{align*} Thus the value assigned to $n$ by a condition in $G$ is unique. [/step] [step:Define the total binary sequence determined by the generic filter] For each $n\in\omega$, choose any $p_n\in G$ with $n\in\operatorname{dom}(p_n)$, whose existence was proved above. Define $c:\omega\to 2$ by \begin{align*} c(n)=p_n(n). \end{align*} The preceding uniqueness step shows that this definition is independent of the chosen condition $p_n$. Since every $n\in\omega$ has such a condition and every value $p_n(n)$ lies in $2$, the map $c:\omega\to 2$ is total. [/step] [step:Identify the sequence with the interpretation of the canonical Cohen name] By the definition of the interpretation of a name by a generic filter and by the definition of $\dot c$, \begin{align*} n\in \dot c^G \iff \text{there exists }p\in G\text{ such that }n\in\operatorname{dom}(p)\text{ and }p(n)=1. \end{align*} If $c(n)=1$, then the chosen condition $p_n\in G$ satisfies $p_n(n)=1$, so $n\in\dot c^G$. Conversely, if $n\in\dot c^G$, then some $p\in G$ satisfies $n\in\operatorname{dom}(p)$ and $p(n)=1$. By uniqueness of the value assigned at coordinate $n$, this forces $c(n)=1$. Hence \begin{align*} c(n)=1 \iff n\in \dot c^G. \end{align*} Since this holds for every $n\in\omega$, $\dot c^G$ determines the total function $c:\omega\to 2$ as claimed. [/step]