Build $\mathbb{R}$ from $\mathcal{P}(\mathbb{N})$

It's well known that $\mathbb{R}$ has the same cardinality as $\mathcal{P}(\mathbb{N})$; but I would fain know if there is a way to construct $(\mathbb{R}, +,\cdot, \leq )$ using only definitions that rely upon $\mathcal{P}(\mathbb{N})$'s elements and their respective properties.

Solution 1:

Although the variant construction I propose below still uses rationals, it is more directly and naturally related to ${\cal P}(\mathbb N)$ (and not to one of its quotients by some equivalence relation) and so gets closer to the "perfect" solution the OP is expecting. As noted in other answers here, it suffices to construct the field structure on the set $\cal S$ of inifinite sequences of zeroes and ones (thanks to characteristic functions).

Consider the following four types of intervals (in $\mathbb Q$) with rational endpoints :

$$ T_1 = {\mathbb Q}, T_2(a) = ]-\infty,a[ \ (a\in {\mathbb Z}), T_3(a) = [a,+\infty [ \ (a\in {\mathbb Z}), T_4(a,b)= [a,b[ \ (a,b\in {\mathbb Q}) $$

Let us denote by $\cal T$ the set of all those intervals. We define a "splitting map" $\sigma=(\sigma_1,\sigma_2)$ from $\cal T$ to ${\cal T}^2$. This map $\sigma$ splits any $I\in \cal T$ into two parts $\sigma_1(I)$ and $\sigma_2(I)$ such that $\sigma_1(I)$ is always "to the left" of $\sigma_2(I)$ (in the sense that $x\leq y$ whenever $x\in {\sigma}_1(I), y\in {\sigma}_2(I)$).

Here is the formal definition of $\sigma$ :

$$ \sigma(T_1) = (T_2(0),T_3(0)), \\ \sigma(T_2(a)) = (T_2(a-1),T_4(a-1,a)), \\ \sigma(T_3(a)) = (T_4(a,a+1),T_3(a+1)), \\ \sigma(T_4(a,b))=(T_4(a,\frac{a+b}{2}),T_4(\frac{a+b}{2},b)) $$

Now for any finite sequence $(a_1, \ldots ,a_n)$ with each $a_k$ in $\lbrace 0,1 \rbrace$, define an interval $I_{(a_1, \ldots ,a_n)}$ inductively~: for the empty sequence $()$ we set $I_{()}=T_1$, and for the others sequences we define $I_{(a_1, \ldots ,a_{n-1},0)}=\sigma_1(I_{(a_1, \ldots ,a_{n-1})})$ and $I_{(a_1, \ldots ,a_{n-1},1)}=\sigma_2(I_{(a_1, \ldots ,a_{n-1})})$.

For any $n\geq 1$ and $\varepsilon \in \lbrace 0,1 \rbrace$ let $J(n,\varepsilon)$ denote the union of all $I_{a}$ where $a$ is an $n$-element sequence ending with $\varepsilon$.

Let $a=(a_1,a_2, \ldots )$ and $b=(b_1,b_2, \ldots )$ be two sequences in $\cal S$. Recall that a (finite) initial segment of $a$ is a sequence of the form $(a_1,a_2, \ldots ,a_k)$ for some $k$.

For any $n\geq 1$, define $c_n$ to be $0$ if $I_{a'}+I_{b'} \subseteq J(n,0)$ for some initials segments $a',b'$ of $a$ and $b$ respectively, and $1$ otherwise. Then the Feanor sum $a+b$ of $a$ and $b$ is the sequence $c=(c_1,c_2, \ldots )$.

For any $n\geq 1$, define $d_n$ to be $0$ if $I_{a'} \times I_{b'} \subseteq J(n,0)$ for some initials segments $a',b'$ of $a$ and $b$ respectively, and $1$ otherwise. Then the Feanor product $a \times b$ of $a$ and $b$ is the sequence $d=(d_1,d_2, \ldots )$.

Finally, define the Feanor order on $\cal S$ as follows $ a \leq b$ iff $a=b$, or $I_{a'}$ is "to the left" of $I_{b'}$ (in the sense that $x\leq y$ whenever $x\in I_{a'}, y\in I_{b'}$) for some initials segments $a',b'$ of $a$ and $b$ respectively.

Then $\cal S$ endowed with Feanor addition, multiplication and order has all the properties required of $\mathbb R$.

Solution 2:

I will construction $\mathbb R$ from sequences in $\mathbb{Z}$. This is not precisely a construction from $\mathcal{P}(\mathbb N)$, but it can be turned into one by observing that every sequence in $\mathbb Z$ can be constructed from $\mathcal{P}(\mathbb{N})$ by mapping

$$\left\{x_1+2^{\sigma_1}1\cdot10^{\lceil\log_{10}x_1\rceil+1},x_2+2^{\sigma_2}11\cdot10^{\lceil\log_{10}x_2\rceil+1},x_3+2^{\sigma_3}111\cdot10^{\lceil\log_{10}x_3\rceil+1},\ldots\right\}\mapsto \left((-1)^{\sigma_n}x_n\right)_{n\in\mathbb N}$$

where each $\sigma_i\in\{0,1\}$ is chosen to determine the sign. This can be done independent of the ordering of the set, as we can tell which element is $x_n+2^{\sigma_n}1\cdots1\cdot10^{\lceil\log_{10}x_1\rceil+1}$ by counting the number of $1$'s or $2$'s before the first $0$. There are of course many other ways this could be done; this is merely the first one I thought of.

I cannot take credit for the method below. It was invented by Steve Schanuel and was given to me in sketch form in a real analysis problem set. Here is the sketch of the method with details filled in by me:

Let $G$ be the collection of sequences $f:\mathbb N\to\mathbb Z$ so that there is a constant $C$ so that $$\forall m,n\in\mathbb N, |f(m)+f(n)-f(m+n)|\leq C.$$ Let $H$ be the collection of all $f:\mathbb N\to\mathbb Z$ so that there is a constant $C$ satisfying $|f(n)|\leq C$ for all $n\in\mathbb N$.

(1) Prove that if $f\in G$, then $\lim\limits_{n\to\infty} f(n)/n$ exists in $\mathbb R$.

Proof: To begin, we note that if $f,C$ are such that $|f(n)+f(m)-f(n+m)|\leq C,\forall n,m\in\mathbb N$, then $|nf(m) - f(nm)|\leq (n-1)C,\forall n,m\in\mathbb N$ by induction on $n$. The base case of this ($n=1$) is trivial, as $|f(m)-f(m)|=0\leq 0,\forall m\in\mathbb N$, while if we have that $|nf(m) - f(nm)|\leq (n-1)C$ then $$\begin{eqnarray}|(n+1)f(m) - f((n+1)m)| &=& |nf(m) + f(m) - f(nm+m)|\\ &\leq& |nf(m) - f(nm)| + |f(nm) + f(m) - f(nm+m)|\\ &\leq& (n-1)C + C = nC.\end{eqnarray}$$ Thus for $f\in G$ and $m>n\geq N\in \mathbb N$ we have that $$\begin{eqnarray}|f(m)/m - f(n)/n| &=& |nf(m) - mf(n)|/(nm)\\ &=& |nf(m) - f(nm) - (mf(n) - f(nm))|/(nm)\\ &\leq& (|nf(m)-f(nm)| + |mf(n)-f(nm)|)/(nm)\\ &\leq& ((n-1)C+(m-1)C)/(nm)\\ &\leq& C(n+m)/(nm)\leq 2C/n\leq 2C/N\end{eqnarray}$$ so given any $\epsilon>0$ we can choose $N$ such that $N > 2C/\epsilon$ thus $|f(m)/m - f(n)/n|\leq 2C/N < \epsilon$. From this we see that the sequence $f(n)/n,n\in\mathbb N$ is Cauchy, so by the completeness of $\mathbb R$ we see that $\lim\limits_{n\to\infty} f(n)/n$ exists in $\mathbb R$.

(2) If $f,g\in G$, show that $\lim\limits_{n\to\infty} f(n)/n=\lim\limits_{n\to\infty} g(n)/n$ if and only if $f-g\in H$.

Proof: Let $f,g\in G$ with the corresponding constants $A$ and $B$. Assume first that $f-g\in H$, so there exists some constant $C$ such that $|f(n)-g(n)|\leq C,\forall n\in\mathbb N$. Since $\lim\limits_{n\to\infty} f(n)/n$ and $\lim\limits_{n\to\infty} g(n)/n$ exist, we have that $\lim\limits_{n\to\infty} f(n)/n - \lim\limits_{n\to\infty} g(n)/n$ equals $\lim\limits_{n\to\infty} (f(n)-g(n))/n$, and since $|(f(n) - g(n))/n|\leq C/n$ and $\lim\limits_{n\to\infty} C/n = C\lim\limits_{n\to\infty} 1/n = 0$ we have $\lim\limits_{n\to\infty} f(n)/n - \lim\limits_{n\to\infty} g(n)/n = 0$ thus $\lim\limits_{n\to\infty} f(n)/n = \lim\limits_{n\to\infty} g(n)/n$. If on the other hand we have $\lim\limits_{n\to\infty} f(n)/n = \lim\limits_{n\to\infty} g(n)/n$ then we have $\lim\limits_{n\to\infty} (f(n)-g(n))/n = 0$. It is important to note that $n,m\in\mathbb N$ $$|f(n) - g(n) + f(m) - g(m) -(f(m+n) - g(n+m))|< A + B$$ which gives us that $$(n-1)(A+B)\geq |n(f(m)-g(m)) - f(nm)+g(nm)| \geq |n|f(m)-g(m)| - |f(nm)-g(nm)||$$ thus $$n|f(m)-g(m)| - (n-1)(A+B) \leq |f(nm)-g(nm)|\leq n|f(m)-g(m)|+(n-1)(A+B)$$ If we have some $m\in\mathbb N$ such that $|f(m)-g(m)| > A + B+1$, then we have $$n|f(m)-g(m)| - (n-1)(A+B) > n(A+B+1) - (n-1)(A+B) = n + A + B > n$$ so $|f(nm)-g(nm)|/(nm) > 1/m,\forall n\in\mathbb N$, contradicting the fact that $\lim\limits_{n\to\infty} (f(n)-g(n))/n = 0$. Hence we must have $|f(m)-g(m)| \leq A + B+1,\forall m\in\mathbb N$ thus $f-g\in H$.

(3) For $f,g\in G$, we put $f\sim g$ if $f-g\in H$. Prove that $\sim$ is an equivalence relation. The set of equivalence classes is denoted by $G/H$. The equivalence class of $f\in G$ is denoted by $[f]$. Show that there is a function $j:G/H\to\mathbb R$ so that $j([f])=\lim\limits_{n\to\infty} f(n)/n$ for all $f\in G$. Show that $j$ is one-to-one.

Proof: Let $f,g,h\in G$ be such that $f\sim g$ and $g\sim h$, thus $f-g,g-h\in H$. Since $|f(n)-f(n)|\leq 0,\forall n\in \mathbb N$ and $|g(n)-f(n)| = |f(n)-g(n)|$ we see that $f-f,g-f\in H$ so $f\sim f,g\sim f$ giving us the reflexive and commutative properties, while the fact that $|f(n) - h(n)|\leq |f(n)-g(n)|+|g(n)-h(n)|$, both of which are bounded, gives us $f-h\in H$ so $f\sim h$ thus $\sim$ is transitive as well. Hence $\sim$ is an equivalence relation. We can define a function $j:G/H\rightarrow \mathbb R$ by setting $j([f]) = \lim\limits_{n\to\infty} f(n)/n$, which is well-defined because $[f] = [g]\implies f - g\in H$ thus $\lim\limits_{n\to\infty} f(n)/n = \lim\limits_{n\to\infty} g(n)/n$. We can also see that $j$ is one-to-one, as if $j([f])=j([g])$ then $\lim\limits_{n\to\infty} f(n)/n = \lim\limits_{n\to\infty} g(n)/n$ thus $f-g\in H$ so $[f] = [g]$.

(4) Prove that $j:G/H\to\mathbb R$ is a bijection.

Proof: We already have that $j$ is one-to-one, so we need only prove that $j$ is surjective. Let $r\in \mathbb R$ be an arbitrary real number and define $f(n)$ as the least integer greater than or equal to $rn$. For $m,n\in\mathbb N$ we have $$|f(n)+f(m)-f(n+m)|\leq 3 + |nr + mr - (n+m)r| = 3$$ so $f\in G$. We also have that $|f(n)/n - r| = |f(n) - rn|/n < 1/n$ so for any $k\in\mathbb N$, $n\geq k\implies |f(n)/n - r| < 1/k$, hence $\lim\limits_{n\to\infty} f(n)/n = r$. This gives us $j([f]) = r$, completing the proof.

Thus we have constructed $\mathbb R$.