Proof of Existence of a Sequence of Positive Real Numbers with Limit Zero

Write $1$ for the multiplicative identity of the underlying field, $-x$ for the additive inverse of $x$, $x-y$ for $x+(-y)$, and $x^{-1}$ for the multiplicative inverse of $x$ when $x\ne 0$. Set $2=1+1$. By claim 8 of Elementary Order Arithmetic in an Ordered Field we have $0<2$, the inverse $2^{-1}$ exists, and for every real $\varepsilon$ with $0<\varepsilon$ both $0<\varepsilon\cdot 2^{-1}$ and $\varepsilon\cdot 2^{-1}<\varepsilon$ hold. Properties of the order on $\mathbb{N}$ are those of Properties of the Order on the Natural Numbers.

Define a sequence $(h_k)_{k\in\mathbb{N}}$ in $\mathbb{R}$ by recursion on the natural numbers:

Step 1 (positivity). We show $0<h_k$ for every $k$ by the principle of induction. For $k=1$, claim 6 of Elementary Order Arithmetic in an Ordered Field gives $0<1=h_1$. If $0<h_k$, then claim 8 of that lemma applied with $\varepsilon=h_k$ gives $0<h_k\cdot 2^{-1}=h_{k+1}$.

Step 2 (strict decrease). For every $k$, Step 1 gives $0<h_k$, so claim 8 of Elementary Order Arithmetic in an Ordered Field applied with $\varepsilon=h_k$ gives $h_{k+1}=h_k\cdot 2^{-1}<h_k$.

Step 3 (monotonicity along the index). If $j,k\in\mathbb{N}$ and $j\le k$, then $h_k\le h_j$. Indeed, if $j=k$ this is reflexivity of $\le$. Otherwise $j<k$, and by claim 7 of Properties of the Order on the Natural Numbers there is $t\in\mathbb{N}$ with $k=j+t$. We induct on $t$. For $t=1$: $h_{j+1}<h_j$ by Step 2, hence $h_{j+1}\le h_j$. If $h_{j+t}\le h_j$, then $h_{j+t+1}<h_{j+t}$ by Step 2, so $h_{j+t+1}\le h_{j+t}\le h_j$ and transitivity gives $h_{j+t+1}\le h_j$.

Step 4 (the infimum). Let $S=\{h_k:k\in\mathbb{N}\}$, a nonempty subset of $\mathbb{R}$. By Step 1, $0\le h_k$ for every $k$, so $0$ is a lower bound for $S$ and $S$ is bounded below. By Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below the greatest lower bound $c=\inf S$ exists, and $0\le c$ because $0$ is a lower bound and $c$ is a greatest lower bound.

Step 5 ($2c$ is also a lower bound). Let $k\in\mathbb{N}$. Since $c$ is a lower bound for $S$ and $h_{k+1}\in S$, we have $c\le h_{k+1}=h_k\cdot 2^{-1}$. As $0<2$ implies $0\le 2$, claim 5 of Elementary Arithmetic in an Ordered Field gives $2c\le 2\,(h_k\cdot 2^{-1})$. By commutativity and associativity of multiplication in a field, together with $2\cdot 2^{-1}=1$ and $h_k\cdot 1=h_k$,

Hence $2c\le h_k$ for every $k$, so $2c$ is a lower bound for $S$.

Step 6 ($c=0$). Since $c$ is the greatest lower bound, Step 5 gives $2c\le c$. By distributivity, $2c=(1+1)c=c+c$. By claim 3 of Elementary Arithmetic in an Ordered Field, $2c\le c$ is equivalent to $0\le c-2c$. Using claim 6 of Additive Cancellation and Elementary Additive Identities in a Field for $-(c+c)=(-c)+(-c)$, together with associativity of addition and claims 3 and 4 of that lemma,

so $0\le -c$. By the sign reversal in claim 4 of Elementary Order Arithmetic in an Ordered Field, this is equivalent to $-(-c)\le -0$, and claims 4 and 5 of Additive Cancellation and Elementary Additive Identities in a Field give $-(-c)=c$ and $-0=0$; hence $c\le 0$. With $0\le c$ from Step 4 and antisymmetry of $\le$, we conclude $c=0$.

Step 7 (limit). Let $\varepsilon$ be a real number with $0<\varepsilon$. By claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ applied to the nonempty set $S$, which is bounded below, there is $s\in S$ with $s<\inf S+\varepsilon=0+\varepsilon=\varepsilon$, the last equalities by Step 6 and the additive identity axiom of a field. By the definition of $S$ there is $K\in\mathbb{N}$ with $s=h_K$, so $h_K<\varepsilon$.

Let $k\in\mathbb{N}$ with $K\le k$. By Step 3, $h_k\le h_K$, and $h_K<\varepsilon$, so the mixed transitivity in claim 2 of Elementary Order Arithmetic in an Ordered Field gives $h_k<\varepsilon$. By claim 4 of Additive Cancellation and Elementary Additive Identities in a Field we have $h_k-0=h_k$, and since $0<h_k$ the absolute value satisfies $|h_k|=h_k$. Hence $|h_k-0|<\varepsilon$ for every $k$ with $K\le k$.

Since $\varepsilon>0$ was arbitrary, Limit of a Sequence of Real Numbers shows that $(h_k)_{k\in\mathbb{N}}$ has limit $0$. Together with Step 1 this exhibits a sequence with the required properties.