Least Upper Bound of R (via Dedekind Cuts)

We'll prove the following:

here's the definition if you need:

But we'll be proving this in the context of dedekind cuts:

Let $\mathcal{A}$ be a nonempty subset of $\mathbb{R}$ which is bounded above. As the picture above shows, if we let $\alpha =\bigcup _{A_i\in \mathcal{A}}{A}_i$, then $\alpha$ will be the least upper bound.

First, let's verify it's a cut:

Proof

1. ($\mathrm{\varnothing }\ne \alpha \ne \mathbb{Q}$). Clearly $\mathcal{A}\ne \mathrm{\varnothing }\Rightarrow A\in \mathcal{A}$ and since $A$ is a cut then $A\ne \mathrm{\varnothing }$. Then $\mathrm{\exists }a\in A$ so clearly $a\in \alpha$ by definition. Similarly, notice that $\alpha \ne \mathbb{Q}$ as follows: let $B$ be an upper bound of $\mathcal{A}$ (as given by the definition of $\mathcal{A}$). Since $B$ is a cut, then $B\ne \mathbb{Q}$ so $\mathrm{\exists }y\notin B$. Then $y\notin \alpha$ because for any arbitrary $a\in \alpha$ then $a\in A_i$ for some $\mathrm{\exists }{A}_i\in \mathcal{A}$. $A_i$ is bounded by $B$, so then $A_i\subseteq B$. Thus then $a\in B$. As a result, then $a\le y$ (as otherwise then $y\in B$ by the definition of a cut, which is a contradiction). Since $a$ was arbitrary, then $\mathrm{\forall }a\in \alpha (a\le y)$, so choose for example $y+1\notin \alpha$ which shows $\alpha \ne \mathbb{Q}$.
2. ($\alpha$ satisfies (2) of a cut): Let $a\in \alpha$ be arbitrary. As such then $\mathrm{\exists }A\in \mathcal{A}$ where $a\in A$. Let $q<a$. We'll show $q\in \alpha$. Since $A$ is a cut, and $q<a$, then $a\in A$. By the definition of $\alpha$ then $q\in \alpha$.
3. ($\alpha$ satisfies (3) of a cut): Let $a\in \alpha$ so $\mathrm{\exists }A\in \mathcal{A}$ s.t. $a\in A$. Since $A$ is a cut then $\mathrm{\exists }p>a$ s.t. $p\in A$. But then that means that $p\in \alpha$. Thus there always is $p$ where $p>a$ and $p\in \alpha$.
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   Now we need to show that $\alpha$ is an upper bound, and that $\alpha$ is the least upper bound.

Proof

1. ($\alpha$ is an upper bound of $\mathcal{A}$): $A\le \alpha \iff A\subseteq \alpha$.
2. ($\alpha$ is the least upper bound of $\mathcal{A}$) for any other upper bound $B$ has $\alpha \le B$ so $\alpha$ is the smallest one.
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• #todo/school : Verify this is the right proof idea. 📅 2024-10-03 ✅ 2024-10-07