# Proof

(related to Theorem: Supremum Property, Infimum Property)

Let $$D$$ be non-empty subset of real numbers with an upper bound, say $$B_0$$. We will to show that $$D$$ has a supremum by constructing it (The existence of an infimum for a non-empty lower-bounded subset of real numbers can be proven in a similar way.)

We consider any $$x_0\in D$$. Then we have $$x_0 < B_0$$, or, equivalently, $$B_0 - x_0 > 0$$. This is because $$B_0$$ is an upper bound, by assumption. We now construct sequences of real numbers. * $$x_0 \le x_1 \le x_2 \le \ldots$$, $$x_i\in D$$ and * $$B_0 \ge B_1 \ge B_2 \ge \ldots$$ of upper bounds of $$D$$

with the following property:

$B_n-x_n\le \frac{B_0-x_0}{2^n}\quad\quad\text{for all }n\in\mathbb N\quad\quad( * )$

in the following recursive way:

1. Having constructed the first $$n$$ sequence members $$x_0 \le \ldots\le x_n$$ and $$B_0 \ge \ldots\ge B_n$$, we set $$M:=\frac{B_n-x_n}{2}$$ as the middle of the closed real interval $$[x_n,B_n]$$.
2. If $$M$$ is an upper bound of $$D$$, then we set $$x_{n+1}:=x_n$$ and $$B_{n+1}:=M$$.
3. Otherwise, there exists an $$x_{n+1}\in D \cap (M,B_n]$$ and we set $$B_{n+1}:=B_{n}$$.
4. We repeat all these steps "infinitely many times".

In all steps, the required properties $$x_n \le x_{n+1}$$, $$B_n \ge B_{n+1}$$, and $$( * )$$ are all preserved. Note that the sequence $$(B_n)_{n\in\mathbb N}$$ is monotonically decreasing with the lower bound $$x_0$$. Therefore, according to the monotone convergence theorem, it is convergent to some limit $$B$$:

$\lim_{n\to\infty}B_n=B.$

Because convergence preserves upper bounds for sequence members, and because for any $$x\in D$$, we have by construction $$x\le B$$, $$B$$ proves to be an upper bound of $$D$$. It remains to be shown that there is no smaller upper bound of $$D$$ than $$B$$. Take any $$B'$$ with $$B' < B$$. We will show that $$B'$$ is no more an upper bound of $$D$$.

With a growing $$n$$, the number $$2^{-n}$$ becomes arbitrarily small. Therefore, we can find an $$n$$ such that for any $$B' < B$$ we get $\frac{B_0-x_0}{2^n} < B - B'.\quad\quad ( * * )$

It follows that

$\begin{array}{rcll} x_n&\ge& B_n-\frac{B_0-x_0}{2^n}&\text{by virtue of }( * )\\ &\ge& B-\frac{B_0-x_0}{2^n}&\text{sequence }(B_n)_{n\in\mathbb N}\text{ is monotonically decreasing}\\ & > & B'&\text{by virtue of }( * * ) \end{array}$

Therefore, there is no smaller upper bound of $$D$$ than $$B$$. Thus, $$B$$ is a supremum.

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983