# Show the equivalence of the following statements (Supremum, Infimum)

Let $$A\subset \mathbb{K}$$, $$s\in \mathbb{K}$$. ($$\mathbb{K}$$ is an ordered field!). Show the equivalence of the following statements:

(i) $$s=\sup A$$

(ii) $$s$$ is an upper bound of $$A$$ and for all $$\epsilon > 0$$ exists a $$x\in A$$ with $$x>s-\epsilon$$

(iii) $$s$$ is an upper bound of $$A$$ and for all $$\epsilon > 0$$ exists a $$x\in A$$ with $$x\geq s-\epsilon$$

I wanted to show the the equivalence like that: $$(i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (i)$$ (don't know what it's called in English).

Definitions:

$$s \in \mathbb{K}$$ is an upper bound of $$A$$ if $$\tilde{s}\geq x$$ for all $$x\in A$$.

An upper bound is called Supremum of $$A$$ if $$s\leq \tilde{s}$$ for all upper bounds $$\tilde{s}$$ of $$A$$.

My attempt:

(i) $$\Rightarrow$$ (ii)

Since, by definition, $$\tilde{s}\geq x \land s\leq \tilde{s}$$, one can deduce that $$s=x$$. It should now be trivial to show that $$x>s-\epsilon$$ if $$\epsilon>0$$

(ii) $$\Rightarrow$$ (iii)

Since we have already proven that there exists a $$x\in A$$ with $$x> s-\epsilon$$, we solely have to prove that there are $$x\in A$$ with $$x=s-\epsilon$$. But I don't know how to.

(iii) $$\Rightarrow$$ (i)

...

• I dont understand: How does $a \ge b$ and $c \le a$ imply $b = c$? You don't say where $\tilde{s}, s,x$ come from so this is not true in general, i.e. consider $a = 4$, $b = 2$ and $c = 3$. – Viktor Glombik May 5 at 12:16

## 1 Answer

(i) $$\implies$$ (ii): It is clear that, if $$s=\sup A$$, then $$s$$ is an upper bound of $$A$$. Take $$\varepsilon>0$$. Since $$s$$ is the least upper bound of $$A$$, $$s-\varepsilon$$ is not an upper bound of $$A$$, which means that there is a $$x\in A$$ sych that $$x\geqslant s-\varepsilon$$.

(ii) $$\implies$$ (iii): There is some $$x\in A$$ such that $$x>s-\varepsilon$$. But then $$x\geqslant s-\varepsilon$$.

(iii) $$\implies$$ (i): Suppose that $$s\neq\sup S$$. Then here is some upper bound $$s'$$ of $$A$$ such that $$s'. Take $$\varepsilon=\frac{s-s'}2$$. There is some $$x\in A$$ such that $$x\geqslant s-\varepsilon=s-\frac{s-s'}2=\frac{s+s'}2>s'$$. This is absurd, since $$s'$$ is an upper bound of $$A$$.

• "(ii) $\Rightarrow$ (iii)": How is that a proof? Does $x>s-\varepsilon$ imply $x\geqslant s-\varepsilon$, and why? – Analysis May 5 at 12:23
• Since $>$ means “larger than” and $\geqslant$ means “larger than or equalt to”, it is obvious that $a>b\implies a\geqslant b$. – José Carlos Santos May 5 at 12:26
• Ahh, you're right...! – Analysis May 5 at 12:32
• But why do you use $\geqslant$ instead of $\geq$? – Analysis May 5 at 12:39
• Since $s'<s$, $\frac{s-s'}2>0$. So, I can take $\varepsilon=\frac{s-s'}2$. And I did take that $\varepsilon$ because, with it, I was able to reach a contradiction. – José Carlos Santos May 5 at 13:09