Infimum and supremum for a set

Question

Suppose $A\subseteq\mathbb{R}$ is bounded from below and $a=\inf(A)$. Show that $$ a=\inf(A)\quad\Leftrightarrow\quad a=\sup\{c\in\mathbb{R}: x>c~\forall x\in A\} $$

Intuitively, if $a$ is the infimum of the set $A$, it is the largest lower bound of $A$ and thus the smallest upper bound, i.e. the supremum, of the set $M:=\{c\in\mathbb{R}:x>c~\forall x\in A\}$, and vice versa.

But formally...

I would do it this way:

$\Rightarrow$:

Let $a=\inf A$ and $c\in M$. Then, $c$ is a lower bound of $A$ and thus smaller than $a$ since $a$ is the largest lower bound, i.e. $c\leqslant a$. But then $a$ is the smallest upper bound on $M$ since suppose there exists $\tilde{a}$ such that $c\leqslant\tilde{a}<a$ for some $c\in M$, then this contradicts $a=\inf A$.

Answer 1

Every element of the set $C:=\{c\in\mathbb R\mid\forall x\in A[x>c]\}$ is by definition a lower bound of $A$.

Also by definition $a$ is the greatest lower bound (=infinum) of $A$ so we must have $c\leq a$ for every $c\in C$.

That makes $a$ an upper bound of $C$.

It now remains to prove that $a$ is the least upper bound (=supremum) of $C$.

For every $x<a$ some $y$ exists with $x<y<a$.

From $y<a$ it follows easily that $y\in C$ and then from $x<y$ it follows that $x$ is not an upperbound of $C$.

Proved is now that $a$ is the least upper bound of $C$.

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Edit:

Conversely let it be that $a=\sup C$.

If $x\in A$ then clearly $x$ is an upper bound of $C$ and we conclude that $a\leq x$.

This tells us that $a$ is a lower bound of $A$.

It remains to be shown that $a$ ist the greatest lower bound of $A$ i.e. that every $y>a$ is not a lower bound of $A$.

If $y>a$ is a lower bound of $A$ then for every $x<y$ we find that $x\in C$.

So taking $y>x>a$ then leads to $x\in C$ and this contradicts $a=\sup C$.

We conclude that condition $y>a$ makes it impossible for $y$ to be a lower bound of $A$.

Proved is now that $a=\inf A$.

Answer 2

you can use the following characterization :$$a=\inf(I) \Leftrightarrow [\forall x\in I, x\geq a \land \forall \epsilon>0,\exists x_{\epsilon} \in I, x_{\epsilon} <a+\epsilon] $$

Answer 3

Let $\alpha=\inf(A).$ By definition, it is the greatest lower bound of $A,$ i.e. $$\alpha=\max(N)\quad\text{where}\quad N=\{c\in\Bbb R\mid\forall x\in A\quad c\le x\}.$$ Now, your $M:=\{c\in\Bbb R\mid\forall x\in A\quad c<x\}$ clearly satisfies $$(-\infty,\alpha)\subseteq M\subseteq N=(-\infty,\alpha]$$ hence $\sup(M)=\alpha.$