# Approve that if there is a lower bound for each $A \subseteq R$ then there is $infA$

I must approve that if there is a lower bound for each $$A\subseteq R$$ then there is $$infA$$. ($$A \neq \emptyset$$)

I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $$sup$$ and $$inf$$.

My solution:

Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that

$$B = \{ x \in R : x$$ is lower bound of A $$\}$$

$$B$$ is not empty because there is a $$m \leqslant x$$ for all $$x\in A$$.

Is $$infA =m$$?

Two cases:

1. Let us consider that $$m \neq infA$$ then $$\exists e > 0$$ such that $$\forall x \in A$$, $$m+e \leqslant x$$.

So, $$m+e \in B$$ and $$m+e>m$$.

We wonder if there is $$supB$$. If there is $$supB$$ then $$supB = infA$$

Let us assume that there isn't $$supB$$. Then there is $$e > 0$$ for each $$x>0$$ so that

$$x \leqslant M - e$$ where $$x \leqslant M$$ for each $$x \in B$$

Let us define $$M = m + e$$, then $$x \leqslant m$$, which is impossible because

$$m+e >m$$. Consequently, $$x > M -e$$.

As a result, $$supB=M=infA$$

2. If $$infA=m$$ then there is $$infA$$

From (1) and (2), there is always $$infA$$

Could you validate my solution ?

EDIT:

1. Can we assume that $$supB = infA$$ ?
2. Can we select the $$M = m + e$$ ?
3. Introducing the above two cases, can we assume that there is always $$infA$$ ?
• Actually, $B$ might be empty. You should use $x \in \mathbb{R}$ instead. – lzralbu Jan 27 at 14:45
• Thanks, you are right. I will edit it. – Dimitris Dimitriadis Jan 27 at 15:41