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 ?


  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 $infA$ ?
  • $\begingroup$ Actually, $B$ might be empty. You should use $x \in \mathbb{R}$ instead. $\endgroup$ – lzralbu Jan 27 at 14:45
  • $\begingroup$ Thanks, you are right. I will edit it. $\endgroup$ – Dimitris Dimitriadis Jan 27 at 15:41

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