Equivalence of 'Existence of maximum/mininum' and 'Axiom of induction'. Peano's Axiom of induction (in simple formulation) states the following:
If $A \subset \mathbb{N}$, such as $0 \in A$ and $ (n \in A) \Rightarrow (n+1 \in A)$,  then $A=\mathbb{N}$.
In one of the books I found a claim (without proof) that the Axiom of induction is equivalent to any of the following two:


*

*Any subset of $\mathbb{N}$ with upper bound contains maximum

*Any subset of $\mathbb{N}$ contains minimum



Regarding equivalence of 1 and 2, it probably can be proven by taking negatives: if $A$ is a subset of $\Bbb{N}$ with lower bound, then $B=\{-a|\; a\in A\}$ has an upper bound, therefore it must contain maximum $m$, whereas $-m$ will be minimum for $A$ and vice versa.

Tried to prove also the equivalence of Axiom of induction and statement 1. 
Assume Axiom of induction is true.
Limit to the set of natural numbers, where any bound set $A$ must have a supremum $s$. Then $s-1 \in A$, therefore from Axiom of induction is follows that $s \in A$, so $s$ is the maximum.
Firstly I an not sure, if the proof above is strict enough. Secondly, I can't think of the way to prove the opposite, that Axiom of induction follows from statement 1.  

Update: using k.stm's idea (+1 to k.stm for that), I am going to prove that Axiom of induction follows from from statement 2 (which is equivalent to 1)
Suppose statement 2 is true and axiom of induction is false. Then there exists a set A so that  $0 \in A$ and $ (n \in A) \Rightarrow (n+1 \in A)$,   while A is a strict subset of $\mathbb{N}$, so the complement $А^c$ is non-empty. According to statement 2, set $A^c$ should contain minimum $m$.  Since $0 \notin A$, then $0 \lt m$, so according to transitivity $0$ must be less than any number in $А$, including $0$ itself which is a contradiction.
Still it doesn't prove the opposite implication.
 A: Your argument does not make sense. The axiom of induction says something about subsets $A ⊆ ℕ$ such that $0 ∈ A$ and $∀n ∈ ℕ \colon~ n ∈ A \implies n+1 ∈ A$.
You are trying to say something about a bounded set $A$. But can you be sure that $0 ∈ A$ and that $∀n∈ ℕ \colon~ n ∈ A \implies n+1 ∈ A$? You are trying to use “$s - 1 ∈ A \implies s ∈ A$” directly, but that’s not given here. So why would that be the case?
Instead, view it this way: The axiom of induction says that some subsets $A ⊆ ℕ$ with certain properties are already all $ℕ$. Equivalently, it says that some subsets $A ⊆ ℕ$ with certain properties have empty complement $A^\mathrm c = ∅$.
Using contraposition, these certain properties in question can be paraphrased as:


*

*Zero can be no minimum of $A^\mathrm c$.

*No other element of $ℕ$ can be a minimum of $A^\mathrm c$.


From this, it’s easy to see that the axiom of induction is equivalent to every nonempty subset $A ⊆ ℕ$ having a minimum (again, this is using contraposition). This, in turn, is easily seen to be equivalent to every bounded, nonempty subset $A ⊆ ℕ$ having a minimum.
A: To show: "Any subset of $\mathbb{N}$ with upper bound contains maximum"  
Let $A\subseteq \mathbb{N}$ such that $\beta\in\mathbb{N}$ is an upper-bound of $A$.
By pigeon-hole principle, $|A|\le \beta$.
Now we can use induction on size of $A$ to show that there exists a maximum.
But, principle of mathematical induction relies on Well-Ordering Principle of $\mathbb{N}$, which in turn, relies on inductive axiom of Peano.  
Also note, that you can not define subset of negative integers like $B=\{-a:a\in A\}$, because $B$ is no more subset of $\mathbb{N}$, so Peano's axioms can't be applied on $B$.  
Additional comment:
Caution: I assume $0\in\mathbb{N}$.

Proposition: If $|A|=m$, for some $m\in\mathbb{N}$, then $\sup{A}\in A$  

Define inductive set $T=\{n:\sup{A}\in A,\text{when }|A|=n\}$.
We need to show that $T=\mathbb{N}\setminus\{0\}$.
But inductive axiom of Peano, is not suitable to prove this. So, we instead define set $S=T\cup\{0\}$ and show that $S=\mathbb{N}$ using inductive ($5^{th}$) axiom of Peano.
$0\in S$, by definition.
Assume $n\in S$, i.e., $\sup{A}\in A$ when $|A|=n$. Let $\sup{A}=\alpha$.
When $|A'|=n+1$, $A'=A\cup\{b\}$, where $b$ is the new element.
Two possible cases for $b\in A'$: either $b>\alpha$ or $b\le \alpha$. ($\because$ Trichotomy).
$\implies b=\sup{A'}$ or $\alpha=\sup{A'}$.
In either case, $\sup{A'}\in A'$.
Hence, $n+1\in S$.
By inductive axiom of Peano, we have shown that for every size $m\in\mathbb{N}$ of set $A$, $\sup{A}\in A$.
