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If $(S, \leq)$ is a partial ordered class and $S$ is finite, how can I prove that there exists a minimal element in $S$? I kind of mix up the the terms "smallest" and "minimal" - that is why I dont really know what the approach is when showing that there exists a minimal element in a set.

Some help would be much appreciated.

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    An element $a$ is *minimal* if there is no *other* element $b$ in $S$ such that $b \leq a$. An element $a$ is the *least* element if for all elements $b$ of $S$, $a \leq b$. – 2017-01-08
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    Then, what is the difference between an element being the smallest and an element being minimal..? – 2017-01-08
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    Suppose $S=\{p,q,r\}$ and your partial order relation is $\{(p,p), (q,q), (r,r), (p,q)\}$. That is, every element is less than or equal to itself and $p \leq q$. Then, both $p$ and $r$ are minimal, but there is no least/smallest element. Dually, both $q$ and $r$ are maximal, but there is no greatest element. Finite *lattices*, which presumably you'll encounter later, have least and greatest elements. – 2017-01-08

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An element $s \in S$ is minimal iff there is no smaller element, that is $$ \forall t \in S \quad t \le s \Rightarrow t = s $$ i. e. the only element smaller or equal to $s$ is $s$ itself.

$s$ is the smallest element, iff it is smaller then all elements, i. e. $$ \forall t \in S \quad s \le t. $$ (Note, that a smallest element is minimal, but not vice versa).

To show that a finite set has a minimal element, use e. g. induction on $|S|$: Pick $s \in S$, if $s$ is minimal, we are done, otherwise consider the non-empty set $\{t \in S : t < s\}$, which by induction has a minimal element. Now show that it is also a minimal element for $S$.

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    Can you give me an example for a set where an element is minimal but not the smallest? I am kind of confused, I cannot distinguish between those two deifnitions.. – 2017-01-08
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    Consider $S := \mathbf N - \{0,1\}$ with the order $m \le n \iff m \mid n$. Then $2$ is minimal, as there is no $m \ne 2$ with $m \mid 2$, but not smallest as $3 \nmid 2$. – 2017-01-08
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    Thank you, that helped me. Kind of tricky at the beginning. – 2017-01-08
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    @martini: I think you mean "not smallest as $2\nmid 3$"? – 2017-01-08
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    Upps, correct, thank you for pointing this out. – 2017-01-08
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    So if $S\setminus \{s\}$ has a minimal element $x$ by induction, why shouldn't it be the minimal element of $S$ if $s$ is not? – 2017-01-08
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    Sure. There is no problem, just write it down, that's all I wanted to say. – 2017-01-08