Smallest Numbers
Consider the natural numbers $\mathbb{N}$, what is the smallest number? Clearly the answer is $0$, as it is smaller than every positive number, and there are no negative numbers in $\mathbb{N}$ so it must have $0$ as the smallest.
Now consider $A={1,2,3}\subset\mathbb{N}$, does this subset have a smallest number? This isn’t a trick question, the answer is pretty clearly $1$.
What about the set of all even numbers $\mathbb{E}={0,2,4,\ldots}$?. Again this is pretty straightforward to see the answer is $0$. This is pretty simple, but it leads to a very important theorem that proves to be quite useful in theoretical mathematics.
Well Ordering Principle: Every non-empty subset of the natural numbers $A$ contains a smallest element.
Proof
We will do this using a proof both by induction and contradiction. Assume that there exists some set $A\subset\mathbb{N}$ where $A\neq\varnothing$ and $A$ does not have a smallest item. If $A$ does not have a smallest item, then $0\not\in A$ because $0$ is the smallest possible natural number, so if $0$ was inside $A$ then it would automatically be the smallest item.Now we will show that $A$ must actually be empty by induction. Assume that for some value $n$, every natural number $< n$ is not in $A$, so $$ \begin{align} 0&\not\in A \\\\ 1&\not\in A \\\\ &\vdots \\\\ n-1&\not\in A \end{align}$$ since none of the values are in $A$, if $n\in A$ that would make $n$ the smallest item. But since $A$ has no smallest item we know that means $n\not\in A$. Continuing our argument we know by induction that $n+1, n+2, n+3\ldots\not\in A$ and since we have our base case $0\not\in A$, we know this means that for every $x\in\mathbb{N}$, $x\not\in A$. This means that $A=\varnothing$ which contradicts our assumption $A$ was non-empty and proves our claimQ.E.D.It turns out you can use the Well Ordering Principle to prove some properties of sets by considering the set of all conter examples, and finding a contradiction involving the smallest element of counterexample.