How is induction related to a well founded set? I read a proof that went along these lines: 
"Let $(X, \le)$ be a woset. Let $E$ be a subset of $X$ such that: 


*

*the minimal element of $X$ is a member of $E$

*for any $x \in X$, if $\forall_y[y \lt x \rightarrow y \in E]$, then $x \in E$


Then, $E = X.$ Proof: 
Suppose that $E \ne X$. Then, let $x$ be the minimal element of $X - E$. Since, it is an element of $X - E$, it can not be an element of $E$. But, any $y \lt x$ would imply $y \in E$, because $x$ is the minimal element of $X - E$, so any element $y$ such that, $y \lt x$ would have to be in $E$. If it wasn't, $x$ wouldn't be the minimal element of $X -E$. Since $y \lt x$ implies $y \in E$, then by the second condition, $x \in E$ and a contradiction arises, so $X = E$."
But, then the claim was made that this being true allows proof by induction on any well founded set. I understand the proof, but I don't see how this theorem and induction are related. Could someone explain?  
 A: Notice that a well-ordered set is precisely a well-founded total ordering, but the proof you quoted makes no use of the total property, so it is valid whenever $X$ is well-founded.
Next, notice that condition 1 is redundant, because it is simply the special case of condition 2 where $x$ is the minimal element of $X$.
(In that case, there simply is no $y \in X$ such that $y < x$, so the antecedent proposition that $y \in E$ for all $y < x$ is vacuously true, therefore the consequence that $x \in E$ is also true.)
Finally, notice that the proof actually works in both directions.  Condition 2 can be rewritten:
2' If $x \in X \setminus E$ then there exists $y < x$ such that $y \in X \setminus E$.
Equivalently:
2'' $X \setminus E$ has no minimal element.
So what was proved was the implication:
"If $E$ is any subset of $X$ such that $X \setminus E$ has no minimal element, then $X \setminus E$ is empty."
Since $E$ is an arbitrary subset of $X$, this is equivalent to the implication:
"If $F$ is any subset of $X$ with no minimal element, then $F$ is empty."
Or to put it another way:
"If $F$ is any non-empty subset of $X$, then $F$ has a minimal element."
Of course this is true, by definition, if and only if $X$ is well-founded.
In  other words, the proposition that induction on $X$ is a valid method of proof can actually be used as an alternative definition of what it means for the poset $X$ to be well-founded.
