Inductive Subset and the Induction Principle I'm reading Dudley's Real Probability and Analysis when I came across the following definitions of Inductive Subset and Induction Principle.

I would like to prove the induction princple just like Dudley has done, but I'm a bit confused about the definition of inductive subset.
I translated the highlighted passage as it follows:
$I$ = set of all inductive subsets
$I = \left \{ Y \subset X | \forall (x \in X \rightarrow y \in Y) \forall (y \in X \rightarrow y < x)(x \in Y) \right \} $
Is it correct?
 A: Let $(X, <)$ a p.o. set and let $Y \subset X$.
We say $Y$ inductive iff

$\forall x \in X \ [\forall y \in X \ (y < x \to y \in Y) \to x \in Y].$


The definition is applied to the proof of the Induction Principle considering a well-ordered set $X$ ordered by $<$.
For the non-trivial case, we have $Y \subset X$ and assume that $X \setminus Y$ is not empty.
Let $y$ the least element in $X$ such that $y \notin Y$ (it exists by well-ordering of $X$ : "every non-empty subset of $X$ has a least element").
Please, note that in the proof the author exchanged the role of $x$ and $y$ with respect to the previous definition of inductive set.
Now $y$ is the least element in $X$ which is not in $Y$; thus, all elements $x$ of $X$ that are "less than" $y$ must belong to $Y$.
$Y$ is inductive, and thus from :

$\forall y \in X \ [\forall x \in X \ (x < y \to x \in Y) \to y \in Y]$

we conclude with : $y \in Y$, contradicting the fact that $y \in X$ and $y \notin Y$.
Thus, we have to refute our assumption that $X \setminus Y$ is not empty and we conclude with :


$Y=X$.



Note on the above "symbolization".
The English text says :

for every $x \in X$ such that $y \in Y$ for all $y \in X$ such that $y < x$, we have $x \in Y$.

The text is a little bit convoluted, but it is clear that "we have $x \in Y$" is the "concusion" of the definition, and thus must be the consequent of the conditional.
Thus, we can try with :

$\forall x \in X \ [\text {such that} \ (y \in Y \ \text {for all} \ y \in X \ \text {such that} \ y < x) \to x \in Y].$

Now we have to "unwind" :

$(y \in Y \ \text {for all} \ y \in X \ \text {such that} \ y < x)$

and we can rephrase it as : $(y \in Y \ \text {if, for all} \ y \in X \ \text {such that} \ y < x)$, i.e. with :

$\forall y \in X (y < x \to y \in Y)$.

A: If $X$ is a set and $R$ is a relation on $X$ then a set $Y\subseteq X$ is $R$-inductive if it has the following property:$$\hat{x}\subseteq Y\Rightarrow x\in Y$$
This for each $x\in X$ and under definition: $$\hat{x}:=\{y\in X\mid yRx\}$$
The property states that being an element of $Y$ is somehow inherited from $R$-predecessors. 
If each $R$-predecessor of $x$ is an element of $Y$ then $x$ is an element of $Y$.
