Note: It seems that the term inductive set has varying definitions. I am not referring (I think) to the definition used here, so this question is not a duplicate of that one.
Specifically, consider the definition found on p. 12 of Dudley, Real Analysis and Probability:
More generally, let $(X, <)$ be any partially ordered set. A subset $Y \subset X$ will be called inductive if, 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 author then goes on to state that the set $( -\infty, 0)$ in $\mathbb{R}$ is inductive (presumably $\mathbb{R}$ with the standard partial order is meant), but:
Question: Is that a typo? I.e. how is $(-\infty, 0)$ inductive using the above definition? It seems like $(-\infty, 0]$ might be inductive using the above definition, but not $(- \infty, 0)$.
Either a correction of my flawed reasoning, or a sanity check that it is indeed a typo, would help.
(Flawed) reasoning: If we take $X = \mathbb{R}$ and $Y = (-\infty, 0)$, then seemingly $0 \in X$ is such that for all $y \in \mathbb{R}$ with $y < 0$, $ y \in (-\infty, 0)$. (That's literally the definition of that half-open interval, right?) So then seemingly by the definition of inductive set, if $(-\infty, 0)$ were inductive, we would have $0 \in (-\infty,0)$, which is obviously untrue.
However, the way the definition is worded in Dudley is very confusing to me and it seems likely that I am messing up the order of logical quantifiers and connectives to get a non-equivalent statement in my mind, in particular how I am thinking of the definition is as follows:
A subset $Y \subset X$ is inductive if, for all $x \in X$ such that ($y<x \implies y \in Y$), $x \in Y$.
(Now that I write it out my wording of the definition, in addition to most likely being wrong, is also rather opaque.) Anyway, $y <0$ clearly implies $y \in (-\infty, 0)$, hence my "reasoning" above.
To clarify the definition, Y subset X is inductive when
for all x in X, if (for all y < x, y in Y), then x in Y.
Implied quantifiers create confusion and though popular, should not be used.
(-oo,0) is not inductive because of 0. (-oo,0] is inductive.