Does $\{x \in \mathcal X:x>a \land x<b+1 \}$ give $(a,b+1)$ or $(a,b]$?

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Ordinal numbers:- An ordinal number is a set $\alpha$ with the following properties:

(1) If $x,y \in \alpha$, then either $x\in y, y\in x, $ or $y=x$

(2)If $y \in \alpha$ and $x\in y$, then $x\in \alpha$.

Theorem H.10 There is an uncountable well-ordered set $\Omega$ of ordinal numbers with maximal element $\omega_1$ having the property if $x\in \Omega$ and $x\neq \omega_1$, then $\{y\in \Omega:y\leq x\}$ is countable.

Definition of interval in the book is given by:-

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Doubt:- By the definition of the interval $\{x \in \mathcal X:x>a \} \cap \{x \in \mathcal X:x<b+1 \}=(a,b+1)$. Am I correct? If there is no element between $b$ and $b+1$. I can agree with the notation in the textbook. Please help me.

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It is important here that we're talking about sets of ordinals, so we know there is no possible element strictly between $b$ and $b+1$.

(Recall that $b+1$ is by definition the set $b\cup\{b\}$, and the standard ordering of ordinals is the same as ordering by $\subseteq$, so there can't be any ordinal strictly between $b$ and $b+1$ because $b+1$ has only one additional element).

Therefore the condition $x<b+1$ is the same as $x\le b$, so $(a,b+1)$ and $(a,b]$ are two ways of writing the same set of ordinals.