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Let $X$ be an inductive set. Let $T = \{x \in X : x \subset X\}$. Then let $t \in T$. My question is why does the following hold: $y \in t \implies y \subset t$?

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It doesn’t necessarily. Let $a=\{\{\varnothing\}\}$, and let $X$ be the inductive closure of $\big\{a,\{a\}\big\}$, so that $X$ is inductive. Let $T=\{x\in X:x\subseteq X\}$; then $\{a\}\in T$. Clearly $a\in\{a\}$, but $a\nsubseteq\{a\}$: the only element of $a$ is $\{\varnothing\}$, which is not equal to $a$, the only element of $\{a\}$.

What is true is that $T$ is inductive.

Suppose that $t\in T$. $X$ is inductive, so $t\cup\{t\}\in X$. If $y\in t\cup\{t\}$, then either $y\in t\subseteq X$, or $y=t\in X$; in either case $y\in X$, so $t\cup\{t\}\subseteq X$, and therefore $t\cup\{t\}\in T$. Thus, $T$ is inductive.

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Why in the case that $y \in t \subseteq X$ does this imply that $y \in X$? – user1770201 Jun 5 '13 at 19:41
@user1770201: $t\subseteq X$ means that every element of $t$ is an element of $X$. – Brian M. Scott Jun 5 '13 at 19:46

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