Proving a set is a subset of another While reading a prove regarding transitive closure, I have problem in explaining the statement below in formal proof:
To show that $\bigcup_{n \in \mathbb N}R^n \subseteq T$, it suffices to show that for all $n\in \mathbb N, R^n \subseteq T$.
Intuitively I understand it because $\bigcup_{n\in \mathbb N}R^n$ means $R\cup R^1 \cup R^2... \cup R^n$, so if I can show that for all $n \in \mathbb N, R^n \subseteq T$, then surely $\bigcup_{n\in \mathbb N}R^n \subseteq T$. But how should I put it in right word?
Thanks in advance.
 A: Proposition. Let $Y$ denote a subset of a larger set $\Omega$, and suppose $X_{i \in I}$ is a family of subsets of $\Omega$. Then
$$\forall i \in I(X_i⊆Y) \iff (\cup i \in I : X_i) \subseteq Y$$
Proof.
$(\Rightarrow).$ Assume $\; \forall i \in I(X_i⊆Y).$ We will show $(\cup i∈I : X_i) \subseteq Y$
Now let $x \in (\cup i∈I : X_i).$ We will show $x \in Y.$ By hypothesis, we can find $j \in I$ with $x \in X_j$. Fix any such $i \in I$. Then $X_i \subseteq Y.$ So $x \in Y$.
$(\Leftarrow).$ Assume $(\cup i∈I : X_i) \subseteq Y.$ We will show $\; \forall i \in I(X_i⊆Y).$
Let $j \in I$ be fixed but arbitary. We will show $X_j \subseteq Y$. We by Lemma 1 (which is yet to be proved) have that $X_j \subseteq (\cup i∈I : X_i) = Y.$ This completes the proof.
Lemma 1. Let $X_{i \in I}$ denote a family of subsets of $\Omega$. Then for any $j \in I$, we have that $X_j \subseteq (\cup i∈I : X_i).$
Proof. Let $j \in I$ be fixed but arbitrary. We will show $X_j \subseteq (\cup i∈I : X_i).$
Let $x \in X_j$ be fixed but arbitrary. We will show $x \in (\cup i∈I : X_i).$ This is the same as saying that there exists $i \in I$ such that $x \in X_i$. But there's an obvious choice, namely $i=j$. This completes the proof.
A: The set $S:=\bigcup_{n\in\mathbb N}R^n$ has (by definition) the property that $x\in S$ implies
$$\exists n\in\mathbb N\colon x\in R^n$$
Thus for arbitrary $x\in S$, let $n_0\in \mathbb N$ be such that $x\in R^{n_0}$.
Then from $R^{n_0}\subseteq T$ we have $x\in T$. That is $x\in S$ implies $x\in T$, in other words $S\subseteq T$.
