# Prove that if $\mathcal{F} \subseteq \mathcal{G}$ then $\cup\mathcal{F} \subseteq \cup\mathcal{G}$

Suppose $\mathcal{F}$ and $\mathcal{G}$ are families of sets. Prove that if $\mathcal{F} \subseteq \mathcal{G}$ then $\cup\mathcal{F} \subseteq \cup\mathcal{G}$

My attempt:

Given $\mathcal{F} \subseteq \mathcal{G}$, writing out logical form of Goal we have
$$\exists A \in F(x \in A) \rightarrow \exists A \in G(x \in A)$$

Now assuming $\exists A \in F(x \in A)$ (putting it to list of givens), our Goal is now to obtain

$$\exists A \in G(x \in A)$$

Now, applying Existensial Instantiation, let us assume there exists $A_{0}$, such that $A_{0}$ $\in F$. So now $x \in A_{0}$ I am feeling stuck here.Thanks

EDIT

• I'd suggest you learn how to typeset your questions correctly. It will make them easier to read and add quality overall. – Daniel W. Farlow Aug 3 '15 at 5:42

If I understand you correctly. You have $(F_i)_{i\in I }$ and $(G_i)_{i\in I}$ are families of set and you want the proof of

$$(\forall i \in I \ \ F_i \subset G_i ) \Longrightarrow \bigcup_{i\in I} F_i \subset \bigcup_{i\in I} G_i$$

Suppose that $\bigcup_{i\in I} F_i \not\subset \bigcup_{i\in I} G_i$ this implies that there exist an element $x \in \bigcup_{i\in I} F_i$ such that $x \not\in \bigcup_{i\in I} G_i$ . At least $x$ is in one $F_i$ for an $i \in I$ and $x$ don't belong of any of $G_i$. and this is contradiction for the fact that $(\forall i \in I \ \ F_i \subset G_i )$.

Here's how I would write it.

Let $x\in \cup F$

Then $\exists f\in F$ s.t. $x\in f$.

Since $F\subseteq G$, $f\in G$.

Thus $x\in f\subseteq G$

So $x\in \cup G$

Thus $\cup F\subseteq \cup G$

Let $\mathcal{F}$ and $\mathcal{G}$ be two families of sets such that $\mathcal{F} \subseteq \mathcal{G}$. Hence for every set $X$ if $X \in \mathcal{F}$ then $X \in \mathcal{G}$. Let $\cup\mathcal{F}$ denote the union of the family $\mathcal{F}$, that is, $x \in \cup\mathcal{F}$ iff $x \in F$ for some $F \in \mathcal{F}$. Since $\mathcal{F} \subseteq \mathcal{G}$, then $F \in \mathcal{G}$ as well and once $x \in F$, $x \in \cup\mathcal{G}$.