Given $\mathcal{F}\subseteq \mathcal{G}$, show $\bigcup\mathcal{F} \subseteq \bigcup\mathcal{G}$

Question

Just proof verification. Given $\mathcal{F}\subseteq\mathcal{G}$, show $\bigcup\mathcal{F} \subseteq \bigcup\mathcal{G}$. Here is my attempt:

Let $A$ be an arbitrary set and $A\in\mathcal{F}$. This implies $A\in\mathcal{G}$. Let $x$ be arbitrary. Suppose $x\in\bigcup\mathcal{F}$. Then $x\in\bigcup_{i=0}^{n} S_i \, \forall S_i \in \mathcal{F}\,\forall i,n\in\mathbb{N}$. Then $\exists i\big|S_i=A$.

It follows that $x\in A$, $A\in\mathcal{G}$ and $x\in \mathcal{G}$. By the definition of $\bigcup\mathcal{G}$, $A\in\bigcup\mathcal{G} \therefore x\in \bigcup\mathcal{G}$. Now, $x\in\bigcup\mathcal{F} \Longrightarrow x\in\bigcup\mathcal{G}$. Since $x$ and $A$ were arbitrary, we can conclude that $\bigcup\mathcal{F} \subseteq\bigcup\mathcal{G}$.

If you notice any places where I can be more concise, if my notation/wording is off, or if my argument/reasoning is wrong, I would appreciate it if you let me know. Thanks.

Answer 1

Suppose $\mathcal{F}\subseteq\mathcal{G}$. If $A$ is a set and $A\in\mathcal{F}$, then $A\in\mathcal{G}$.

Fix $x\in\bigcup\mathcal{F}$. Then $x\in A$ for some $A\in\mathcal{F}$. By the above, we have $x\in\bigcup\mathcal{G}$ since $A\in G$. Therefore $\bigcup\mathcal{F}\subseteq\bigcup\mathcal{G}$.

Answer 2

More concise:

Given that $\mathcal F \subseteq \mathcal G$, show $\bigcup \mathcal F \subseteq \bigcup \mathcal G$.

Proof. Let $x \in \bigcup \mathcal F$. Then there is $A \in \mathcal F$ such that $x \in A$, but since $\mathcal F \subseteq \mathcal G$, we have $A \in \mathcal G$. Thus, we are showed that for $x \in \bigcup \mathcal F$ there exists $A \in \mathcal G$ so that $x \in A$, which means nothing more that $x \in \bigcup \mathcal G$, and since $x$ was arbitrary, it follows $\bigcup \mathcal F \subseteq \bigcup \mathcal G$. $\blacksquare$

Answer 3

Let $A$ be an arbitrary set and $A \in \mathcal F$.

$A$ can't both be "arbitrary" and "in $\mathcal{F}$." By being in $\mathcal{F}$, $A$ is not arbitrary anymore.

Then $x\in\bigcup_{i=0}^{n} S_i \, \forall S_i \in \mathcal{F}\,\forall i,n\in\mathbb{N}$.

This sentence is just weird. You're saying that $x$ belongs to all finite unions $\bigcup_{i=0}^{n} S_i$ for any sets $S_i \in \mathcal{F}$ and $n \in \mathbb{N}$ (and you should not say $\forall i \in \mathbb{N}$ because $i$ isn't any element of $\mathbb{N}$, $i$ is between $0$ and $n$). So if we can take any $n$, then we can take $n = 0$ and you're claiming $x \in S_0$ for any $S_0 \in \mathcal{F}$. This does not describe the union of $\mathcal{F}$ but rather the intersection.

$\exists i\big|S_i=A$

$A$ is just any element of $\mathcal{F}$, and $x$ any element of $\bigcup \mathcal{F}$, which means that $x$ belongs to some element of $\mathcal{F}$, not necessarily the $A$ that you chose at the beginning.

$A\in\bigcup\mathcal{G} \therefore x\in \bigcup\mathcal{G}$

$A$ is an element of $\mathcal{G}$ which makes it a subset of $\bigcup \mathcal{G}$, not an element.

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The ideas are there but: 1) your using the incorrect notation which isn't conveying what you need to say 2) you should avoid too many symbols—in particular, you should avoid using $\forall, \exists, \implies, \therefore$ and should instead write these out in words.