Is Proof of A Subset of Union of Family Valid?

[bubblescript]

Homework Statement

If $\mathcal{F}$ is a family of sets and $A \in \mathcal{F}$, then $A \subseteq \cup \mathcal{F}$.

Homework Equations

$A \subseteq \cup \mathcal{F}$ is equivalent to $\forall x(x \in A \rightarrow \exists B(B \in \mathcal{F} \rightarrow x \in B))$.

The Attempt at a Solution

Suppose $A \in \mathcal{F}$. Let $x$ be arbitrary and suppose $x \in A$. Clearly $\exists B \in \mathcal{F}$ is true if we say that $B=A$. Therefore we conclude that if $A \in \mathcal{F}$, then $A \subseteq \cup \mathcal{F}$.

 

[fresh_42]

Homework Statement

If $\mathcal{F}$ is a family of sets and $A \in \mathcal{F}$, then $A \subseteq \cup \mathcal{F}$.

Homework Equations

$A \subseteq \cup \mathcal{F}$ is equivalent to $\forall x(x \in A \rightarrow \exists B(B \in \mathcal{F} \rightarrow x \in B))$.

The Attempt at a Solution

Suppose $A \in \mathcal{F}$. Let $x$ be arbitrary and suppose $x \in A$. Clearly $\exists B \in \mathcal{F}$ is true if we say that $B=A$. Therefore we conclude that if $A \in \mathcal{F}$, then $A \subseteq \cup \mathcal{F}$.

If $\cup \mathcal{F} = \cup_{B\in \mathcal{F}}B$ then yes, $B=A$ does the job.

 

[bubblescript]

If $\cup \mathcal{F} = \cup_{B\in \mathcal{F}}B$ then yes, $B=A$ does the job.

I'm unfamiliar with that notation. By $\cup \mathcal{F}$ I mean the union of all the sets that are in $\mathcal{F}$.

In other words $x \in \cup \mathcal{F}$ is equivalent to $\exists B \in \mathcal{F}( x \in B)$.

 

[fresh_42]

I'm unfamiliar with that notation. By $\cup \mathcal{F}$ I mean the union of all the sets that are in $\mathcal{F}$.

Yes, that's what I said: $$\cup \mathcal{F}=\bigcup_{B \in \mathcal{F}} B$$
Union of all sets $B$, i.e. $\cup B$ where all sets from $\mathcal{F}$ are taken, i.e. $\cup_{B \in \mathcal{F}}B$.

If you simply call a list of sets a family, then $\cup \mathcal{F}$ could have meant $\{\{B\}\,\vert \,B \in \mathcal{F}\}=\mathcal{F}$ in which case the union would be a set of sets, which doesn't contain the single elements of its sets.

The notation $\cup \mathcal{F}$ is a bit sloppy, as it denotes only one set: $\cup \mathcal{F} = \mathcal{F}$. What you really mean is the union of all sets in $\mathcal{F}$.

E.g. $$\bigcup_{i=1}^2 A_i = A_1 \cup A_2 = \bigcup_{i \in \{1,2\}}A_i = \bigcup_{A_i \in \{A_1,A_2\}}A_i$$
and the difference to the above case is only that $\{1,2\}$, resp. $\{A_1,A_2\}$ is replaced by $\mathcal{F}$.

 

[bubblescript]

Ok that makes a lot of sense, thanks.