Proof that a set $B$ is a subset of the union of a family of sets. I am having trouble even starting a proof for the following: for every $B\in\mathcal{A}$, $B\subseteq \displaystyle \bigcup_{A \in \mathcal{A}} A$. Here is my initial thoughts, of course assume that there is a $B\in\mathcal{A}$ (where $\mathcal{A}$ is a family of sets). Then, there exists an element $x\in \displaystyle\bigcup_{A \in \mathcal{A}}A$. Then by definition of a union over $\mathcal{A}$, $x\in A$. Then, by our assumption, $x\in B$. (And that is all I have).
I may be missing something inbetween my logic, or I may be completely and utterly wrong... My main issue is determining everything I have to show since it is a "for every" statement. Any help is appreciated, thank you!
 A: Good effort! Your proof is kind of "the wrong shape". There are some fairly rigid rules for how you prove things about sets, which I will try and spell out a bit.
In particular, the way you prove a statement of the form "for all $x \in X$, the property $P(x)$ holds", is to start your proof by saying "let $x$ be an arbitrary element of $X$". From that point on you can think of $x$ as essentially being "fixed".
The way you prove the statement "$X \subseteq Y$" is to prove the statement "for all $x$ in $X$, we have $x \in Y$" (this is actually the definition of $\subseteq$. Try to convince yourself that this is the same thing as $X \subseteq Y$!). Looking at the previous paragraph, this means we have to take an arbitrary $x \in X$ and show we have $x \in Y$.
The way you prove the statement "there is some element $x$ in $X$ such that the property $P(x)$ holds" is to explicitly name an element $x_0 \in X$, and to show that $P(x_0)$ holds for that particular element.
The way you prove the statement $x \in \bigcup_{A \in \mathcal A} A$ is to prove that there is an $A$ in $\mathcal A$ such that $x \in A$ (this is the definition of a union. Again convince yourself that this is a sensible definition.). So by the last paragraph, you have to explicitly tell me an $A_0 \in \mathcal A$ and prove that $x \in A_0$.
Now we can put this all together to prove that "for all $B \in \mathcal A$, we have $B \subseteq \bigcup_{A \in \mathcal A} A$". We will keep applying the previous paragraphs "recursively" to whittle down the statement we are trying to prove.
To start with, the statement is of the form "for all $B \in \mathcal A$, ...". So let $B$ be an arbitrary element of $\mathcal A$.
Now the statement says "$B \subseteq \bigcup_{A \in \mathcal A} A$". So we know we have to say let $x$ be an arbitrary element of $B$.
Now the statement says "$x \in \bigcup_{A \in \mathcal A} A$". So we have to choose some particular element of $\mathcal A$ that has $x$ as an element. If we look back at the current "situation", we know that $B$ is an element of $\mathcal A$, and $x$ is an element of $B$. Perfect! Consider $A_0 = B$. We have $B \in \mathcal A$ and $x \in B$ by assumption. Hence $x \in \bigcup_{A \in \mathcal A} A$, and therefore we have proved that $B \subseteq \bigcup_{A \in \mathcal A} A$".
Here I have italicised the actual sentences that I think you should stitch together to obtain the proof you want. I think you really do need about this level of detail - in my opinion, the other answer just restates what you're trying to prove, and then says it's true because it's obvious. The key thing is that you have to start with an element of $B$ and show it's in the union, not other way around, which is what it looks like your proof does!
In particular it is a bit confusing that $\mathcal A$ is a set whose elements are also sets - it takes some care to deal with this fact properly.
The rules I spelled out for proving things about sets are not a complete totality - there will sometimes be other approaches, especially as you start proving and quoting lemmas and theorems. But in particular in simple proofs like this, it should always boil down to doing these things.
A: Suppose that $\mathcal{A}=\{A_{i}\}_{i\in I}$ with $A_{i_0}=B$. You want to show that $B\subseteq \bigcup_{A\in \mathcal{A}}A$ or equivalently $A_{i_0}\subseteq \bigcup_{i\in I}A_i$, but this containment since $i_0\in I$.
Does it make sense now?
