I'm new to thinking about families of sets and am struggling with the following problem.

Compute and prove $\cap \mathcal A$ and $\cup\mathcal A$ if $\mathcal A=\{A\subseteq \Bbb R$ | if $A\neq \emptyset $ then $0\in A$}

I'm thinking that $\cap \mathcal A$ might be either $\emptyset$ or 0, but I'm unsure which. I'm thinking it's the former because if it was 0 then $0\in A \forall A\in \mathcal A$ but $\emptyset$ is a subset of the real line right, so $0\notin \ A$ when $A=\emptyset$.

As for $\cup \mathcal A$ I'm also unsure. I'm thinking it's either $\Bbb R$ or 0.

I think the reason I'm struggling with this is that on my brief experience in doing this before, the set A was an interval so you could define $A_1, A_2,...$ but here you can't do that...any help on getting me on the right track to thinking about this would be appreciated!

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    $\begingroup$ To get braces in mathjax, precede them with a backslash \$\{\}\$ $\endgroup$ Sep 5 '15 at 23:31
  • $\begingroup$ @PaulSinclair thank you. I'll try to fix that now $\endgroup$ Sep 5 '15 at 23:32
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    $\begingroup$ Note that $\emptyset \in \mathcal A$ and $\Bbb R \in \mathcal A$. That should clear up any confusion as to what the intersection and union are! $\endgroup$ Sep 5 '15 at 23:35
  • $\begingroup$ @PaulSinclair So the intersection is $\emptyset$ and the union is \Bbb R? Now just to prove it...thank you! $\endgroup$ Sep 5 '15 at 23:39

Well, first recall the definitions:

  • $x\in\bigcap\cal A$ if and only if for every $A\in\cal A$, $x\in A$.
  • $x\in\bigcup\cal A$ if and only if there is some $A\in\cal A$ such that $x\in A$.

And two basic properties, if $A\in\cal A$, then $\bigcap\mathcal A\subseteq A\subseteq\bigcup\cal A$.

Now let us look at $\cal A$. $A\in\cal A$ when the following is true, if $A\neq\varnothing$, then $0\in A$. Note that both $\varnothing$ and $\Bbb R$ are in $\cal A$. Use this and the above to find the wanted sets.


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