Set Theory Q: Show A∩⋃ⁿᵢ=1Bᵢ = ⋃ⁿᵢ=1(A∩Bᵢ)

[jaejoon89]

Letting A, B_1, B_2, ..., B_n subsets of X, then show

A\cap\bigcup_{n}^{i=1}B_{i} = \bigcup_{n}^{i=1}\left(A \cap\right B_{i})

----

Is it sufficient to say...

By De Morgan law, we have

\left(A\cap \right B_{i})\cup\left(A \right \cap\ B_{2})\cup\ --- \cup\left(A\cap \right\ B_{n}) = \bigcup_{i=1}^{n}\left(A\cap \right B_{i})

Is that sufficient, or is there a better/more complete way to do it?

 

[hatsoff]

Letting A, B_1, B_2, ..., B_n subsets of X, then show

A\cap\bigcup_{n}^{i=1}B_{i} = \bigcup_{n}^{i=1}\left(A \cap\right B_{i})

----

Is it sufficient to say...

By De Morgan law, we have

\left(A\cap \right B_{i})\cup\left(A \right \cap\ B_{2})\cup\ --- \cup\left(A\cap \right\ B_{n}) = \bigcup_{i=1}^{n}\left(A\cap \right B_{i})

Is that sufficient, or is there a better/more complete way to do it?

Well, that looks fine to me. But the other way to do it, if you prefer to avoid ... type notation, is to choose

x\in A\cap\bigcup_{n}^{i=1}B_{i},

and say that therefore x\in A and x\in B_i for some i\in\mathbb{N}. Therefore

x\in A\cap B_i\subseteq\bigcup_{n}^{i=1}\left(A \cap\right B_{i})

\Rightarrow x\in\bigcup_{n}^{i=1}\left(A \cap\right B_{i}).

Then choose y\in\bigcup_{n}^{i=1}\left(A \cap\right B_{i})

\Rightarrow y\in A\cap B_i\subseteq A\cap\bigcup_{n}^{i=1}B_{i}

\Rightarrow y\in A\cap\bigcup_{n}^{i=1}B_{i}.

So A\cap\bigcup_{n}^{i=1}B_{i}= A\cap\bigcup_{n}^{i=1}B_{i}. \blacksquare

Obviously, though, that way requires a bit more writing. It's a matter of preference which you wish to use; both are valid.

 

[jaejoon89]

For the last line, don't you mean

<br /> A\cap\bigcup_{n}^{i=1}B_{i}=\bigcup_{n}^{i=1}\left(A \cap\right B_{i})

But A and B_1, B_2, ... , B_n are subsets of X, so does the Demorgan law really apply here (since A and B_i are in the same family)? I thought they had to be distinct.