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

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

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.