Set theory question

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  • #1
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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?
 

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  • #2
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Letting [tex]A, B_1, B_2, ..., B_n[/tex] subsets of [tex]X[/tex], then show

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

----

Is it sufficient to say...

By De Morgan law, we have

[tex]\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})[/tex]

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 [tex]...[/tex] type notation, is to choose

[tex]x\in A\cap\bigcup_{n}^{i=1}B_{i}[/tex],

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

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

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

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

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

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

So [tex]A\cap\bigcup_{n}^{i=1}B_{i}= A\cap\bigcup_{n}^{i=1}B_{i}[/tex]. [tex]\blacksquare[/tex]

Obviously, though, that way requires a bit more writing. It's a matter of preference which you wish to use; both are valid.
 
  • #3
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For the last line, don't you mean

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

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.
 
Last edited:

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