# Basic sets

#### Guest

##### Active member
If $G$ and $\left\{F(n): n \in \mathbb{K}\right\}$ are a family of sets, show that $\displaystyle G \cap \cap_{n \in \mathbb{K}}F(n) = \cap_{n \in \mathbb{K}}(F(n) \cap B).$

I said if $b$ is an ement of $\displaystyle G \cap \cap_{n \in \mathbb{K}}F(n)$ then $b$ is in both $G$ and $F(n)$ for every $n \in \mathbb{K}$. Say for example that there are $m$ lots of $n$ in $\mathbb{K}$ and denote these by $n_{1}, n_{2}, \cdots n_{m}$. We can pair each of $f(n_1), f(n_2), f(n_3), \cdots, f(n_m)$ with $G$ i.e. $b \in G \cap F(n_{1})$, $b \in G \cap F(n_{2})$, $b \in G \cap F(n_{3})$, $\cdots$, $b \in G \cap F(n_{m}).$ That's $b \in \cap_{n \in \mathbb{K}}(F(n) \cap B).$ Working backwards, implies $b \in G \cap F(n_{1})$, $b \in G \cap F(n_{2})$, $b \in G \cap F(n_{3})$, $\cdots$, $b \in G \cap F(n_{m})$. And we just get rid of the pairing - $b \in G$ and $b \in [f(n_1), f(n_2), f(n_3), \cdots, f(n_m)]$ i.e. $b \in G \cap \cap_{n \in \mathbb{K}}F(n)$. Therefore $\displaystyle G \cap \cap_{n \in \mathbb{K}}F(n) = \cap_{n \in \mathbb{K}}(F(n) \cap B).$

I'm pretty sure the above is pretty crap. Could someone please post up a proof - I'm completely lost on how to write maths proofs. I can prove bit more difficult stuff, but the fact that it appears something so obvious puts me off completely (I'm not sure how much to leave out etc).

#### ThePerfectHacker

##### Well-known member
Here is a suggestion. When you have two sets $A$ and $B$ and you want to show $A=B$ you show that any element of $A$ is in $B$ and any element of $B$ is in $A$. Try using this approach.

You want to prove,
$$A \cap \left( \bigcap_{n=0}^{\infty} B_n \right) = \bigcap_{n=0}^{\infty} (A\cap B_n)$$
If,
$$x\in A \cap \left( \bigcap_{n=0}^{\infty} B_n \right) \implies x \in A \text{ and }x \in \bigcap_{n=0}^{\infty} B_n$$
Then, as $x\in \bigcap_{n=0}^{\infty} B_n$ it means $x\in B_n$ for every $n$.

Thus, $x\in A$ and $x\in B_n$, so $x\in (A\cap B_n)$ for every $n$. In particular,
$$x\in \bigcap_{n=0}^{\infty} (A\cap B_n)$$
Now do the argument in reverse, if you know what I am sayin'.

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#### Guest

##### Active member
Here is a suggestion. When you have two sets $A$ and $B$ and you want to show $A=B$ you show that any element of $A$ is in $B$ and any element of $B$ is in $A$. Try using this approach.

You want to prove,
$$A \cap \left( \bigcap_{n=0}^{\infty} B_n \right) = \bigcap_{n=0}^{\infty} (A\cap B_n)$$
If,
$$x\in A \cap \left( \bigcap_{n=0}^{\infty} B_n \right) \implies x \in A \text{ and }x \in \bigcap_{n=0}^{\infty} B_n$$
Then, as $x\in \bigcap_{n=0}^{\infty} B_n$ it means $x\in B_n$ for every $n$.

Thus, $x\in A$ and $x\in B_n$, so $x\in (A\cap B_n)$ for every $n$. In particular,
$$x\in \bigcap_{n=0}^{\infty} (A\cap B_n)$$
Now do the argument in reverse, if you know what I am sayin'.
Thank you very much. That's what I've been trying to do, but failed horribly.

So suppose $x\in \bigcap_{n=0}^{\infty} (A\cap B_n)$, then $x \in (A \cap B_n)$ for every $n$, i.e. $x \in A$ and $x \in B_n$ for every $n$, that's $x\in A \cap \bigcap_{n=0}^{\infty} B_n.$

#### Guest

##### Active member
In an attempt to really understand the proof, I replaced one of the intersections with complement.

$$A \setminus \left( \bigcap_{n=0}^{\infty} B_n \right) = \bigcap_{n=0}^{\infty} (A\setminus B_n)$$
If,
$$x\in A \setminus \left( \bigcap_{n=0}^{\infty} B_n \right) \implies x \in A \text{ and }x \not \in \bigcap_{n=0}^{\infty} B_n$$
Then, as $x \not \in \bigcap_{n=0}^{\infty} B_n$ it means $x \not \in B_n$ for every $n$.

Thus, $x\in A$ and $x \not \in B_n$, so $x \in (A\setminus B_n)$ for every $n$. And

$$x\in \bigcap_{n=0}^{\infty} (A\setminus B_n)$$

And we do the argument in reverse. I can't find my error. #### ThePerfectHacker

##### Well-known member
If $x\in \bigcap_{n=0}^{\infty} B_n$ it means $x\in B_n$ for all $n$.

Thus,

If $x\not \in \bigcap_{n=0}^{\infty} B_n$ it means $x\not \in B_n$ for some $n$.

#### Guest

##### Active member
Oh, I see. Thanks. Is this right now then?

If,
$$x\in A \setminus \left( \bigcap_{n=0}^{\infty} B_n \right) \implies x \in A \text{ and }x \not \in \bigcap_{n=0}^{\infty} B_n$$
Then, as $x \not \in \bigcap_{n=0}^{\infty} B_n$ it means $x \not \in B_n$ for some $n$.

Thus, $x\in A$ and $x \not \in B_n$, for some $n$, so $x \in (A\setminus B_n)$ for some $n$. In particular,

$$x\in \bigcup_{n=0}^{\infty} (A\setminus B_n).$$

Doing the reverse, let $$x\in \bigcup_{n=0}^{\infty} (A\setminus B_n).$$ Then $x \in (A\setminus B_n)$ for some $n$. That's $x\in A$ and $x \not \in B_n$ for some $n$ and the latter implies $\displaystyle x \not \in \bigcap_{n=1}^{\infty}B_n$, so $\displaystyle x \in A \setminus \bigcap_{n=1}^{\infty}B_n$.

Therefore $\displaystyle A \setminus \bigcap_{n=1}^{\infty}B_n = \bigcup_{n=1}^{\infty} (A \setminus B_n)$.

Very good.