# Started learning proofs - need some feedback

1. May 25, 2013

### 4Fun

Hello guys,

this is my first post on this forum. I want to learn advanced/pure mathematics basically just because I find it really interesting and challenging and I have started to learn about proofs. I'm currently reading Velleman's book and I have reached the part in which you actually start to learn writing proofs. Since Velleman only offers solution for some of the proofs I dont know whether my proofs are actually valid. I would really appreciate if someone would be willing to quickly take a look at some proofs I write and give me some feedback.

Proposition: Proove that if F is a family of sets and A $\in$ F, then $\cap$ F $\subseteq$ A.

Givens: A $\in$ F
Goal: $\cap$ F $\subseteq$ A

$\cap$ F $\subseteq$ A is equivalent to $\forall$ x (x $\in$ $\cap$ F -> x $\in$ A).

Now I let x be an arbitrary element.

Question here: Does x have to be an element or a set? Because $\cap$F consists only of sets right?!

Then I assume that x $\in$ $\cap$ F.

Givens: A $\in$ F, x $\in$ $\cap$ F
Goal: x $\in$ A

Now x $\in$ $\cap$ F means that $\forall$ A $\in$ F (x $\in$ A) for some A.

So basically that for every element ( or set of F, since F is a family of sets) x is an element of that set. Since A $\in$ F, x is also an element of A.

Now the formal proof:

Let x be arbitrary. Suppose that x $\in$ $\cap$ F, which means that for all sets of F, x is an element of each of those sets. Since A is one of those sets, it follows that x is an element of A. Since x was arbitrary it follows that in general if A $\in$ F then $\cap$ F $\subseteq$ A.

Now although I think that my scratch work was correct, I think the formal proof still sounds incorrect. Could anybody please give my some feedback?

2. May 25, 2013

### reenmachine

I just started to learn proofs as well so take my comments with a grain of salt and wait for the official mentors/helpers before jumping to conclusions , but everything in mathematic is a set.So if $F=\{A,B,C\}$ , then $F$ is simply a set with multiples sets as it's elements so $A \in F$ , $B \in F$ and $C \in F$.

So $A$ , $B$ , $C$ are elements of $F$ , but that doesn't mean that they aren't sets as well.

Suppose $A=\{1,2\}$ , $B=\{1,3\}$ and $C=\{1,4\}$.Then $\bigcap F = \{1\}$ and it's true that $\{1\} \subseteq A$.The logic will work even if there's no intersection , like for exemple if $A=\{1\}$ , $B = \{2\}$ and $C = \{3\}$.Then $\bigcap F = \varnothing$.This is the empty set.You can safely assume that the empty set is a subset of ALL sets.So $\varnothing \subseteq A$.

Last edited: May 25, 2013
3. May 25, 2013

### micromass

Staff Emeritus
That seems alright. The formal proof is close to something I would write.

What you did was: let $x$ be arbitrary. Assume that $x\in \bigcap F$. That is alright, but it doesn't tell us what $x$ is. So you might call it a bit vague. It might be better to contract this into one sentence: "Let $x$ be an arbitrary element of $\bigcap F$."

4. May 25, 2013

### verty

Actually, there are variants of set theory that have classes as well as sets, and variants that have so-called urelements or individuals that are not sets but can be members of sets. This allows one to talk of things like sets of elephants or sets of apples. There is even a variant with individuals and a universal set, called NFU.

I find the history of modern ideas pretty interesting.