
#1
May2513, 09:52 AM

P: 10

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 [itex]\in[/itex] F, then [itex]\cap[/itex] F [itex]\subseteq[/itex] A. Ok I'll start with my scratch work: Givens: A [itex]\in[/itex] F Goal: [itex]\cap[/itex] F [itex]\subseteq[/itex] A [itex]\cap[/itex] F [itex]\subseteq[/itex] A is equivalent to [itex]\forall[/itex] x (x [itex]\in[/itex] [itex]\cap[/itex] F > x [itex]\in[/itex] A). Now I let x be an arbitrary element. Question here: Does x have to be an element or a set? Because [itex]\cap[/itex]F consists only of sets right?! Then I assume that x [itex]\in[/itex] [itex]\cap[/itex] F. Givens: A [itex]\in[/itex] F, x [itex]\in[/itex] [itex]\cap[/itex] F Goal: x [itex]\in[/itex] A Now x [itex]\in[/itex] [itex]\cap[/itex] F means that [itex]\forall[/itex] A [itex]\in[/itex] F (x [itex]\in[/itex] 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 [itex]\in[/itex] F, x is also an element of A. Now the formal proof: Let x be arbitrary. Suppose that x [itex]\in[/itex] [itex]\cap[/itex] 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 [itex]\in[/itex] F then [itex]\cap[/itex] F [itex]\subseteq[/itex] 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
May2513, 10:54 AM

PF Gold
P: 520

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##. 



#3
May2513, 11:09 AM

Mentor
P: 16,698





#4
May2513, 11:53 AM

HW Helper
P: 1,373

Started learning proofs  need some feedbackI find the history of modern ideas pretty interesting. 


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