My Math Forum Set Theory proof issue

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 December 31st, 2016, 02:27 PM #1 Newbie   Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0 Set Theory proof issue Hey all, can you help me out with this set theory proof? Gush I can't seems to get this proof part! Where do I start? Is there a common place ground I can start with? Here is the set relation: $\displaystyle A \bigtriangleup B \subseteq (A \bigtriangleup D) \cup (B \bigtriangleup D)$ I need to proof if a proper set exists in the relation between the left hand side and the right hand side (?) I can use a truth table to show that, I think... and maybe to get a generic approach with Venn diagrams, but these are not acceptable solutions by my professor... but how do I go about a general proof?
December 31st, 2016, 07:09 PM   #2
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Quote:
 Originally Posted by shanytc Gush I can't seems to get this proof part! Where do I start? Is there a common place ground I can start with?
Excellent question. How to get started with a problem like this.

The very first thing to do is always to make sure you understand exactly what the question is asking. And to clearly write down what the question is looking for. Often just writing down the question clearly is 90% of the way to the answer.

Since your phrasing of the question is confusing, I'll take the liberty of being picky about it in order to clarify some things.

Quote:
 Originally Posted by shanytc I need to proof
I need to prove. A pet peeve of mine. Also, a math proof isn't just symbols, it's a story in words. Spelling and grammar enhance clarity.

Quote:
 Originally Posted by shanytc if a proper set
I have a problem here, I never heard of a proper set.

There's a proper class, that's a collection that's too big to be a set. The class of all sets is a proper class.

And there's a proper subset, that's a subset that's not the whole set.

But there's no proper set.

Do you mean just set?

Quote:
 Originally Posted by shanytc exists in the relation between the left hand side and the right hand side
I got lost here for a while. "exists in the relation" doesn't parse. And the relation between the LHS and the RHS is just the usual subset relation $\subseteq$, right? But even so, it doesn't mean anything for some set to "exist in the subset relation."

I'm guessing that we are supposed to do three things, basically:

* Prove that this expression isn't always true;

* Prove that it's not always false;

* Find $A$, $B$, and $D$ such that it's true.

But that's a guess. Possibly this isn't an exact transcription of the problem; and also your understanding of the question is a little shaky.

Quote:
 Originally Posted by shanytc (?)
I see you already share my concerns

And second, because this is the single universal truth I know about getting started on math problems. Always write down an extremely clear, detailed explanation of what they are asking you for.

If you do that one single thing in every math class, it will change your life. Writing down a clear statement of the question organizes your mind around the problem.

Quote:
 Originally Posted by shanytc (?) I can use a truth table to show that, I think... and maybe to get a generic approach with Venn diagrams
I completely agree. Use pictures, truth trees, symbolic expressions, everything you've got. Especially visual diagrams and pictures.

Quote:
 Originally Posted by shanytc (?) but these are not acceptable solutions by my professor...
The prof wants a set-theoretic style answer. But how you get to it can and should involve Venn diagrams and truth tables and anything else that works for you.

Quote:
 Originally Posted by shanytc (?) but how do I go about a general proof?
Figure out exactly what is being asked, then write it down as clearly as you can.

January 1st, 2017, 05:30 AM   #3
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Quote:
 Originally Posted by Maschke Excellent question. How to get started with a problem like this. The very first thing to do is always to make sure you understand exactly what the question is asking. And to clearly write down what the question is looking for. Often just writing down the question clearly is 90% of the way to the answer. Since your phrasing of the question is confusing, I'll take the liberty of being picky about it in order to clarify some things. I need to prove. A pet peeve of mine. Also, a math proof isn't just symbols, it's a story in words. Spelling and grammar enhance clarity. I have a problem here, I never heard of a proper set. There's a proper class, that's a collection that's too big to be a set. The class of all sets is a proper class. And there's a proper subset, that's a subset that's not the whole set. But there's no proper set. Do you mean just set? I got lost here for a while. "exists in the relation" doesn't parse. And the relation between the LHS and the RHS is just the usual subset relation $\subseteq$, right? But even so, it doesn't mean anything for some set to "exist in the subset relation." I'm guessing that we are supposed to do three things, basically: * Prove that this expression isn't always true; * Prove that it's not always false; * Find $A$, $B$, and $D$ such that it's true. But that's a guess. Possibly this isn't an exact transcription of the problem; and also your understanding of the question is a little shaky. I see you already share my concerns I'm going on about this for two reasons: One, your next step is to write down exactly what the question is asking of you. And second, because this is the single universal truth I know about getting started on math problems. Always write down an extremely clear, detailed explanation of what they are asking you for. If you do that one single thing in every math class, it will change your life. Writing down a clear statement of the question organizes your mind around the problem. I completely agree. Use pictures, truth trees, symbolic expressions, everything you've got. Especially visual diagrams and pictures. The prof wants a set-theoretic style answer. But how you get to it can and should involve Venn diagrams and truth tables and anything else that works for you. Figure out exactly what is being asked, then write it down as clearly as you can.
Dear Mashke,
I apologize for my lack of clear English. Please let me rephrase:

I've been given this problem to solve:

"Prove that for all sets A, B, C, exists: A△B⊆(A△D)∪(B△D)

Is A△B also a proper subset? Elaborate"

So basically I need to prove whether A△B is also a proper subset (or not).

I hope it is more clear?

Last edited by shanytc; January 1st, 2017 at 05:32 AM.

 January 1st, 2017, 09:45 AM #4 Senior Member   Joined: Feb 2012 Posts: 144 Thanks: 16 mmh... a proper subset of what?
January 1st, 2017, 10:25 AM   #5
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Quote:
 Originally Posted by mehoul mmh... a proper subset of what?
The right hand side

 January 1st, 2017, 04:11 PM #6 Senior Member   Joined: Aug 2012 Posts: 1,919 Thanks: 533 I just drew out a diagram of all the logical possibilities and convinced myself that it's always true. If the prof wouldn't accept that I'd just transform it to a laundry list of "If x is in A and not in B then this else that else if this then that etc." Tedious but you can read it right off your diagram. On the other hand by asking if the inclusion is proper, I'm wondering if I did something wrong! Last edited by Maschke; January 1st, 2017 at 04:33 PM.
 January 1st, 2017, 08:54 PM #7 Senior Member   Joined: Aug 2012 Posts: 1,919 Thanks: 533 This problem is a little tricky. I haven't got a full theory of what's going on yet, but I have an example of where the LHS is a proper subset of the RHS. First I made myself this diagram. It shows that if $x$ is in the set at the top, it must be in the set at the bottom. But, you can not always go up! For example, let $A = \{1, 2, 3\}$ $B = \{4, 5, 6\}$ $D = \{7\}$ Now the set $A \bigtriangleup B = \{1, 2, 3, 4, 5, 6\}$ As you follow any one of these numbers from top to bottom, it always ends up in the bottom row. However if you start at the bottom, $A \bigtriangleup D = \{1, 2, 3, 7|$ and $A \bigtriangleup D = \{4, 5, 6, 7|$. Their union is $\{1, 2, 3, 4, 5, 6, 7\}$. But now if you let $x = 7$, you can never make it up to the top row. You get stuck at the second line from the top. You can only make one AND clause true in each parent group, so their OR is False. In this case, the LHS is a proper subset of the RHS. That's not the end of it. There are probably cases where equality holds. Or maybe it's always a proper subset. I haven't figured that out yet. Have you made any progress? Honestly, one of the secrets to problems is that you just keep working at them till you figure them out. Like Edison said, it's 1% inspiration and 99% perspiration. ps -- What if all the sets are empty? Then every expression is empty and the LHS = RHS. So that's at least one example of equality. Next research question: Are there examples with some or all of the sets non-empty that lead to equality? Under exactly what circumstances do you get equality and when is the inclusion proper? I think the prof wants you to analyze the problem like this. Thanks from shanytc Last edited by skipjack; January 1st, 2017 at 09:35 PM.
 January 1st, 2017, 09:55 PM #8 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out. Just in case it's of any convenience, Wolfram has a nice widget. Thanks from Maschke
 January 3rd, 2017, 12:00 PM #9 Senior Member   Joined: Feb 2012 Posts: 144 Thanks: 16 Take D=empty set. Then A delta D = A and B delta D = B so that the right-hand side is AuB. Now if AnB is not empty the inclusion is strict. Thanks from shanytc

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