From: nguyen@drutx.ATT.COM (Christopher Chu)
Newsgroups: soc.culture.vietnamese
Subject: Re: [LOGIC] Empty Set Problem
Summary: Few notes to DDK
Date: 30 Jul 92 16:59:38 GMT
Organization: AT&T Bell Laboratories - Denver, Colorado

Dear DDK,
+----------------------------------------------------------------------------+
| From: vibama@cs.hw.ac.uk (Scott Doan)                                      |
| Newsgroups: soc.culture.vietnamese                                         |
| Subject: [LOGIC] Empty Set Problem                                         |
| Message-ID: <1992Jul30.101650.25820@cs.hw.ac.uk>                           |
| Date: 30 Jul 92 10:16:50 GMT                                               |
|                                                                            |
| ----------------------------------------------------------------------     |
| Problem: prove that any empty set is a subset of a non-empty set.          |
| -----------------------------------------------------------------------    |
| Solution:                                                                  |
| --------                                                                   |
| The above statement can be restated as follows:                            |
|                                                                            |
| Prove that A => B is true where                                            |
| A: x is a member of the empty set,                                         |
| B: x is a member of a non-empty set                                        |
| for all x.                                                                 |
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We could immediately state the logical statements to be proven because we
are familiar with the concept of set containment. This ability may not be
available in people who do not know math.
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| A is obviously false since an empty set has no element,                    |
| x can't be a member of any empty set (*).                                  |
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There is only one empty set like I have mentioned in my post. Suppose that
A and B are both empty sets. Then, A is a subset of B and B is a subset of
A. It follows immediately A and B are identical.
+----------------------------------------------------------------------------+
| Therefore A => B is always true regardless of B's validity.                |
| -----------------------------------------------------------------------    |
| The proof core lies at (*). Once you get it then you can prove it.         |
+----------------------------------------------------------------------------+
I think the the proof also lies in the fact that, in mathematical logic,
A => B is always true if A is false; this is NOT OBVIOUSLY INTUITIVE to
people who do not have a background in logic and that was why I indicated
that ~A v B is logically equivalent to A => B and used the former instead. (*)
To translate this definition of logic, intuitively, may I say, one's
argument is always valid if he she started out with nonsense. You see, you
can't even argue with the *beginning* of the argument :-)
+----------------------------------------------------------------------------+
| So it is quite possible to prove it in one second. I met this problem      |
+----------------------------------------------------------------------------+
It may be obvious for one to see why the problem statement to be proven is
obviously true right after reading the first few words of the problem
statement. I think that many did. However, it may take hours to explain
why and how we could figure the answer in a few seconds to ourselves and
people who may have different background.
+----------------------------------------------------------------------------+
| not too long ago as it was set as an excercise in the maths lesson         |
| at 10th grade (in VN, BTW, I haven't left VN for too long)                 |
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That's true! I have made a mistake on my post. I said this problem was for
11th grade student. At the beginning of the 10th grade we started to learn
logic, functions, sets ... 
+----------------------------------------------------------------------------+
| Of course everyone on the net will be able to prove                        |
| it if they have time (5-10 minutes) but not many people can see the above  |
| trick in one moment. The  problem is not designated to test any netters as |
+----------------------------------------------------------------------------+
I disagree with the above. I am sure that most people ho sudy morndern
algebra know the problem you stated by heart and so do CS theorists.
+----------------------------------------------------------------------------+
| it is so trivial but it tests Long Nguyen himself to see if he has a logic |
| cell ... He is the only one who knows if he can solves it in one           |
| moment. Judge for yourself Long Nguyen.                                    |
|                                                                            |
| Sincerely,                                                                 |
|                                                                            |
| DDK                                                                        |
| PS: I am not too sure if someone has posted the solution already.          |
| If so, please accept my apologies !!!                                      |
+----------------------------------------------------------------------------+
Yes, I did, and I separated logic from set theory in my argument as follows:

(1) Set theory tells you what logical statement to be proven when one
    wants to show containment among sets.
(2) Mathematical Logic tells you by definition that you may say A => B is
    always true if A is false.

Also, I have used (*) in my argument and I have explained why I did it.

Sincerely Yours,

Chris