If A, B, and C are sets, how would you prove that:::

if A is a subset of B, then A - C is a subset of B - C?

Not really sure of how you can bring in C into it.

2/25/2007 7:59:51 PM

first you assume A is a subset of B. So if x is in A then x is in B.

Next you suppose you have some element of A-C and show that the same element is in B-C. Also, be sure to include the possibility of empty sets.

[Edited on February 25, 2007 at 8:08 PM. Reason : .]

2/25/2007 8:07:53 PM

So would it be right to say...


Proof: Let A,B, and C be sets. Assume A is a subset of B.
Then for all x, if x exists in A, then x exists in B.
Suppose A - C, then x exists in A and not C.
Suppose B - C, then x exists in B and not C.
Since x exists in A and B, A - C is a subset of B - C.
Thus, if A is a subset of B, then A - C is a subset of B - C.
End Proof.

I feel like im kinda throwing that set C in there, and I also didnt include anything about the case of an empty set, do I NEED to put that in there? and if so, where?

But also, If you were to say A={2,4} B={2,4,6} and C={2,4,6,8}...then B-C would be empty set, and so would A -C. Empty set isnt a subset of empty set...so that messes things up.

[Edited on February 25, 2007 at 10:28 PM. Reason : e]

2/25/2007 10:11:29 PM

Quote :

"Suppose A - C, then x exists in A and not C. "

No. "Suppose A-C" is a fragment. Suppose A-C what?

Quote :

"Empty set isnt a subset of empty set...so that messes things up. "

The Empty set is a subset of every set, including itself.

2/25/2007 10:30:22 PM

not according to our teacher

2/25/2007 10:31:08 PM

OK I'll get you started.

Let A,B, and C be sets.
Assume A is a subset of B. Thus if x is in A then x is in B.

If A-C is empty then A-C is a subset of B-C.
Suppose A-C is non-empty.
Now, x in A-C implies x is in A and x is not in C.
.
.
.
Therefore x is in B-C.
Thus A-C is a subset of B-C
QED.


Now I set the problem up and finished the conclusion. You just need to connect the dots.

2/25/2007 10:36:53 PM

^^ You must have misunderstood him/her. Please read your book!

[Edited on February 25, 2007 at 10:38 PM. Reason : .]

2/25/2007 10:37:26 PM

i get exactly what you are saying, but what if B - C is empty?

2/25/2007 10:38:47 PM

^ You should be able to answer that yourself. Think about it....

2/25/2007 10:39:51 PM

the teachers reasoning for saying empty set isnt a subset up empty set is b/c "the empty set has no elements", im reading straight from her notes. but i see what you are saying if what she told us is wrong.

2/25/2007 10:40:50 PM

The ONLY subset of the empty set is the empty set itself..

Who the fuck is your teacher?

2/25/2007 10:42:42 PM

haha, Williams, i appreciate the help

2/25/2007 10:43:12 PM

Quote :

"but what if B - C is empty?"

Think about this. If you assume A subset of B, and A-C non-empty is it possible to have B-C empty?

2/25/2007 10:43:50 PM

^^ If you are serious, I am going to contact Dr. Paur about this tomorrow.

I have never heard of WIlliams is that a grad student?

2/25/2007 10:46:01 PM

she isnt a grad student, maybe she just told us wrong in the notes, but i kinda find that hard to beleive if she explained her reasoining in our notes also.

2/25/2007 10:47:47 PM

well I don't see her in the faculty listing. I don't know man. Trust me or read you book. I am right.


Any way the point here is moot. Its impossible to have B-C empty under the assumptions I stated.
Namely, A-C non-empty

[Edited on February 25, 2007 at 10:52 PM. Reason : .]

2/25/2007 10:49:53 PM

has to be this person:

Brenda B Williams
Mathematics
Lecturer
bdburns@unity.ncsu.edu


syllabus says B. Burns Williams

2/25/2007 10:53:05 PM

Quote :

"Lecturer"

yeah, if you are sure she said this then you should confront here about it. Read your book and find that I am right, then I would politely point this out to her.

2/25/2007 10:54:53 PM

thanks for the help

2/25/2007 10:55:13 PM

np

2/25/2007 10:55:23 PM

Just thought I'd add one thing: clalias is right, the empty set is a subset of itself.

I could be wrong, but I tend to doubt your teacher was wrong about this--it's a pretty fundamental thing. Is it possible that your teacher said something else and you're misremembering or misunderstanding it?

For example, she may have said that the empty set is NOT an element of itself (which is correct--the empty set has no elements). Starting out, it's easy to get the concepts of "element of" and "subset of" confused. The empty set has no elements, but it DOES have exactly one subset (itself).

2/26/2007 12:44:04 AM

doesnt look like it. she started the section of by calling it "Sets" and then proceeded by 4 examples all with the subset symbol, not even talking about elements. she had to of just told us wrong, ill address her tomorrow about it

2/26/2007 1:24:57 AM

Well, if that's what she said, definitely bring it up. And if she continues to claim that the empty set is not a subset of itself, just mention the following quick proof:

If A is any set (including the empty set), the empty set is a subset of A:

Proof:
For all x, if x is an element of the empty set, x is also an element of A. (This is vacuously true, since there are no elements in the empty set). Therefore, the empty set is a subset of A.

2/26/2007 2:28:57 AM

while we're at it why not throw in arbitrary unions and intersections of the empty set with itself. Those would all be subsets of the empty set as well.

2/26/2007 2:41:07 AM

Sure, but I don't see that they're really worth mentioning, since an arbitrary union or intersection of the empty set with itself is still the empty set. There's only one subset of the empty set.

2/26/2007 3:08:58 AM

so i asked her about it today, said she never said that yet people had it in their notes. She said it is false that empty set isnt a subset of empty set, yet had it in the notes with the subset information, not elements. so whoknows.

2/26/2007 4:25:38 PM

The confusion may be in the difference between a subset and a proper subset, if your prof has mentioned those two concepts separately. The empty set is a subset of any set, including the empty set, but is a proper subset of any non-empty set.

Anyway, clalias is spot on about the way you should be thinking about this proof.

2/26/2007 8:45:45 PM