Proving Subsets: A Venn Diagram Approach

In summary: But technically, as you say, the proof "collapses" or "fails" if we try to use x from the empty set. So I should have started with "If x \in X", as you say.
  • #1
leilei
8
0
Proof subset?

Given three sets A, B, and C, set X = (A-B) U (B-C) U (C-A) and
Y = (A∩B∩C) complement C. Prove that X is subset of Y. Is Y necessarily a subset of X? If yes, prove it. If no, why?
---When I draw the two venn diagrams X and Y, they are the same, but I don't know how to prove it...

Can someone help me out here...
Thanks in advance!
 
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  • #2
The usual proof for such a statement is: Let x be an element from X and try to prove that it is also an element of Y. So if x is in X, you know that it is in A but not in B, and/or it is in B but not in C, and/or it is in C but not in A. You want to show that it must be in (A∩B∩C)C. If it is in (A∩B∩C) then it would be in A and B and C, so it being in the complement means it is at least not in A or not in B or not in C. So you could suppose it is both in A and B and show that it cannot be in C.

So let me write this out in the right order:
Let [itex]x \in X[/itex]. Suppose that [itex]x \in A, x \in B[/itex]. Then definitely, [itex]x \not\in A - B, x \not\in C - A[/itex], because if it is in A it cannot be in any set from which we remove all elements of A (and similarly for B). But it must be in one or more of (A-B), (B-C) and (C-A), so it must be in (B - C). That is, x is in B (which we knew) but not in C. So if x is not in C, it cannot be in the intersection of C with whatever set you make up. In particular, it is not in [itex]A \cap B \cap C[/itex]. Therefore, it must be in the complement of that set, which is called Y.

Now try to do the same reasoning for [itex]Y \subset X[/itex]. You have already shown by your Venn diagram that if x lies in Y, it must lie in X. So try to prove it in the same way as I just did.
 
  • #3
Thanks a lot !
 
  • #4
There is, however, a technical problem with "Let [itex]x \in X[/itex]"- the proof collapses is X is empty. Far better to start "IF [itex]x \in X[/itex]". That way, if X is empty, the hypothesis is false and the theorem is trivially true.
 
  • #5
You are right, obviously the empty set is a subset of any set S (vacuously, all its elements are also in S).
 

Related to Proving Subsets: A Venn Diagram Approach

What is a Venn diagram?

A Venn diagram is a visual representation of sets and their relationships. It consists of overlapping circles or shapes that represent different sets and the areas where they intersect represent elements that are common to both sets.

How can a Venn diagram be used to prove subsets?

A Venn diagram can be used to visually show the relationship between two sets and determine if one set is a subset of another. If all the elements in Set A are also present in Set B, then Set A is a subset of Set B.

What are the advantages of using a Venn diagram in proving subsets?

A Venn diagram provides a clear and visual representation of the relationship between sets, making it easier to understand and prove subsets. It also allows for quick identification of common elements and differences between sets.

What are the limitations of using a Venn diagram in proving subsets?

A Venn diagram may not be able to accurately represent complex subsets with multiple sets and overlapping elements. It also does not provide a numerical or algebraic proof, which may be necessary in some cases.

How can errors be avoided when using a Venn diagram to prove subsets?

To avoid errors, it is important to accurately label the sets and their elements in the Venn diagram. It is also important to clearly understand the definition of a subset and to check for all the elements of the subset in the larger set before concluding it is a subset.

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