Set Theory 2 | Scoins.net | DJS

Set Theory 2

Set Theory 2:  using sets a little

This is quite difficult to type and you may need to change your choice of browser to read this page. I have used Unicode characters to maximise the number of browsers which will show the characters that I want you to read. The information in the two pages on set theory has a non-empty intersection. ¹

Using capitals for sets and lower case for elements of sets, then set A is a subset of B means that A is contained in or is equal to B, written  A⊆B. The number of distinct subsets within a set is rapidly large.

For set A={a,b,c} the possible sets are {a}. {b}, {c}, {a, b}, {b,c}, {c,a}, {a,b,c} while assuming the order of elements in the set does not matter. So we might recognise that if n(A)=N then n(subsets of A)=N!

There is the concept of the empty set, which has no members, written {} or ø. Is this contained in set A? Under some circumstances yes, and under others, no; this means we need extra adjectives, such as describing a general set as non-zero or not empty. Just when you thought it was safe....

The previous page introduced the ideas of union and intersection. You should have come across Venn diagrams by now, which are a very good graphic way of working with sets. They are, however, hard for me to reproduce, so I’ll restrict myself to other aspects of set theory and point you to some websites that do explain (and draw) Venn diagrams fairly well. Go find them.


Consider just two overlapping sets, A and B. Let’s think about (A∪B)’, those elements ‘not’ in both A and B.
I claim this is also A∩B’.
Evidence: let ξ be positive integers below ten, and let A={2,3,5,9} and B={1,3,4,5,7,9}. Then A∪B={1234579} and (A∩B)’={6,8}. A’={1,4,6,8} and B’={2,6,8} so A’∩B’={6,8}.

Similarly, I claim that (A∩B)’=A’∪B’. A∩B={3,5,9} so (A∩B)’={1,2,4,6,8}. But, using the lists above A’∪B’ = {1,2,4,6,8} which is the same thing.

Once you are convinced this is so, you will accept that the general rule is that the ‘not’ treatment reverses, inverts or swaps the ∩ and ∪ symbols.

What about distributive features? Could (A∪B)∩C be written differently? Well of course it can, but you may not say different is simpler. Using the sets above, add C={1,4,8,9} and let’s look at (A∪B)∩C, which is {1,2,3,4,5,7,9}∩{1,4,8,9}={4,9}. Meanwhile A∩C={9} and B∩C={4,9} so (A∪B)∩C=(A∩C)∪(B∩C).


Exercise:

Students doing this for revision: convert the expressions in Q1-6 so that the bracket disappears: if necessary, check using the sets in the example. 

To students doing this topic for the first time, for Q1-6 write out the set that fits the description and see if you can find another way to describe those elements without the brackets.

1   (A’∩B)‘                                              2    (A∪B’)’

3   (A∩B)∪C                                          4    (A∪B’)’∪C

5   (A∩B)’∪C                                          6   (A∪B)’∩C’    

7.   Draw the venn diagram for yourself that represents the example used above; treat this as an exercise in following instructions :-  Draw (and label A,B,C) three equal sized overlapping circles and draw a rectangle around them. Insert the numbers in the (eight) spaces. In my diagram, only {3,5} belong in the same space and no space is empty.

Write an expression that identifies:

8       only the {7}                                   9    {1, 4, 9}

10     {3, 5, 9}                                          11    only the {9}

12     {6, 8}                                             13    {2, 6, 7, 8}

DJS 20130429





 

 1 "The information in the two pages on set theory has a non-empty intersection."  Meaning: some stuff is repeated, but that was test to see if you understood the first page.     20210618

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